# Silicon and Germanium Transistor Biasing - Part 5

### Part 5: Germanium:

This article builds on part 1, part 2, part 3, and part 4 of the Silicon and Germanium Transistor Biasing series. If you have not read these articles, please do so first.

#### Germanium Basics

Before we get deeper into the different circuit design approach with germanium transistors, we’ll first take a closer look at the differences of concern when comparing silicon transistors to germanium. There are more things to consider when using germanium transistors. The new equations for calculating bias for both germanium and silicon transistors will necessarily involve more complex math. We will simplify them where possible, but the full equations are also used to drive our Fuzz Face Bias Calculator. The calculator can be used to easily approximate the bias in this circuit for any combination of circuit variables using either germanium or silicon transistors (or a combination), without requiring any math.

When using germanium, concerns that are of particular importance to common audio applications are:

• Low beta (typically)
• Lower $V_{BE}$
• Leakage current
• Temperature sensitivity (largely due to leakage current)

This is not a complete list, but these are the most important differences when comparing silicon to germanium. It is also more common to find PNP versions of germanium transistors, though there are plenty of NPN types as well. In this article, all germanium circuits will be drawn as NPN circuits for uniformity. We will look more closely at the role each of these factors play, but a summary of the differences can be found below:

#### Figure 1: Modern silicon and vintage germanium $h_{FE}$ (beta) measurements - higher beta is more typical of modern silicon types

You can find high beta ($h_{FE}$) germanium transistors and low beta silicon transistors, but typically, germanium transistors will have a notably lower beta than modern silicon types which must be accounted for.

#### Figure 2: Modern silicon and vintage germanium $V_{BE}$ graphs

So far, we’ve been using a $V_{BE}$ of 0.6 - 0.7V for silicon transistors. With germanium transistors, this voltage drop is lower and can also vary more substantially. It is typically stated as being around 0.3V for germanium transistors, but is often closer to 0.1V or less at the low base currents you are likely to find in pedal circuits. This change alone can affect the bias substantially depending on the circuit. For example, in part 4 we determined that with stock Fuzz Face values, high gain silicon transistors are likely to result in a bias voltage around 3V. If we made the same calculations but simply changed the $V_{BE}$ values to 0.1V (which is unrealistic for silicon but merely for demonstration), the expected bias voltage would be >8V.

#### Leakage Current

Leakage current is present in both silicon and germanium transistors, but in silicon transistors it is typically much smaller (orders of magnitude). In a BJT biased in the active region, the collector-base junction is reverse-biased. In an ideal BJT, this junction being reverse-biased means that no current should flow from base to collector or from collector to base. In real world examples, the reverse-biased collector-base junction does not perfectly block current from flowing, and some amount of current will flow from collector to base. What effect this would have depends on how the transistor is used in the circuit (which we’ll see shortly), but this leakage current is something that must be taken into consideration when biasing germanium transistors.

#### Temperature Sensitivity

Transistor parameters can change depending on temperature. Within reasonable operating conditions, most effects are often quite miniscule. One characteristic that does change substantially is leakage current. The leakage current in both silicon and germanium transistors is highly temperature sensitive. It does not take a large temperature increase to double the leakage current. In silicon transistors, the leakage is typically orders of magnitude smaller than leakage currents found in germanium transistors, so even at high temperatures it is usually negligible and can be ignored for silicon.

Unlike silicon transistors, small to moderate increases in temperature can cause drastic bias changes when using germanium transistors due to their leakage currents. When the temperature increases 18°F, the leakage current roughly doubles for both silicon and germanium. Depending on how much the leakage current contributes to the transistor’s quiescent current, doubling it can result in a large fluctuation in the bias voltage.

### Substituting silicon for germanium

The original Fuzz Face circuit is interesting in that typical modern silicon transistors can usually be dropped into the germanium circuit and it will operate fairly well. There is only a very minor difference between the silicon and germanium circuit values, and many modern silicon transistor types will work fairly well in the circuit. The bias voltage may be in a different range, and the output may not sound identical, but the result will be a similarly functioning fuzz unit despite the substantial differences between modern silicon and vintage germanium transistors. The ability to easily make this conversion depends on both the design and the component values used. Though the Fuzz Face circuit works well with many modern silicon transistors, that’s not true of the germanium Schaller Fuzz, which uses the exact same circuit topology but different component values.

#### Figure 3: Germanium Schaller Fuzz circuit which typically does not work well with silicon transistor types

There was a later silicon Schaller Fuzz unit, but the circuit was changed to accommodate the silicon transistors. If we use the component values from the germanium Schaller Fuzz and calculate the bias voltages using BC108Bs (with beta = 300) and our approximations from part 4 again, we get 7.73V on the collector of $Q_2$, which is very near the 9V rail and a poor bias voltage.

We’d like for this voltage to be closer to 6V, which is a ballpark bias voltage found in original units. Let’s see how we can achieve that via transistor selection (no resistance or $V_{CC}$ changes). We’ll use our part 4 approximations for now. Our formula for $V_{C2}$ is:

$$V_{C2} = V_{CC} - (I_{C2} * R_{C2})$$

We do not want to change $V_{CC}$ or $R_{C2}$, so we need to modify $I_{C2}$ to adjust the bias. First, let’s calculate what $I_{C2}$ needs to be to achieve 6V on the collector of $Q_2$:

$$I_{C2} = \frac{V_{CC} - V_{C2}}{R_{C2}}$$$$I_{C2} = \frac{9 - 6}{1470}$$$$I_{C2} = 2.04\text{mA}$$

The ideal $I_{C2}$ is 2.04mA, which is 4 to 5 times higher than the current needed to properly bias the silicon Fuzz Face circuit, and it needs to be increased substantially from its calculated value (863μA). The equation for $I_{C2}$ is:

$$I_{C2} = \frac{I_{E2} * \beta_2}{\beta_2 + 1}$$

We can see from this equation that $Q_2$ beta affects $I_{C2}$. However, $I_{C2}$ can never be larger than $I_{E2}$. $Q_2$ beta effectively controls how much of $I_{E2}$ flows through the collector (vs. the base). For high beta, it’s very nearly all of it ($I_{C2} \approx I_{E2}$). We want to increase our $I_{C2}$, and our beta is already high. Increasing $Q_2$ beta further will make a negligible difference, so we must increase $I_{E2}$ in order to achieve the necessary $I_{C2}$ increase. Our formula for $I_{E2}$ is:

$$I_{E2} = \frac{V_{E2}}{R_{SH}}$$

We do not want to modify $R_{SH}$, so we must adjust $V_{E2}$:

$$V_{E2} = V_{BE1} + (I_{B1} * R_{FB})$$

While $V_{BE1}$ is a transistor spec that varies, it does not vary much in silicon transistors. It tends to be 0.6-0.7V. It can vary more with germanium transistors, but it stays in the same rough range (0.05-0.3V), and is not something that can be reliably used to adjust the $V_{C2}$ voltage. We don’t want to change the $R_{FB}$ resistance, so we need to adjust $I_{B1}$ to change $V_{E2}$.

$$I_{B1} = \frac{I_{C1}}{\beta_1}$$

$I_{B1}$ can be adjusted with $Q_1$ transistor beta. A decrease in $Q_1$ beta will increase $I_{B1}$. Increasing $I_{B1}$ will result in a larger voltage drop across $R_{FB}$ and thus a larger $V_{E2}$.

#### Figure 4: Voltage and current changes for a $Q_1$ beta decrease and a $Q_1$ beta increase

This larger $V_{E2}$ results in more current flowing out of the emitter ($I_{E2}$, approximately set with $\frac{V_{E2}}{R_{SH}}$), which results in more current flowing into the collector ($I_{C2}$). This increased collector current causes more voltage to drop across $R_{C2}$ ($V_{RC2}$), decreasing $V_{C2}$. So a decrease in $Q_1$ beta results in a decrease in $V_{C2}$, which is what we need here. The way these changes interact with a change in $Q_1$ beta can be seen in Figure 4.

In order to achieve 6V on the collector of $Q_2$, we determined that we needed 2.04mA to flow into the collector. Let’s see what beta value we would need to achieve this. Since it is the $Q_1$ beta that affects this value, we will keep $Q_2$ beta at 300 and adjust $Q_1$ only. Since $I_{C2} \approx I_{E2}$ for high $Q_2$ beta, we will assume $I_{C2} = I_{E2}$ for a reasonable approximation. That means that we need 2.04mA to flow out of the emitter. We’ll determine what our $V_{E2}$ needs to be to achieve this.

$$V_{E2} = I_{E2} * R_{SH}$$$$V_{E2} = 2.04\text{m} * 1000$$$$V_{E2} = 2.04\text{V}$$

And we will determine the base current needed to achieve $V_{E2}$ = 2.V:

$$V_{E2} = V_{BE1} + (I_{B1} * R_{FB})$$$$I_{B1} = \frac{V_{E2} - V_{BE1}}{R_{FB}}$$$$I_{B1} = \frac{2.04 - 0.7}{100000}$$$$I_{B1} = \frac{1.37}{100000}$$$$I_{B1} = 13.7\text{uA}$$

Last, we’ll determine the $\beta_1$ value required to achieve this base current. First, we should approximate $I_{C1}$:

$$I_{C1} = \frac{V_{CC} - V_{C1}}{R_{C1}}$$$$I_{C1} = \frac{V_{CC} - (V_{E2} + V_{BE2})}{R_{C1}}$$$$I_{C1} = \frac{9 - (2.07 + 0.7)}{15000}$$$$I_{C1} = \frac{9 - 2.74}{15000}$$$$I_{C1} = 417\text{μA}$$

Now we can use this to determine $\beta_1$:

$$I_{B1} = \frac{I_{C1}}{\beta_1}$$$$\beta_1 = \frac{I_{C1}}{I_{B1}}$$$$\beta_1 = \frac{417\text{μ}}{13.7\text{μ}}$$$$\beta_1 = 30.43$$

30.43 is a very low beta, and in this case, it is a decent ballpark figure, but should only be considered a rough approximation. It is true that the Schaller Fuzz requires quite a low beta in $Q_1$ to bias well, which is why most modern silicon types will result in a very poor bias in that circuit. However, our formulas used some approximations which greatly simplified the required calculations, but decreased the accuracy of the approximations for low beta transistors. Clearly, we need to be able to properly calculate the results with low beta transistors as well. We will derive some equations for doing so shortly, but to get the full picture, we first need to look at transistor leakage.

### Transistor Leakage

Leakage current in a transistor is current that flows in a real transistor where it should not in an ideal/theoretical transistor. In an ideal transistor, with nothing connected to the base, no current should flow through the transistor. An external source of base current should be necessary for current to flow in the transistor. In reality, this is not the case. Some amount of leakage current will flow through the reverse-biased junctions in the transistor. This is true of both silicon and germanium transistors, though the amount of leakage current in silicon types is usually orders of magnitude smaller and can be considered negligible in most applications.

In germanium transistors, there is often enough leakage current present that it must be taken into consideration.

#### Figure 5: Operation of transistor leakage measurements $I_{CBO}$ and $I_{CEO}$

$I_{CEO}$ is the leakage current which flows from collector to emitter with the base left open. It can sometimes be found in datasheets, although it is much more common to find $I_{CBO}$ in datasheets, which is the leakage that flows from collector to base with the emitter left open. When you hear pedal builders refer generally to a “leakage” figure, they typically mean $I_{CEO}$, but if transistor beta is known, we can find $I_{CBO}$ from $I_{CEO}$ and vice-versa.

First, let’s look at $I_{CBO}$ in Figure 5. With emitter floating, there is a leakage current that flows through the reverse-biased base-collector junction. 0-10μA are common figures you might find for $I_{CBO}$ in small signal germanium transistors. If we instead open the base terminal as seen in the second arrangement, the current that was flowing from collector to base in the first configuration is still flowing, but instead of flowing through an external return path, that current now flows through the base-emitter junction of the transistor. Like any base current, this gets multiplied by transistor beta, causing an additional $I_{CBO} * \beta$ to flow through the collector. In other words:

$$I_{CEO} = I_{CBO} * (\beta + 1)$$

You’ll notice neither configuration seen in Figure 5 is something we would typically use in a real application. $I_{CEO}$, for example, only tells us how the transistor behaves with the base open, which is typically not how we will be using the transistor. A transistor with $I_{CEO}$ of 200μA might be described as “having 200μA of leakage”, but depending on the circuit, there may actually be notably less than that flowing into the collector. $I_{CEO}$ at a certain temperature should be considered a maximum amount of collector current that could flow due to leakage at that temperature, but that amount can be decreased depending on the circuit, as will be demonstrated shortly.

#### Figure 6: In this arrangement, an ideal transistor is in cutoff, while a leaky transistor might be biased in the active region.

In the circuit seen in Figure 6, no current will be flowing with an ideal BJT, and our collector voltage would be the same as our supply (9V), because with no current flowing there is no voltage drop across the resistor. The amp will not function. However, if $I_{CEO}$ is 230μA, which is a fairly common leakage measurement you might find in a germanium transistor, 2V will drop across the collector resistor at that current, and our collector voltage becomes 7.0V, which puts the BJT in the active region for proper signal amplification.

We mentioned that depending on the circuit, the actual leakage current flowing through the collector can be notably less than the $I_{CEO}$ measurement. This happens because in most circuit configurations with a base connection, some of the $I_{CBO}$ leakage current will flow out of the base terminal instead of through the base-emitter junction. $I_{CBO}$ current which flows out of the base terminal does not contribute back to collector/emitter current.

However, in the arrangement seen in figure 6, the full $I_{CEO}$ current will flow through the collector. The reason for this is that $I_{CEO}$ is the leakage current measurement with the base left open, and with the only external base connection being made through a series capacitor here, the base terminal is effectively open for DC signals (capacitors are effectively open-circuit for DC analysis). $I_{CBO}$ is direct current and cannot flow through the input capacitor, so it must flow through the base-emitter junction, causing the full amount of $I_{CEO}$ to flow through the collector.

#### Figure 7: With no external DC return path for the base node, $I_B = I_{CBO}$. With an external return path, $I_B < I_{CBO}$, as not all of $I_{CBO}$ flows into the transistor base.

When there is a separate current return path out of the base node, some of the $I_{CBO}$ current will leave the base node via this path rather than through the transistor’s base, causing less of $I_{CBO}$ to flow into the base and causing $I_C$ to be less than $I_{CEO}$.

The ability to bias transistors using only leakage (no external base bias resistors necessary) as seen in Figure 7 may appear like a distinct advantage of germanium transistors, but it is a poor way to bias a transistor stage. Transistor leakage can fluctuate and vary substantially, and if a leakage value is listed in the transistor datasheet, it is often given as a maximum value, leaving a wide range of potential real leakage values. It’s also an additional factor that needs to be considered for proper biasing of the circuit, adding complexity to our calculations and more variance to the results.

The above Figure 7 does not show the full picture either, as leakage current is highly temperature-dependent. It roughly doubles when temperature increases 10°C (18°F). This would result in large bias swings with temperature changes, as the bias is entirely dependent on this leakage current.

From a circuit design standpoint, leakage current is pretty much undesirable across the board. However, germanium transistors with moderate leakage are somewhat sought-after still because they can be biased this way. It is generally a very poor method for biasing a transistor stage due to the fluctuation and unpredictability described above, but because some iconic effects used transistor stages which were biased this way, it is fairly common for builders to seek out leakier transistor types for those uses. Two real examples can be seen in Figure 8.

#### Figure 8: Leakage-biased stages found in the Tonebender Mk2 and Tonebender Mk1

Figure 8 shows the Tonebender Mk2 input stage and the Tonebender Mk1 input stage. The Mk2 stage is a common emitter amplifier. The Mk1 stage is a common collector unity gain buffer (also called an emitter-follower). They operate quite differently, but in both stages, base has a pulldown resistor to GND, unlike our previous examples. When leakage is negligible, no DC current can flow in or out of the base node. Emitter, base, and GND are all at the same potential, and the BJT stages will not work.

The usual way to bias this stage for a silicon transistor is to also add a resistor from base to +9V. This will allow current to flow from 9V through the resistor and then through the pulldown resistor and the base-emitter junction, biasing the transistor. In addition to allowing silicon here, this method also allows for the use of non-leaky germanium types.

Using leaky germanium transistors, this stage can still work with the resistor left out because the leakage currents allow current to flow through the transistor base still.

#### Figure 9: Tonebender Mk2 input stage biased for low/no leakage, and the original leakage-biased arrangement

Figure 9 shows the Tonebender MkII input stage modified with a resistor connected between base and 9V, and the original configuration where leakage current biases the transistor. Note that although the $I_{CBO}$ leakage is drawn externally here, leakage current happens inside the transistor. The leakage drawn here flows through the reverse-biased base-collector junction. Note also that while leakage current comes from the transistor’s collector, the additional resistor in the first arrangement is connected to $V_{CC}$ instead. You may see examples online of efforts to simulate leakage by connecting a resistor from base to collector in a common emitter amp.

#### Figure 10: “Simulated” leakage current taken from the collector also introduces negative AC feedback

This can work for providing the necessary current to the base. The problem with simulating leakage this way is that the collector is not just a source of a DC voltage, the way $V_{CC}$ is. The collector of the transistor is also our output in the common emitter amp configuration. By connecting a resistor from collector to base, we introduce a path not only for DC to flow, biasing the transistor, but also for some AC to flow from the output (collector) to the input (base), introducing negative feedback. Negative feedback can be desirable in many cases, and it is similarly implemented in the Zonk II, Fuzzrite, Big Muff, and (in a somewhat different implementation) in the Fuzz Face circuit, but introducing it here can notably change the performance, which is not the goal.

At a stable temperature, actual leakage current remains quite constant, and effectively functions like direct current. Simulating leakage with a resistor connected to the DC source VCC will introduce the necessary direct current without adding negative feedback.

#### Figure 11: Currents flowing into and out of the base node in the Mk2 input stage

Figure 11 shows a closeup of the base node in the Tonebender Mk2’s leakage-biased common emitter amp. Using Kirchoff’s Current Laws, which state that the current flowing into a node must equal the current flowing out of it, we can determine the relationship:

$$I_B = I_{CBO} - I_{RB1}$$

In this case, current flows into the base node via the leakage current through the reverse-biased base-collector junction ($I_{CBO}$). Current flows out of the base node via the 100kΩ $R_{B1}$ resistor ($I_{RB1}$) and also via the forward-biased base-emitter junction ($I_B$). We can calculate the current flowing through $R_{B1}$ if we know $V_{BE}$. For example, let’s assume we have a transistor with the following specs:

$$\beta = 70$$$$I_{CBO} = 4\text{μA}$$$$V_{BE} = 0.1\text{V}$$

Measuring leakage using something like the R.G. Keen method[1], this transistor might be shown to have 284μA of “total” leakage ($I_{CEO}$) because of the following relationship:

$$I_{CEO} = I_{CBO} * (\beta + 1)$$$$I_{CEO} = 4\text{μA} * 71$$$$I_{CEO} = 284\text{μA}$$

If you know $I_{CEO}$ rather than $I_{CBO}$, you can find $I_{CBO}$ with $\beta$ and $I_{CEO}$:

$$I_{CBO} = \frac{I_{CEO}}{\beta + 1}$$

With a known $I_{CBO}$, it’s fairly easy to calculate the base current here:

$$I_B = I_{CBO} - I_{RB1}$$$$I_B = I_{CBO} - \frac{V_B}{R_{B1}}$$$$I_B = 4\text{μA} - \frac{0.1}{100000}$$$$I_B = 4\text{μA} - 1\text{μA}$$$$I_B = 3\text{μA}$$

With 3μA flowing into the base, we can calculate $I_C$ with the following:

$$I_C = I_B * \beta$$$$I_C = 3\text{μA} * 70$$$$I_C = 210\text{μA}$$

Assuming $V_{CC}$ = 9V, we can easily calculate $V_C$:

$$V_{C} = V_{CC} - (I_C * R_C)$$$$V_{C} = 9 - (210\text{μA} * 10\text{kΩ})$$$$V_{C} = 9 - 2.1\text{V}$$$$V_{C} = 6.9\text{V}$$

This bias method is very similar to the bias seen in Figure 6. It may appear that we’ve decreased the transistor’s leakage here as $I_C < I_{CEO}$, but we’ve really just shunted some of the $I_{CBO}$ to GND via the base resistor so that it does not get amplified by the transistor’s current gain. This method of biasing still depends entirely on the transistor’s leakage, and will still fluctuate substantially with leakage fluctuations. We will take a closer look at the temperature sensitivity of germanium and how it can be mitigated some, but this is still a very poor way to bias a transistor stage, even if it is used in the Tonebender MkII.

### Replacing Leakage Bias Current

We’d like to be able to modify this transistor stage to work for low/no leakage transistors. Doing so will reduce the temperature sensitivity, allow us to use silicon transistors here, and allow for a wider range of germanium transistor specs.

If we could replace the transistor used in our previous example with one that had identical specs except for $I_{CBO}$ = 0μA (no leakage), re-biasing would be very simple. Because beta is the same, we would need the same amount of current flowing into the base to achieve the same collector current and voltage. Our equation for base current when our transistor had leakage was:

$$I_B = I_{CBO} - I_{RB1}$$

If $I_{CBO}$ = 0 (no leakage), no current will flow into the base. We need to introduce a new source of current to make up for the lack of $I_{CBO}$. This is where the additional base resistor comes into play, as seen in Figure 9. With a resistor ($R_{B2}$) connected from base to $V_{CC}$, current will flow through the resistor to the base node. Our new equation for current flowing into the base is:

$$I_B = I_{RB2} - I_{RB1}$$

With the same $V_{BE}$, $I_{RB1}$ has not changed, so to achieve the same base current we simply need to set $I_{RB2}$ equal to our old $I_{CBO}$, which was 4μA. The equation for $I_{RB2}$ is:

$$I_{RB2} = \frac{V_{CC} - V_B}{R_{B1}}$$

Rearranging to solve for $R_{B1}$, setting $I_{RB2}$ = 4μA:

$$R_{B1} = \frac{V_{CC} - V_{B}}{I_{B2}}$$$$R_{B1} = \frac{9 - 0.1}{4\text{μ}}$$$$R_{B1} = \frac{8.9}{4\text{μ}}$$$$R_{B1} = 2225000\text{Ω}$$$$R_{B1} \approx 2.2\text{MΩ}$$

The ideal $R_{B2}$ is 2,225,000Ω, which we’d typically round to the nearest common resistor value of 2.2MΩ.

#### Figure 12: Tonebender Mk2 first stage modified for a transistor with low or no leakage

Figure 12 shows what this redesigned amp would look like with the 2.2MΩ resistor added from base to $V_{CC}$. With the right low leakage transistor, this will function nearly identically to the leakage-biased stage, only with less temperature sensitivity. Though we’ve improved the temperature sensitivity substantially, this stage is still temperature sensitive (due to $V_{BE}$, beta, and $r_{e’}$ changes) and still requires fairly precise transistor beta selection. This is largely due to the fact that the emitter is grounded, which is generally not recommended unless there is some amount of negative feedback present (which would provide more bias stability).

The temperature sensitivity of this stage is no longer due to leakage nor is it unique to germanium (as $V_{BE}$, Beta, and $r_{e’}$ are also temperature-sensitive in silicon), so we will leave it here for now. We will modify this stage to be less temperature sensitive and more tolerant of varying transistor specs for both silicon and germanium in Part 6.

### Approximating Fuzz Face Bias with Leakage

Now that we have looked at the operation of transistor leakage, we can again look at the Fuzz Face circuit topology and this time come up with better approximations for calculating bias currents and voltages for any transistor specs, not just low leakage and high beta types.

#### Figure 13: Germanium Fuzz Face circuit topology - DC Analysis

A general overview of how we will derive these formulas is that we start by seeing if there are any known values in the circuit. For example, from a simple glance at the Fuzz Face circuit, we can see that $Q_{1E}$ will always be 0V, as it is grounded. For both transistors, we know that.

$$V_B = V_E + V_{BE}$$

Because $V_{E1}$ is 0, we also know

$$V_{B1} = V_{BE1}$$

If we knew any one transistor pin current, it would be easy to calculate most if not all of our other unknown values for given transistor specs. The problem is that even if we can determine some of the pin voltages, the currents are still unknown. So we’ll start by simply selecting an unknown current and deriving an equation with no unknown dependencies. We do this by using the known current relationships of transistors and fairly rudimentary electronics principles like Ohm’s Law and Kirchoff’s Current Laws.

Once we’ve solved for one of the currents, it’s very easy to solve for the rest with simple equations. This section necessarily contains more math and some unwieldy equations. Understanding the process is more important than following the math completely, and we will leave you with the useful equations we’ve derived for calculating Fuzz Face bias.

The voltage of most interest when biasing the Fuzz Face is the $Q_2$ collector voltage ($V_{C2}$). The equation for calculating $V_{C2}$ (with Ohm’s law) is:

$$V_{C2} = V_{CC} - (I_{C2} * R_{C2})$$

Here, $I_{C2}$ is unknown, but we’ll work our way backwards to see if we can solve for our unknown values. We can determine $I_{C2}$ with $I_{E2}$:

$$I_{C2} = I_{E2} * \frac{\beta_2}{\beta_2 + 1}$$

Note that this relationship does not change for a transistor with leakage. $I_{E2}$ is unknown as well, but from Kirchoff’s current laws, we know it is:

$$I_{E2} = I_{RSH} + I_{RFB}$$

Where $I_{RSH}$ is the current through the shunt resistor and

$$I_{RSH} = \frac{V_{E2}}{R_{SH}}$$

This makes our $I_{E2}$ equation:

$$I_{E2} = \frac{V_{E2}}{R_{SH}} + I_{RFB}$$

$V_{E2}$ is:

$$V_{E2} = V_{B1} + (I_{RFB} * R_{FB})$$$$V_{E2} = V_{BE1} + (I_{RFB} * R_{FB})$$

Substituting this into $I_{E2}$ makes our $I_{E2}$ equation the following:

$$I_{E2} = \frac{V_{BE1}}{R_{SH}} + \frac{I_{RFB} * R_{FB}}{R_{SH}} + I_{RFB}$$$$I_{E2} = \frac{V_{BE1}}{R_{SH}} + I_{RFB} * (\frac{R_{FB}}{R_{SH}} + 1)$$

$R_{SH}$ and $R_{FB}$ are known values, and $V_{BE1}$ will typically not change much, so we can approximate that or take it from the transistor measurement. The value that does change substantially here depending on transistor specs is $I_{RFB}$. The $Q_2$ emitter current, and thus the $Q_2$ collector current and bias voltage, are largely determined by $I_{RFB}$, the current through the feedback resistor.

#### Figure 14: It appears from the schematic that $I_{RFB} = I_{B1}$, which is not accurate

Note that in our previous examples using high beta silicon transistors, we had considered $I_{RFB} = I_{B1}$. That may appear to be the case here as well, but it is not accurate, and it is one of the most critical ways that leakage affects the bias of the Fuzz Face circuit. With zero leakage, the only way for current to enter the base of the transistor is through $R_{FB}$. That means that $I_{B1}$ and $I_{RFB}$ are equal. But remember that for a transistor with leakage, the leakage current $I_{CBO}$ flows from the collector to the base node. This provides an additional source of base current. With leakage, the equation for base current is:

$$I_{B1} = I_{RFB} + I_{CBO1}$$

#### Figure 15: A visualization of the additional base current that flows due to $I_{CBO}$

For the bias voltages of $Q_1$, it makes little difference whether the necessary $Q_1$ base current comes from $I_{CBO}$, $I_{RFB}$, or some combination of the two. But for the bias of $Q_2$, whether or not the $Q_1$ base current comes from $I_{CBO}$ or $I_{RFB}$ makes a large difference. Remember that our $I_{E2}$ (and thus our $Q_2$ collector voltage) depends largely on $I_{RFB}$. The more current flowing through $R_{FB}$, the larger the voltage drop across it. As the voltage across $R_{FB}$ increases, $V_{E2}$ increases. This increases $I_{E2}$ and $I_{C2}$, which decreases our bias voltage $V_{C2}$.

If we used a transistor in $Q_1$ with no leakage, all of the base current must flow through $R_{FB}$. If we replaced $Q_1$ with a transistor that had the same specs but some non-negligible amount of leakage current, some of that base current would instead be supplied via $I_{CBO}$, causing less current to flow through $R_{FB}$, decreasing $V_{E2}$ and increasing our bias voltage.

Knowing $I_{RFB}$ is very important for determining our bias voltage. We can calculate it with a known $I_{CBO1}$ and $I_{B1}$, but $I_{B1}$ is unknown.

There are equations we can use for deriving the other unknown transistor pin currents if we just knew one of them, but initially we do not know what any of the pin currents will be. We need to find a way to solve for one of the currents (base, collector, or emitter for either $Q_1$ or $Q_2$) to be able to solve for the others. In our approximation for high beta silicon transistors, we were able to do this by ignoring the unknown variables with a negligible effect on the resulting bias. For germanium transistors with lower beta, those same unknown variables will not necessarily be negligible, so we cannot ignore them.

It is very easy to determine $I_{RFB}$ with $I_{B1}$, so we’ll start there:

$$I_{B1} = \frac{I_{C1}}{\beta_1}$$

We’ll design for a measured or target beta, so that will not be an unknown variable, but $I_{C1}$ is unknown. Using Kirchoff’s current laws, we might determine that $I_{C1}$ is:

$$I_{C1} = I_{RC1} - I_{B2}$$

But this is incorrect. Remember that there is also leakage current flowing from each transistor’s collector to base. $Q_{1C}$ is common with $Q_{2B}$, so the $Q_2$ $I_{CBO}$ leaks into the $Q_{1C}$ / $Q_{2B}$ node. This must be included as one of the currents flowing into the node in our calculation:

$$I_{C1} = I_{RC1} + I_{CBO2} - I_{B2}$$

Where $I_{RC1}$ is the current through $R_{C1}$. Using ohm’s law, we can determine that $I_{RC1}$ is:

$$I_{RC1} = \frac{V_{CC} - V_{C1}}{R_{C1}}$$

Substituting this back into our $I_{C1}$ equation, we get:

$$I_{C1} = \frac{V_{CC} - V_{C1}}{R_{C1}} + I_{CBO2} - I_{B2}$$

And substituting this into our $I_{B1}$ equation, we get:

$$I_{B1} = \frac{\frac{V_{CC} - V_{C1}}{R_{C1}} + I_{CBO2} - I_{B2}}{\beta_1}$$

We have two unknowns here, $V_{C1}$ and $I_{B2}$. We’ll address $V_{C1}$ first. We used Ohm’s law to rewrite $I_{RC1}$ in terms of $V_{C1}$. If we tried to use Ohm’s law to rewrite $V_{C1}$, we would simply undo that work and end up with $I_{RC1}$ as an unknown again. Fortunately, there is another fairly simple way we can solve for $V_{C1}$. Note that $Q_1$ collector is common with $Q_2$ base, so

$$V_{C1} = V_{B2}$$

We know that $V_{B2}$ (and thus $V_{C1}$) is:

$$V_{C1} = V_{B2} = V_{E2} + V_{BE2}$$

The base voltage is the emitter voltage plus the base-emitter voltage drop. We’ll rewrite the unknown $V_{E2}$ using the following:

$$V_{E2} = V_{BE1} + (I_{RFB} * R_{FB})$$$$V_{E2} = V_{BE1} + ((I_{B1} - I_{CBO}) * R_{FB})$$

This makes our $V_{C1}$ equation:

$$V_{C1} = V_{BE2} + V_{BE1} + ((I_{B1} - I_{CBO}) * R_{FB})$$

Substituting this back into our $I_{B1}$ equation, we get:

$$I_{B1} = \frac{\frac{V_{CC} - (V_{BE2} + V_{BE1} + ((I_{B1} - I_{CBO}) * R_{FB}))}{R_{C1}} + I_{CBO2} - I_{B2}}{\beta_1}$$

$I_{B2}$ is still unknown, but we can rewrite that in terms of $I_{B1}$ as well. Remember that the relationship between $I_{B2}$ and $I_{E2}$ is:

$$I_{B2} = \frac{I_{E2}}{\beta_2 + 1}$$

And that $I_{E2}$ is:

$$I_{E2} = I_{RFB} + I_{RSH}$$$$I_{E2} = I_{B1} - I_{CBO1} + \frac{V_{E2}}{R_{SH}}$$

$V_{E2}$ is unknown, but we can rewrite that as well. Remember that $V_{E2}$ is:

$$V_{E2} = V_{BE1} + (I_{RFB} * R_{FB})$$

Substituting this back into our original $I_{B2}$ equation gives us:

$$I_{B2} = \frac{I_{E2}}{\beta_2 + 1}$$$$I_{B2} = \frac{I_{B1} - I_{CBO1} + \frac{V_{E2}}{R_{SH}}}{\beta_2 + 1}$$$$I_{B2} = \frac{I_{B1} - I_{CBO1} + \frac{V_{BE1} + (I_{RFB} * R_{FB})}{R_{SH}}}{\beta_2 + 1}$$$$I_{B2} = \frac{I_{B1}}{\beta_2 + 1} - \frac{I_{CBO1}}{\beta_2 + 1} + \frac{V_{BE1}}{(R_{SH} * \beta_2) + R_{SH}} + \frac{I_{RFB} * R_{FB}}{(R_{SH} * \beta_2) + R_{SH}}$$

Writing $I_{B2}$ in terms of $I_{B1}$ becomes a fairly complex equation, but $I_{B1}$ is now the only unknown. If we substitute this back into our $I_{B1}$ equation for $I_{B2}$, the only unknown value is the one we are solving for, $I_{B1}$:

$$I_{B1} = \frac{V_{CC} - (V_{BE2} + V_{BE1} + (I_{B1} - I_{CBO1})*R_{FB})}{\beta_{1}*R_{C1}} + \frac{I_{CBO2}}{\beta_{1}} - \frac{I_{B1}}{\beta_{1}*(\beta_{2}+1)}$$$$+ \frac{I_{CBO1}}{\beta_{1}*(\beta_{2}+1)} - \frac{V_{BE1}+I_{RFB}*R_{FB}}{\beta_{1}*R_{SH}*(\beta_{2}+1)}$$

We should also simplify this so $I_{B1}$ is not found on the right-hand side of the equation. We are not going to show that process because it’s a lot of work to arrive at the same equation simply written more neatly, but the result is below:

$$I_{B1}=\frac{\frac{(\beta_{2}+1)(V_{CC}+I_{CBO1}*R_{FB}-I_{CBO2}*R_{C1}-V_{BE1}-V_{BE2})}{(\beta_{2}+1)*(R_{FB}+R_{C1}*\beta_{1})+R_{C1}}\color{red}{-\frac{V_{BE1}*R_{C1}-I_{CBO1}*R_{FB}*R_{C1}}{R_{SH}*(R_{C1}*(\beta_{2}*\beta_{1}+\beta_{1}+1)+R_{FB}*(\beta_{2}+1))}-\frac{I_{CBO1}*R_{C1}}{(\beta_{2}+1)*(R_{FB}+R_{C1}*\beta_{1})+R_{C1}}}}{1+\color{red}{\frac{R_{FB}+R_{C1}}{R_{SH}*(R_{C1}*(\beta_{2}*\beta_{1}+\beta_{1}+1)+R_{FB}*(\beta_{2}+1))}}}$$

This is obviously quite a complex equation, and will give the most accurate $I_{B1}$ result. Fortunately, the factors in red are extremely small compared to the others. They can effectively be considered negligible for a very accurate approximation for all realistic circuit values. Our Fuzz Face Bias Calculator uses the full equation, but if you are calculating $I_{B1}$ by hand, you may want to use the simplified approximation:

$$I_{B1}\approx\frac{(\beta_{2}+1)(V_{CC}+I_{CBO1}*R_{FB}-I_{CBO2}*R_{C1}-V_{BE1}-V_{BE2})}{(\beta_{2}+1)*(R_{FB}+R_{C1}*\beta_{1})+R_{C1}}$$

Note that when $I_{CBO1}$ and $I_{CBO2}$ are 0, this becomes even simpler. This is a very slightly less accurate approximation, but it remains accurate for all realistic transistor specs and circuit values, rather than only working for high beta silicon like our Part 4 equations. With the expected variations in transistor specs and component tolerance, this approximation is going to be accurate enough for this application.

Now that we have a method for calculating $I_{B1}$, we can also calculate the other values.

$$I_{RFB} = I_{B1} - I_{CBO1}$$$$I_{C1} = I_{B1} * \beta_1$$$$I_{E1} = I_{B1} * (\beta_1 + 1)$$$$V_{E2} = V_{BE1} + (I_{RFB} * R_{FB})$$$$I_{E2} = \frac{V_{E2}}{R_{SH}} + I_{RFB}$$$$I_{B2} = \frac{I_{E2}}{\beta_2 + 1}$$$$I_{C2} = I_{B2} * \beta_2$$$$V_{C1} = V_{E2} + V_{BE2}$$$$V_{B2} = V_{C1}$$$$V_{C2} = V_{CC} - (R_{C2} * I_{C2})$$$$V_{B1} = V_{BE1}$$$$V_{E1} = 0$$

### Biasing the Schaller Fuzz with Transistor Selection

Let’s look back at our Schaller Fuzz example again, this time using our new formulas. Remember that our goal was to take the germanium Schaller Fuzz circuit with the stock resistor values, and achieve proper bias with transistor selection.

Note that we are mainly looking at one voltage ($V_{C2}$), and this voltage is affected by beta, leakage, and $V_{BE}$ of both transistors. There is not one set of transistor specifications that can achieve a certain bias voltage, but many. To narrow down the number of variables present, we’ll use 0.1V as a general germanium transistor $V_{BE}$ approximation, and we’ll design for transistors with negligible leakage ($0μA) to minimize heat sensitivity. This still leaves two variables that control$V_{C1}$,$\beta_1$and$\beta_2$. Though$\beta_1$has a much larger effect on the bias voltage, different combinations of$\beta_1$and$\beta_2$can be used for the same$V_{C2}$, which means there is no one solution, but many combinations of$\beta_1$and$\beta_2$which will work. We will address that shortly. Since$V_{C2}$is the voltage of most interest for biasing, we start with our$V_{C2}$formula: $$V_{C2} = V_{CC} - (R_{C2} * I_{C2})$$ From here, we are going to insert the known values where we can, and substitute our known formulas in for any unknown values. We’ll then repeat the process with the new formula until we have written$V_{C2}$in terms of$I_{B1}$,$\beta_2$, and known values. $V_{C2} = 9 - (1470 * I_{C2})V_{C2} = 9 - (1470 * I_{B2} * \beta_2)$(Using$I_{C2}$equation)$V_{C2} = 9-(1470*I_{E2}*\frac{\beta_{2}}{\beta_{2}+1})$(Using$I_{B2}$equation)$V_{C2} = 9-(1470*(\frac{V_{E2}}{1000}+I_{RFB})*\frac{\beta_{2}}{\beta_{2}+1})$(Using$I_{E2}$equation)$V_{C2} = 9-(1470*(\frac{0.1+(I_{RFB}*100000)}{1000}+I_{RFB})*\frac{\beta_{2}}{\beta_{2}+1})$(Using$V_{E2}$equation)$V_{C2} = 9-(1470*(\frac{0.1+(I_{B1}*100000)}{1000}+I_{B1})*\frac{\beta_{2}}{\beta_{2}+1})$(Using$I_{RFB}$equation)$V_{C2} = 9-((\frac{147+(147,000,000*I_{B1})}{1000}+(1470*I_{B1}))*\frac{\beta_{2}}{\beta_{2}+1})$(Simplifying)$V_{C2} = 9-((0.147+(147,000*I_{B1})+(1470*I_{B1}))*\frac{\beta_{2}}{\beta_{2}+1})$(Simplifying)$V_{C2} = 9-((0.147+(148,470*I_{B1}))*\frac{\beta_{2}}{\beta_{2}+1})$(Simplifying) We could now substitute our$I_{B1}$equation in here, but a simpler way to solve this is to do it in two parts. First, we’ll determine what our$I_{B1}$needs to be, then we’ll use our$I_{B1}$equation to determine how we can achieve the necessary$I_{B1}$. $$V_{C2} = 9 -\frac{(0.1470 + (I_{B1} * 148470)) * \beta_2}{\beta_2 + 1}$$ First, we need a target$V_{C2}$. From reports online as well as the measurements we have taken of an original unmodified germanium Schaller Fuzz (from which the schematic we are using was drawn),$V_{C2}$typically hovers around 6V in original units. We will use that as our target. $$6 = 9 - ((I_{B1} * 148470 + 0.147) * \frac{\beta_2}{\beta_2 + 1})$$$$3 = (I_{B1} * 148470 + 0.147) * \frac{\beta_2}{\beta_2 + 1}$$ Rewriting this in terms of$I_{B1}$: $$I_{B1} = \frac{3 * \frac{\beta_2 + 1}{\beta_2} - 0.147}{148470}$$$$I_{B1} = 20.2\text{μ} * \frac{\beta_2 + 1}{\beta_2} - 0.99\text{μ}$$ The exact$I_{B1}$value required to achieve 6V depends on$\beta_2$. Until we know or decide on a$\beta_2$value, we won’t know the exact$I_{B1}$necessary. As we mentioned earlier,$\beta_1$and$\beta_2$both affect$V_{C2}$. But let’s take a look at how$\beta_2$affects the values here. Its effect is much smaller than it may initially appear. Note that one of the factors is multiplied by $$\frac{\beta_2 + 1}{\beta_2}$$ As$\beta_2$approaches infinity, this value approaches 1, and for most reasonable values of$\beta_2$, it is very close to 1. Table 1 shows what this value will be for various$\beta_2$values: Table 1:$\frac{\beta2 + 1}{\beta2}$calculations for various$\beta_2$values$\beta_2\frac{\beta_2 + 1}{\beta_2}$31.333 51.200 101.100 201.050 301.033 401.025 501.020 601.017 701.014 801.013 901.011 1001.01 2001.005 3001.003 Unless$\beta_2$is abnormally low, its effect on the necessary$I_{B1}$will be minimal. We can choose a common germanium$\beta$value like 70 as our$Q_2$target, then solve for$I_{B1}$. If we end up using a$Q_2$with a different$\beta$value, our bias voltage will change very little from our target for any reasonable$\beta_2$value. Using $$\beta_2 = 70$$ let’s solve for$I_{B1}$: $$I_{B1} = (20.2\text{μ} * 1.014) - 0.99\text{μ}$$$$I_{B1} = 19.4928\text{μA}$$ Now let’s use our$I_{B1}$equation to determine the$Q_1$transistor specs we need to achieve this in the stock Schaller circuit. We’ll use the simplified version of our$I_{B1}$equation for a very close approximation which is much easier to calculate by hand: $$I_{B1}\approx\frac{(\beta_{2}+1)(V_{CC}+I_{CBO1}*R_{FB}-I_{CBO2}*R_{C1}-V_{BE1}-V_{BE2})}{(\beta_{2}+1)*(R_{FB}+R_{C1}*\beta_{1})+R_{C1}}$$$$I_{B1}\approx\frac{(70+1)(9+0-0-0.1-0.1)}{(70+1)*(100000+15000*\beta_{1})+15000}$$$$I_{B1} = \frac{624.8}{7114000 + (1065000 * \beta_1)}$$ Rewriting this to solve for$\beta_1$: $$\beta_1 = \frac{\frac{624.8}{I_{B1}} - 7115000}{1065000}$$$$\beta_1 = \frac{\frac{624.8}{19.4928\text{μ}} - 7115000}{1065000}$$$$\beta_1 = \frac{32052860 - 7115000}{1065000}$$$$\beta_1 = \frac{24937860}{1065000}$$$$\beta_1 = 23.42$$$Q_1$needs the low beta value of 23.42 to achieve the proper bias voltage. If we use our more complex$I_{B1}$equation, the more accurate result is 23.23, which is extremely close to our approximation. If we were to use our calculations from part 4, the results would be less accurate due to the low$V_{BE}$of the germanium transistors and the low beta requirement. We have discussed previously how beta can vary even among transistors of the same type. Trying to recreate the Schaller circuit with a$Q_1$that has that exact calculated beta value would be a futile exercise - particularly since beta changes some with temperature. A more reasonable approach would be to consider$23 the approximate beta measurement, aim for beta values in that ballpark, and live with some bias voltage variations. Original Schaller units would not all have had the exact same beta or bias either.

Note also that for bias purposes, the $\beta_1$ target of $23 is much more important than the$\beta_2$target of 70. A wider range of transistors can be used for$Q_2~ without causing much variation in the bias voltage.

### Further Applications

Knowing the above, we can plug in any combination of transistor specs and resistor values to determine what the bias voltage should be, and it remains very accurate for both germanium and silicon. Remember that this equation now accounts for leakage current, which is quite temperature sensitive. The more leakage current, the more of a role it plays in the bias, making the resulting bias voltage more sensitive to temperature. An easy way to minimize temperature sensitivity in a germanium Fuzz Face (and other circuits using the same topology) is to select low leakage transistors and use the above equations to bias them.

In part 6, we’ll take a more in-depth look at the functionality of other classic transistor circuits, including finishing the design of a more stable Tonebender Mk2 input stage. We’ll take a closer look at the heat-sensitivity caused by transistor leakage and see how different transistor specs affect the temperature stability of real circuits. We’ll also take a look at the cause and potential solutions for some of the other common quirks found in classic fuzz circuits, such as their sensitivity to where they are placed in the signal chain, poor “Fuzz” control range, and the effects that transistor Beta has on the circuit beyond the bias conditions.