# Silicon and Germanium Transistor Biasing - Part 3

### Part 3: Designing an improved, stabilized common emitter amp:

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

In Part 2, we designed a simple common emitter amp with no emitter resistor and with the base bias current being set with an approximation of a current source. The result was a functional amplifier with widely varying bias voltages due largely to variations in the transistor beta ($\beta$). In this article, we’ll redesign our common emitter amplifier using an emitter resistor and a voltage divider on the base, as seen in Figure 1.

#### Figure 1: Common emitter amp with emitter resistor and voltage divider on the base

Note that like our example in Part 2, for DC analysis the input and output capacitors can be considered open circuit as in Figure 2.

#### Figure 2: DC analysis of a common emitter amp with emitter resistor

When we have an emitter resistor in place, the effective emitter resistance ($R_E$) is the internal resistance of the emitter junction ($r_{e’}$) in series with the emitter resistor ($R_e$). The equation for voltage gain (A) becomes:

$$A = -\frac{R_C}{R_E}$$$$A = -\frac{R_C}{R_e + r_{e’}}$$

When $R_e$ is larger than $r_{e’}$, which it typically is, it plays a larger role in the voltage gain of the circuit. Unlike $r_{e’}$, which depends on temperature and is inversely proportional to emitter current, $R_e$ remains static. As $R_C$ is increased, the negative voltage gain will increase. $r_{e’}$ will still counteract that to some extent even with an emitter resistor in place, but unless the emitter resistor is a very small resistance value, the effects of $r_{e’}$ will be small or negligible.

As we mentioned in part 1, using an emitter resistor helps with stable bias. It also lowers gain and provides more linearity to the amp. If it is bypassed by a capacitor to ground, the emitter is effectively grounded for AC signals. This results in a large voltage gain for AC signals, while retaining the stabilization of DC bias voltages. However, when bypassing the emitter resistor, the linearity benefits are lost. Let’s now take a look at how we can design for the right bias in a common emitter amp with an emitter resistor.

In the case of a common emitter amp with no emitter resistor, the effective emitter resistance is the internal resistance of the transistor emitter ($r_{e’}$), which is given by the equation:

$$r_{e’} = \frac{25\text{mV}}{I_E}$$

For $I_E$ of 1mA, the internal emitter resistance will be 25Ω. It increases as $I_E$ decreases, but generally we select an emitter resistor which is substantially larger. The emitter resistor decreases the available headroom of the common emitter amp, so we do not want to make it too large either. An emitter resistor dropping $10% of the supply voltage is a reasonable compromise. Here, we will aim for$R_E$to drop 1V as it is$11.1% of our $V_{CC}$ and gives us a nice even number to work with.

Using an $R_e$ which drops 1V will substantially stabilize the bias voltage without cutting into the headroom much. We saw in our previous example that with a 4.7k collector resistor and 9V supply, $I_C$ should be about 957μA to achieve a bias voltage of $4.5V on the collector. With$R_e$dropping voltage, the ideal collector bias voltage now occurs with$R_C$dropping$0.5 * (V_{CC} - V_{RE})$rather than$0.5 * V_{CC}$. Since 1V is dropping across$R_E$, we want$R_C$to drop 4V, leaving 5V on the collector (9.0V - 4.0V). We can now calculate the ideal collector current for this new bias voltage: $$I_C = \frac{V_{RC}}{R_C}$$$$I_C = \frac{4.0}{4.7\text{k}}$$$$I_C = 851\text{ μA}$$ Assuming a BC108B transistor with typical$\beta$of 300, we can calculate the ideal base current for achieving$I_C = 851\text{ μA}$: $$I_B = \frac{I_C}{\beta}$$$$I_B = \frac{851\text{ μA}}{300}$$$$I_B = 2.84\text{ μA}$$ #### Figure 3: Common emitter amp with known base and collector currents When using a base voltage divider for bias, the current flowing through the divider circuit ($R_{B1}$and$R_{B2}$) can be “loaded” by current flowing into the base. The base of the transistor is connected to the output of the base voltage divider. Without this connection to the transistor’s base, the equation for the voltage divider would be: $$V_{DIV} = V_{CC} * \frac{R_{B2}}{R_{B1} + R_{B2}}$$ However, with the base of the transistor connected to the output of the voltage divider, the effective resistance into the base of the transistor is in parallel with$R_{B2}$. This changes the output of our voltage divider. We could attempt to compensate for the loading effects of the transistor, but the degree to which the divider is loaded depends on transistor specs which can vary. To minimize the loading effect for a wide range of transistor specifications, we’d like the current flowing through$R_{B2}$to be substantially larger than that flowing into the base. The transistor will still load the divider, but because its current draw is substantially smaller than the current flowing through the other leg of the divider, the loading effect is minimal. A good rule of thumb is to make the current flowing through$R_{B2}$at least ten times larger than that flowing into the base. Knowing$I_B = 2.84\text{μA}$, we want 28.4μA ($I_B * 10$) to flow through$R_{B2}$. To calculate the$R_{B2}$resistance value required, we’ll use ohm’s law. Assume a$V_{BE}$of 0.7V for a silicon transistor, and remember that$R_E$is dropping 1V: $$R_{B2} = \frac{V_B}{I_{RB2}}$$$$R_{B2} = \frac{V_B}{28.4\text{μA}}$$$$R_{B2} = \frac{V_{RE} + V_{BE}}{28.4\text{μA}}$$$$R_{B2} = \frac{1 + 0.7}{28.4\text{μA}}$$$$R_{B2} = \frac{1.7}{28.4\text{μA}}$$$$R_{B2} = 59,859\text{Ω}$$ The current that flows into the base of the transistor as well as the current flowing through$R_{B2}$must also flow through$R_{B1}$. In other words: $$I_{RB1} = I_{RB2} + I_B$$ Because$I_{RB2} = I_B * 10$, we know $$I_{RB1} = I_B * 11$$$$I_{RB1} =$31.2\text{ μA}$$

The voltage drop across $R_{B1}$ can be found by subtracting the base voltage from the supply voltage:

$$V_{RB1} = V_{CC} - V_B$$$$V_{RB1} = 9\text{V} - 1.7\text{V}$$$$V_{RB1} = 7.3\text{V}$$

Now we can calculate the required RB1 resistance value knowing the voltage drop and current through RB1:

$$R_{B1} = \frac{V_{RB1}}{I_{RB1}}$$$$R_{B1} = \frac{7.3\text{V}}{31.24\text{ μA}}$$$$R_{B1} = 233,674Ω$$

Last, we can calculate the emitter resistor value. We know

$$I_E = I_C + I_B$$

and with our known $I_C$ and $I_B$ values we can solve for the emitter current:

$$I_E = 2.84\text{ μA} + 851\text{ μA}$$$$I_E = 853.84\text{ μA}$$

Using ohm’s law and with a known emitter current and voltage drop (1V) across the emitter resistor, we can solve for $R_E$:

$$R_E = \frac{1\text{V}}{853.84\text{ μA}}$$$$R_E = 1,171Ω$$

#### Figure 4: Common emitter amp with target resistances, currents, and voltages

It is not reasonable to aim for the exact resistance values seen in Figure 4, so we will round these resistances to the nearest common value:

• $R_C = 4.7\text{ kΩ}$
• $R_E = 1.2\text{ kΩ}$
• $R_{B1} = 240\text{ kΩ}$
• $R_{B2} = 62\text{ kΩ}$

Using an emitter bypass capacitor gives us high voltage gain while retaining the bias stabilizing benefits of the emitter resistor. $C_e$ tends to be selected so that as much of the desired signal receives full amplification as reasonably possible. $C_e$ forms a high pass filter, so for audio signals it is often selected so that signals of $20Hz and above will pass with full amplification. However,$C_e$can also be intentionally selected to trim some low end if desired. We can think of$C_e$as having a certain amount of resistance for a particular signal frequency. For high enough signals, this resistance is effectively zero, which is why a capacitor is considered a short circuit in AC analysis. At DC, the resistance is effectively infinite, which is why a capacitor is considered an open circuit in DC analysis. For AC signals of a low enough frequency,$C_e$will start to have a non-negligible amount of effective resistance. The amount of effective resistance (known as capacitive reactance) can be calculated with the following formula: $$Z = \frac{1}{2\pi * f * C_E}$$ Where$f$is the frequency of interest. When considering a signal at just one frequency,$C_e$can effectively be considered to be a resistor with resistance$Z$: #### Figure 11: Common emitter amp with emitter bypass cap viewed as a resistor for a set signal frequency For high frequency signals,$Z$will effectively be zero and we can again view the circuit as having its emitter grounded for those signals. But for low enough frequencies,$Z$is non-negligible. The emitter will not be grounded for that frequency, and the voltage gain formula changes to the following: $$A = -\frac{R_C}{r_{e’} + R_e || Z}$$ Where$R_e||Z$is the value of$R_E$in parallel with$Z$. If$Z$is much higher than$R_E$, it will have little effect on the effective resistance, and the gain will be approximately$-\frac{R_C}{R_e}$or approximately -3.8. If it is equal to$R_E$, the gain will be higher, but still substantially lower than the gain of$-162. The gain of the signal approaches -162 as $Z$ approaches zero.

Remember that $Z$ depends on frequency, and its value increases as the frequency decreases. In order to allow all important signals to pass at full gain, we’ll attempt to select a capacitance which results in a low $Z$ for the lowest frequency of interest. For audio circuits, that’s usually a minimum of 20Hz. In order to limit the effects that the capacitor’s reactance has on the gain, a good rule of thumb is to select $C$ so that $Z$ at the lowest frequency is at least an order of magnitude smaller than $R_E$.

Before we calculate $C_e$, let’s first look at how a $Z$ of $\frac{R_e}{10}$ will affect our gain.

$$A = -\frac{R_C}{r_{e’} + R_e || Z}$$$$A = -\frac{-4700}{29 + 1200||120}$$$$A = -\frac{4700}{138}$$$$A = -34.06$$$$A \simeq 30\text{ dB}$$

As you can see, even with a $Z$ of $\frac{R_e}{10}$, our gain is quite a bit lower at the lowest frequencies of interest. But let’s go ahead and consider the required capacitance here. To do so, we’ll set $f = 20\text{Hz}$ and $Z = \frac{R_e}{10}$, and solve for $C_e$. From the formula:

$$Z = \frac{1}{2\pi * f * C_E}$$

We can re-arrange this to:

$$C_e = \frac{1}{2\pi * Z * f}$$

Inserting $Z$ = 120 and $f$ = 20:

$$C_e = \frac{1}{2\pi * 120 * 20}$$$$C_e = \frac{1}{15079.64}$$$$C_e = 0.0000663$$$$C_e = 66.3\text{ μF}$$

Increasing capacitance will increase gain for low frequencies, so we’ll round this value up to 100μF.

#### Figure 12: Common emitter amp with emitter bypass cap selected for $Z$ = $\frac{R_e}{10}$ at 20Hz

Let’s now look at the case where we aim for a $Z$ of $\frac{R_e}{100}$ for our lowest frequencies. First, we’ll calculate the gain:

$$A = -\frac{R_C}{r_{e’} + R_e || Z}$$$$A = -\frac{-4700}{29 + 1200||12}$$$$A = -\frac{4700}{40.88}$$$$A = -115$$$$A \simeq 41.2\text{ dB}$$

As you can see, this is still notably lower than -162. Because $Z$ is inversely proportional to capacitance and we know that with a $Z$ of $\frac{R_e}{10}$, $C_e$ = 66.3μF, we know that with $\frac{R_e}{100}$, $C_e$ = 663μF. In other words, to divide $Z$ by 10, we must multiply $C_e$ by 10. 663μF is a very large capacitance value, and we still have not quite reached our target voltage gain. This is a clear downside of using an emitter bypass capacitor - even with very large capacitance values, it is difficult to achieve uniformly high gain for all frequencies unless our lowest frequencies of interest are high or the emitter bypass capacitor is large.

Fortunately, things are not as bad as they initially appear. Because humans perceive audio on a logarithmic scale, a gain of 162 is perceived very similarly to a gain of 115. Voltage gain is often written in decibel form because it more closely aligns with how it will be perceived. Our gains in decibels are:

• $A (Z = 0) \implies 44\text{dB}$
• $A (Z = \frac{R_e}{100}) \implies 41.2\text{dB}$
• $A (Z = \frac{R_e}{10}) \implies 30\text{dB}$

The lowest frequencies will not reach full gain, but it is very subtle with $Z$ = $\frac{R_e}{100}$, less than 3dB. This is usually considered an acceptable amount of signal attenuation, though it depends on the application. The amount of attenuation is more noticeable for $Z$ = $\frac{R_e}{10}$.

Though humans can perceive audio down to about 20Hz, the fundamental frequency of the low E string of a guitar in standard tuning is $82Hz. To see what our gain is at this frequency with a$C_e$of 100μF, we’ll first determine$Z$at 82Hz: $$Z = \frac{1}{2\pi * f * C_e}$$$$Z = \frac{1}{2\pi * 82 * 0.0001}$$$$Z = 19.4\text{Ω}$$ Now we will plug$Z~ into our gain formula:

$$A = -\frac{R_C}{r_{e’} + R_e || Z}$$$$A = -\frac{-4700}{29 + 1200||19.4}$$$$A = -\frac{4700}{48.1}$$$$A = -97.7$$$$A \simeq 39.8\text{ dB}$$

With our 100μF emitter bypass capacitor, the gain at the lowest fundamental frequency of a guitar in standard tuning is -97.7, which is 39.8dB - much closer to the full gain of 44dB. Also keep in mind that shaving off some low end is often desired in dirt pedals. The fundamental frequency of lower notes can have a large amplitude which generates a large number of harmonics when clipped, dominating the sound. Attenuating the low end can give the sound more bite and less perceived “muddiness”, so it’s fairly common to see emitter bypass capacitors which do not allow the full range of frequencies through when looking at common emitter amps in guitar circuits.

In part 4 of the Silicon and Germanium Transistor Biasing series and the following parts, we will use what has been learned so far to take a clear look at common emitter amplifiers in popular guitar circuits to see the various ways in which they are commonly used.