# Silicon and Germanium Transistor Biasing - Part 4

### Part 4: Fuzz Face Topology:

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

Single transistor stages can be useful on their own, and there are quite a few highly-regarded boost pedals which are not much more than a common emitter amplifier in a box. However, it’s also common to see transistor amplifiers used as part of larger circuits or combined with other transistor amplifiers. In previous sections of this article, it was mentioned that for a single common emitter amp stage, increased voltage gain comes at the cost of linearity. Two or more common emitter amps in series can be used for increased overall voltage gain with reduced non-linearity compared to a single common emitter amp stage. It is fairly common to cascade two or more lower gain common emitter amps. Depending on how they are arranged, sometimes the DC bias conditions of the transistors are completely isolated from each other and each transistor can be analyzed separately. An example of that can be seen in Figure 1.

In this example, both amps are biased separately, with the capacitor between them preventing DC voltage from passing from one stage to the next (the capacitors at the input and output prevent DC from entering the circuit via the input or output connection). The biasing of one amplifier in this circuit has no effect on the bias of the other, as the DC voltage is blocked by the capacitor. Each transistor stage in Figure 1 can be analyzed separately for DC purposes. If the DC voltages are not isolated, this is not true. The system must be looked at as a whole.

One very common example of a two-stage amp where DC is not isolated between stages is the Fuzz Face circuit. The schematic for the silicon variant can be seen in Figure 2. The bias of the two transistors can be interactive (particularly, the bias of ~Q_1~ affects the bias of ~Q_2~).

Let’s take a look at how the bias can be determined in this circuit, starting with silicon transistors. With the capacitors of the Fuzz Face circuit considered to be open circuit for DC analysis, the circuit looks like that seen in Figure 3. The paths where DC flows between stages have been indicated in the schematic.

When current is flowing, there will be a ~0.7V voltage drop from base to emitter for both transistors if they are silicon. We can use this to approximate a few voltages in the circuit. Since the emitter of ~Q_1~ is at 0V, its base should be at approximately 0.7V. We also know that the base of ~Q_2~ and collector of ~Q_1~ (which are common to each other) should be approximately 0.7V higher than the emitter of ~Q_2~ because of its base-emitter voltage drop, but right now the emitter voltage of ~Q_2~ is unknown.

We can make a very close ~V_{E2}~ approximation with the info we have. First, note that with a 33k resistor on the collector of ~Q_1~, with a 9V power supply, we can use ohm’s law to determine that only 273μA of current is needed to drop 9V, putting the transistor in saturation. In order for the transistor to be in the active region, less than 273μA must be flowing through ~Q_{1C}~, so we’ll consider 273μA to be the maximum ~I_{C1}~. For silicon transistors which typically have high betas, this amount of current will result in a very small amount of base current. With a known max collector current, we can determine a max base current if we know transistor beta (~\beta~), due to the relationship:

$$I_B = \frac{I_C}{\beta}$$For this example, we will again assume that two BC108Bs are used with a beta of 300. With a beta of 300, our max ~I_{B1}~ is

$$I_{B1} = \frac{273\text{μA}}{300}$$which is <1μA.

This small ~Q_1~ base current results in a very small voltage drop across the 100k feedback resistor, so we can consider the voltage drop to be negligible for a reasonable approximation of ~V_{E2}~. That means that

$$V_{E2} \approx V_{B1}$$or

$$V_{E2} = 0.7\text{V}$$We will improve the accuracy of our ~V_{E2}~ value shortly, but for now we will use 0.7V which is a rough approximation.

Because

$$V_{C1} = V_{E2} + 0.7$$we can also approximate ~V_{C1}~ as

$$V_{C1} \approx 1.4\text{V}$$~Q_{2B}~ is common with ~Q_{1C}~, so ~1.4V is also the ~Q_{2B}~ voltage. The actual voltage drop at the base-emitter junction can vary among transistors, and it also depends on the amount of current flowing through it. We are also ignoring ~V_{FB}~ in our ~V_{C1}~ equation, so this ~Q_{1C}~ voltage is a rough approximation, but that is good enough for our needs here.

Knowing ~V_{C1}~, we can also make an approximation of the current that flows into ~Q_{1C}~. Using ohm’s law with a 7.6V voltage drop across a 33k resistor, we can determine that 230μA will be flowing through ~R_{C1}~. This current then flows into ~Q_{1C}~ and ~Q_{2B}~ (~I_{C1}~ and ~I_{B2}~). This means that the current flowing through ~R_{C1}~ (~I_{RC1}~) is the sum of ~I_{C1}~ and ~I_{B2}~.

#### Figure 7: Using Kirchoff’s Current Laws, we can see that ~I_{RC1}~ is the sum of ~I_{C1}~ and ~I_{B2}~

We know

$$I_{RC1} \approx 230\text{μA}$$but in order to solve for ~I_{C1}~, we’d also need to know ~I_{B2}~. We could solve for ~I_{B2}~ (and we will later in this article), but the equations start to get quite complex. To keep things simpler for now, we’ll use an approximation here. We can approximate the max ~I_{B2}~ with the max ~I_{C2}~, like we did for ~Q_1~.

Our ~V_{E2}~ is approximately 0.7V. This leaves 8.3V as the maximum voltage that can drop across ~R_{C2}~. We can determine the current this happens at with Ohm’s Law, knowing

$$R_{C2} = 8530\text{Ω}$$Therefore:

$$MAX(I_{C2}) = \frac{MAX(V_{RC2})}{R_{C2}}$$$$MAX(I_{C2}) = \frac{8.3\text{V}}{8530\text{Ω}}$$$$MAX(I_{C2}) = 973\text{μA}$$Using the relationship

$$I_{B2} = \frac{I_{C2}}{\beta}$$We can solve for the max base current as well with a beta of 300:

$$MAX(I_{B2}) = \frac{MAX(I_{C2})}{\beta}$$$$MAX(I_{B2}) = \frac{973\text{μA}}{300}$$$$MAX(I_{B2}) = 3.24\text{μA}$$We know less than 3.3μA will be flowing into ~Q_{2B}~, which is very small compared to ~I_{RC1}~, so we will consider

$$I_{C1} \approx I_{RC1}$$With an approximate ~I_{C1}~, we can approximate ~I_{B1}~ with a known ~Q_1~ beta:

$$I_{B1} = \frac{I_{C1}}{\beta}$$$$I_{B1} = \frac{\sim230\text{μA}}{300}$$$$I_{B1} = 0.77\text{μA}$$Because we have an approximation of ~I_{B1}~ now, we can recalculate ~V_{E2}~ for more accuracy. Instead of ignoring ~V_{FB}~, we will include it:

$$V_{E2} = V_{BE1} + V_{FB}$$$$V_{E2} = V_{BE1} + (I_{B1} * R_{FB1})$$$$V_{E2} = 0.7\text{V} + (0.77\text{μA} * 100\text{kΩ)}$$$$V_{E2} = 0.777\text{V}$$Our ~V_{C1}~ value also depended on our previous ~V_{E2}~ approximation in which ~I_{B1}~ was ignored. We could re-calculate ~V_{C1}~ as well now that we have a more accurate ~V_{E2}~, but for simplicity’s sake we will stop here. The true (non-approximate) calculation for ~V_{C1}~ depends on ~I_{B1}~, and the true (non-approximate) calculation for ~I_{B1}~ depends on ~V_{C1}~:

$$V_{C1} = V_{BE1} + V_{BE2} + (I_{B1} * R_{FB})$$$$I_{B1} = \frac{V_{C1}}{R_{C1}} - I_{B2}$$At first glance, it looks like there is no simple way to solve for either of these without knowing the other. We got around this interdependency by ignoring the (relatively small) voltage that ~I_{B1}~ contributes to ~V_{C1}~ in our initial ~V_{C1}~ estimate, using the following formula instead:

$$V_{C1} = V_{BE1} + V_{BE2}$$This gave us a close enough approximation for ~V_{C1}~, which we used to determine a more accurate ~I_{B1}~. Now that our ~I_{B1}~ has changed, ~V_{C1}~ is slightly inaccurate. But because these formulas depend on each other, we could keep re-calculating them forever. The result of doing so is a smaller and smaller marginal accuracy increase each time they are re-calculated, but as long as we re-calculate ~V_{E2}~ with our ~I_{B1}~ approximation rather than ignoring ~I_{B1}~, the numbers will be very accurate already. We will come up with a better way to handle this in Part 5 of this series of articles, but this approximation is the simpler way, and the resulting error is negligible for high beta silicon.

At this point, we know or have approximations of all relevant ~Q_1~ voltages, but we still have an unknown ~V_{C2}~. We know the max realistic values for ~Q_2~ base and collector currents, but we do not know the actual value for any of the relevant ~Q_2~ currents at this point. We need to know ~I_{C2}~ to solve for ~V_{C2}~:

$$V_{C2} = V_{CC} - (I_{C2} * R_{C2})$$~R_{C2}~ (33kΩ) and ~V_{CC}~ (our supply voltage, 9V) are both known. If we know ~I_{B2}~ or ~I_{E2}~, we can solve for ~I_{C2}~ using beta:

$$I_{C2} = I_{B2} * \beta_2$$$$I_{C2} = \frac{I_{E2} * \beta_2}{(\beta_2 + 1)}$$We do not know ~I_{B2}~ or ~I_{E2}~ at this point, but we can easily solve for ~I_{E2}~. The current flowing out of ~Q_{2E}~ must flow through ~R_{SH}~ and ~R_{FB}~:

#### Figure 9: Using Kirchoff’s Current Laws, we can see that ~I_{E2}~ is the sum of ~I_{RSH}~ and ~I_{RFB}~ (~I_{RFB} = I_{B1}~)

The current flowing through ~R_{FB}~ is ~I_{B1}~, for which an approximation is known (0.77μA), and we can easily solve for the current flowing through ~R_{SH}~ with ohm’s law:

$$I_{RSH} = \frac{V_{E2}}{R_{SH}}$$$$I_{RSH} = \frac{0.777\text{V}}{1000\text{Ω}}$$$$I_{RSH} = 777\text{μA}$$The total emitter current is 777.77μA. ~I_{B1}~ can reasonably be ignored here when using high beta silicon transistors. It is ~0.1% of the total emitter current in this example, and with the approximations we are using we have no need for that level of precision, so we will simply use 777μA for ~I_{E2}~.

We can now solve for ~I_{C2}~ as mentioned previously:

$$I_{C2} = \frac{I_{E2} * \beta_2}{\beta_2 + 1}$$$$I_{C2} = \frac{777\text{μA} * 300}{301}$$$$I_{C2} = 775.2\text{μA}$$We can solve for ~I_{B2}~ as well:

$$I_{B2} = \frac{I_{C2}}{\beta_2}$$$$I_{B2} = \frac{775.2\text{μA}}{300}$$$$I_{B2} = 2.58\text{μA}$$With a known ~I_{C2}~, we can also solve for ~V_{C2}~:

$$V_{C2} = V_{CC} - (I_{C2} * R_{C2})$$$$V_{C2} = 9\text{V} - (777\text{μA} * 8530\text{Ω})$$$$V_{C2} = 9\text{V} - 6.63\text{V}$$$$V_{C2} = 2.37\text{V}$$We’d like our ~V_{C2}~ voltage to be around 5V, so this is a little low. We can see from the above equation that ~V_{C2}~ is highly dependent on ~R_{C2}~, so the simplest way to re-bias ~V_{C2}~ is to modify ~R_{C2}~. To determine the ideal ~R_{C2}~ for a better bias voltage, we’ll re-arrange the above equation, set ~V_{C2}~ = 5, and solve for ~R_{C2}~:

$$V_{C2} = V_{CC} - (I_{C2} * R_{C2})$$$$R_{C2} = \frac{V_{CC} - V_{C2}}{I_{C2}}$$$$R_{C2} = \frac{9\text{V} - 5\text{V}}{775.2\text{μA}}$$$$R_{C2} = 5160\text{Ω}$$To get our bias voltage closer to 5V, we must lower the value of ~R_{C2}~. This tracks with what can be found in later silicon Fuzz Face variants - the Crest Audio Fuzz Face, for example, uses a 6.2kΩ resistor in series with a 330Ω resistor for ~Q_{2C}~ (6530Ω), which should bias up to ~4V with a 9V battery.

When determining the optimal ~R_{C2}~ value for biasing ~Q_{2C}~ to approximately 5V, we did so using the stock Fuzz Face resistors and a transistor with a known typical beta. We’d like to derive some general formulas for determining the bias conditions for any resistor values and transistor types using the same concepts we used to solve for the Fuzz Face values with BC108B. This will allow us to approximate the bias conditions for varying transistor specs and resistor values, and also make it easy to determine proper resistor values for re-biasing a poorly-biased Fuzz Face.

With ~V_{E1}~ grounded, it is always 0V:

$$V_{E1} = 0\text{V}$$As demonstrated in our example, because of the base-emitter diode drop, we know:

$$V_{B1} = V_{E1} + V_{BE1}$$$$V_{B1} = 0\text{V} + V_{BE1}$$$$V_{B1} = V_{BE1}$$We also know:

$$V_{C1} = V_{BE1} + V_{BE2} + V_{FB}$$Where ~V_{FB}~ is the voltage drop across the feedback resistor. In order to avoid interdependence with ~I_{B1}~, we’ll ignore ~V_{FB}~ for a reasonably accurate ~V_{C1}~ approximation:

$$V_{C1} = V_{BE1} + V_{BE2}$$Using an approximate ~V_{C1}~, we can approximate ~I_{C1}~:

$$I_{C1} = \frac{V_{CC} - V_{C1}}{R_{C1} - I_{B2}}$$We determined previously that ~I_{B2}~ is negligible in this calculation if we use high beta, so we will simplify:

$$I_{C1} = \frac{V_{CC} - V_{C1}}{R_{C1}}$$With ~\beta_1~ and ~I_{C1}~, we can solve for ~I_{E1}~ and ~I_{B1}~. Because:

$$\beta_1 = \frac{I_{C1}}{I_{B1}}$$we know:

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

$$I_{E1} = I_{B1} + I_{C1}$$We now have a formula for an approximation of each pin’s voltage and current for ~Q_1~:

$$V_{C1} = V_{BE1} + V_{BE2}$$$$V_{B1} = V_{BE1}$$$$V_{E1} = 0\text{V}$$$$I_{C1} = \frac{V_{CC} - V_{C1}}{R_{C1}}$$$$I_{B1} = \frac{I_{C1}}{\beta_1}$$$$I_{E1} = I_{B1} + I_{C1}$$We’ll now take a look at ~Q_2~. ~V_{E2}~ is equal to the sum of the ~Q_1~ base-emitter voltage drop and the voltage drop across the feedback resistor:

$$V_{E2} = V_{BE1} + V_{FB}$$$$V_{E2} = V_{BE1} + (I_{B1} * R_{FB})$$Using Kirchoff’s current laws, we know that:

$$I_{E2} = I_{RSH} + I_{B1}$$And we can use ~V_{E2}~ to solve for ~I_{RSH}~:

$$I_{RSH} = \frac{V_{E2}}{R_{SH}}$$Because ~I_{B1}~ is negligible compared to ~I_{RSH}~, we can consider

$$I_{E2} \approx I_{RSH}$$$$I_{E2} = \frac{V_{E2}}{R_{SH}}$$Because we know

$$I_{C2} = I_{B2} * \beta$$and

$$I_{E2} = I_{C2} + I_{B2}$$we can calculate ~I_{C2}~ and ~I_{B2}~ using ~I_{E2}~ and beta:

$$I_{E2} = \frac{V_{E2}}{R_{SH}}$$$$I_{B2} = \frac{I_{E2}}{\beta_2 + 1}$$$$I_{C2} = \frac{I_{E2} * \beta_2}{\beta_2 + 1}$$We know ~V_{B2}~ is the ~Q_2~ base-emitter diode drop above ~V_{E2}~:

$$V_{E2} = V_{BE1} + (I_{B1} + R_{FB})$$$$V_{B2} = V_{BE1} + (I_{B1} * R_{FB}) + V_{BE2}$$~V_{C2}~ is equal to the supply voltage minus the voltage drop across the collector resistance of ~R_{C2}~. In a standard Fuzz Face, ~R_{C2}~ consists of two resistors in series which we will treat as one resistance value. We calculate ~V_{RC2}~ by using ohm’s law, ~I_{C2}~, and ~R_{C2}~:

$$V_{RC2} = I_{C2} * R_{C2}$$~V_{C2}~ is simply:

$$V_{C2} = V_{CC} - V_{RC2}$$$$V_{C2} = V_{CC} - (I_{C2} * R_{C2})$$Our equations for approximating all transistor currents and voltages are:

$$V_{C1} = V_{BE1} + V_{BE2}$$$$V_{B1} = V_{BE1}$$$$V_{E1} = 0\text{V}$$$$I_{C1} = \frac{V_{CC} - V_{C1}}{R_{C1}}$$$$I_{B1} = \frac{I_{C1}}{\beta_1}$$$$I_{E1} = I_{B1} + I_{C1}$$$$$$$$V_{C2} = V_{CC} - (I_{C2} * R_{C2})$$$$V_{B2} = V_{E2} + V_{BE2}$$$$V_{E2} = V_{BE1}$$$$I_{C2} = \frac{I_{E2} * \beta_2}{\beta_2 + 1}$$$$I_{B2} = \frac{I_{E2}}{\beta_2 + 1}$$$$I_{E2} = \frac{V_{E2}}{R_{SH}}$$Note that the above approximations remain accurate with silicon transistors and large beta values (>100, the larger the more accurate). With lower beta, the error will increase. The reason for this is that with large betas, the base currents that we are currently ignoring in certain places are very small relative to the other currents flowing through the circuit, allowing for them to be ignored in certain cases for a very accurate approximation while keeping the calculations much simpler. Including these base currents in the calculations adds a substantial amount of complexity.

With lower betas, more base current needs to flow to maintain proper bias currents, and with larger base currents they start to become less and less negligible, and we need to consider those currents in these calculations to maintain accuracy. We will cover how to handle this scenario in Part 5 when we discuss considerations for germanium transistor biasing, where low beta values are much more common.