Silicon and Germanium Transistor Biasing - Part 2

Part 2: Designing a simple common emitter amp:

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

In part 1, we took a general look at common emitter amplifiers and how various circuit components contribute to the bias voltage of common emitter amps. Now, we’ll take a deeper look at how a common emitter amp can be designed with a target bias voltage, voltage gain, etc. We will be required to get a little deeper into the math involved here, but we’ll keep things simple as much as possible.

We are designing these amps with silicon transistors. Everything in this article also applies to germanium transistors, though they have a smaller ~V_{\text{BE}}~ (~0.1V) and leakage current, which must be considered. We will discuss the additional considerations that must be made when using germanium in Part 5.

We’ll start with a very simple common emitter amp circuit with the emitter grounded (no emitter resistor). When designing a common emitter amp, usually there would be some design constraints which govern the selection of the load/collector resistor ~R_{\text{C}}~. The load resistor determines the amplifier’s output impedance and its ability to drive the following stage, and it also can have a large effect on the amp’s overall current draw and voltage gain.

Figure 1: Simple common emitter amp with emitter grounded

Figure 1: Simple common emitter amp with emitter grounded

In Figure 1, the capacitors labeled ~C_{\text{in}}~ and ~C_{\text{out}}~ block DC voltage from entering or exiting, which is necessary to properly bias this transistor stage. Their capacitance value can also determine how much of the low frequency content makes it into or out of the circuit. For now, we will assume their values are selected to let all audio frequencies through.

The bias of a transistor stage refers to its DC voltage. When considering the bias voltages, we are only interested in the DC properties of the circuit. When considering the DC properties of a circuit, capacitors are considered open-circuit. With that in mind, for DC analysis the circuit can be considered to be equivalent to that shown in Figure 2.

Figure 2: DC analysis of the common emitter amp seen in Figure 1

Figure 2: DC analysis of the common emitter amp seen in Figure 1

The collector current (~I_{\text{C}}~) typically plays the largest role in the overall current draw of a common emitter amplifier, and it depends largely on the ~R_{\text{C}}~ collector resistor. In pedal circuits, low current draw was often ideal for long battery life. In an Arbiter Fuzz Face, for example, both transistors should have less than 1mA flowing into the collector. If the common emitter amp needs to drive a large load, it is necessary for the amp to have a larger ~I_{\text{C}}~ and thus a smaller ~R_{\text{C}}~.

Let’s say we want to design a pedal boost that draws a small amount (~ 1mA) of current on a 9V battery supply. This would provide hundreds of hours of use for a 9V battery. We’ll want to bias the collector to

$$\sim 0.5 * V_{\text{cc}}$$$$\simeq 4.5 \text{V}$$

This is so that our output signal can swing close to 9V point-to-point (0V to 9V centered around 4.5V) for maximum clean headroom. Biasing the collector closer to either rail will result in lower headroom for one half of the waveform. This results in a lower clipping threshold and asymmetrical clipping. Some degree of asymmetrical clipping can be desirable in dirt pedals, but for a clean boost ~0.5 * V_{\text{cc}}~ is ideal.

~R_{\text{C}}~ will drop voltage according to the ohm's law formula:

$$V_{\text{RC}} = I_{\text{C}} * R_{\text{C}}$$

We want it to drop ~0.5 * V_{\text{cc}}~ or 4.5V. We’ll set ~I_{\text{C}}~ to 1 mA, the target current draw of our pedal. Note that with an ~I_{\text{C}}~ of 1 mA the actual current draw of this pedal would be slightly over 1 mA, since there must be a (much smaller) current into the base as well. The base current is typically very small compared to ~I_{\text{C}}~, so we will consider ~I_{\text{C}}~ to be the approximate total current draw of the pedal. With a known ~V_{\text{RC}}~ and ~I_{\text{C}}~, we can solve for ~R_{\text{C}}~:

$$R_{\text{C}} = \frac{V_{\text{RC}}}{I_{\text{C}}}$$$$R_{\text{C}} = \frac{0.5 * V_{\text{CC}}}{I_{\text{C}}}$$$$R_{\text{C}} = \frac{4.5}{0.001}$$$$R_{\text{C}} = 4500$$$$R_{\text{C}} = 4.5 \text{kΩ}$$

Since 4.5kΩ is not a common resistor value, we’ll settle for 4.7kΩ instead. ~V_{\text{CC}}~ does not change, so we can solve for our actual expected ~I_{\text{C}}~ with the new ~R_{\text{C}}~ value:

$$I_{\text{C}} = \frac{V_{\text{RC}}}{R_{\text{C}}}$$$$I_{\text{C}} = \frac{0.5 * V_{\text{CC}}}{4700}$$$$I_{\text{C}} = \frac{4.5}{4700}$$$$I_{\text{C}} = 957\text{ μA}$$

Clearly, this is different from our target ~I_{\text{C}}~ of 1mA. Some degree of variation is to be expected from rounded component values and the tolerance or range of component specifications.

Figure 3: DC analysis of common emitter amp with known <span class="tex2jax">~R_{\text{C}}~</span>

Figure 3: DC analysis of common emitter amp with known ~R_{\text{C}}~

As we outlined in part 1, the relationship between ~I_{\text{C}}~ and ~I_{\text{B}}~ depends on the transistor parameter ~\beta~ (beta), where

$$\beta = \frac{I_{\text{C}}}{I_{\text{B}}}$$

In other words, ~\beta~ is the ratio of output (collector) current to input (base) current. We’ll select BC108B for our transistor, for which a ~\beta~ of 300 is typical. With a selected ~\beta~, the target base current can be solved for:

$$I_{\text{B}} = \frac{I_{\text{C}}}{\beta}$$$$I_{\text{B}} = \frac{957\text{ μA}}{300}$$$$I_{\text{B}} = 3.19\text{ μA}$$

We’ll set the base current using only ~R_{\text{B1}}~, and leave ~R_{\text{B2}}~ out of the circuit. We’ll do so by using ~R_{\text{B2}}~ connected to ~V_{\text{CC}}~ as an approximate current source. A voltage source in series with a high value resistor is a rough approximation of a current source. For example, consider the circuit in Figure 4.

Figure 4: Current source approximation

Figure 4: Current source approximation

The resistor labeled ~R_{\text{B1}}~ here is a very high value, 5.6 megaohms. The maximum current that can flow can be determined with Ohm’s law:

$$I = \frac{9\text{V}}{5.6\text{MΩ}}$$$$I = 1.6\text{ μA}$$

This is how much current would flow with a 0 ohm load. As impedance of the load increases, the amount of current flowing will decrease, which is not ideal behavior for a current source. However, given the extremely large ~R_{\text{B1}}~ value, the effect that ~R_{\text{LOAD}}~ has on current is negligibly small for load impedances that are much smaller than ~R_{\text{B1}}~. For example, if the load was a 10k resistor, there would be about a 0.125% decrease in current. For smaller load impedances, that decrease is even smaller.

When using a similar arrangement to set the current into the base of a common emitter stage with no emitter resistor, the load seen at the base of the transistor is of small enough impedance that it can be disregarded for a very accurate approximation. The formula for current into the transistor base then becomes the following:

$$I_{\text{B}} = \frac{V_{\text{CC}} - V_{\text{BE}}}{R_{\text{B1}}}$$

We will use 9V for our source voltage and 0.7V for a general silicon BJT ~V_{\text{BE}}~ approximation. We know we’d like ~I_{\text{B}} = 3.19\text{}~ to get 957μA to flow through the collector, so we insert this value and solve for ~R_{\text{B1}}~:

$$R_{\text{B1}} = \frac{V_{\text{CC}} - V_{\text{BE}}}{I_{\text{B}}}$$$$R_{\text{B1}} = \frac{9 - 0.7}{3.19μ}$$$$R_{\text{B1}} = \frac{8.3}{3.19 * 10^{-6}}$$$$R_{\text{B1}} = \text{2,601,880Ω}$$

This is roughly 2.6MΩ, but the nearest commonly-available resistor value is 2.7MΩ, so we will use that instead. We’ll approximate our base current with the 2.7MΩ resistor:

$$I_{\text{B}} = \frac{V_{\text{CC}} - V_{\text{BE}}}{R_{\text{B2}}}$$$$I_{\text{B}} = \frac{9 - 0.7}{2.7\text{ MΩ}}$$$$I_{\text{B}} = \frac{8.3}{2.7 * 10^{-6}}$$$$I_{\text{B}} = \frac{8.3}{\text{2,700,000}}$$$$I_{\text{B}} = 3.074\text{ μA}$$

Now we will use the base current to determine the collector current:

$$I_{\text{C}} = \beta * I_{\text{B}}$$$$I_{\text{C}} = 300 * {3.074\text{ μA}}$$$$I_{\text{C}} = 922\text{ μA}$$

And now we will use the collector current to determine the voltage drop across RC:

$$V_{\text{RC}} = I_{\text{C}} * R_{\text{C}}$$$$V_{\text{RC}} = 922\text{μA} * 4.7\text{kΩ}$$$$V_{\text{RC}} = 4.33\text{V}$$

With 4.33V dropping across ~R_{\text{C}}~, that leaves 4.67V on the collector.

Figure 5: DC analysis with common emitter amp and known IC and IB

Figure 5: DC analysis with common emitter amp and known IC and IB

This is not exactly our ideal 4.5V, but it is very close. Again, some degree of variation is to be expected in transistor circuits. Bias will depend on available resistor values, component tolerance, supply voltage variations, and transistor parameters. Even among transistors of the same type, ~\beta~ is a parameter that tends to have quite a wide range. For example, though we used an hFE of 300 in the example calculations, BC108B has an hFE range of 200-450. Let’s look at how the bias would vary in this circuit for transistors on the outside of that range:

$$V_{\text{C}} = V_{\text{CC}} - V_{\text{RC}}$$$$V_{\text{C}} = V_{\text{CC}} - (I_{\text{C}} * R_{\text{C}})$$$$V_{\text{C}} = V_{\text{CC}} - (I_{\text{B}} * \beta * R_{\text{C}})$$

With a transistor ~\beta~ of 200:

$$V_{\text{C}} = 9 - (3.07\mu * 200 * 4700)$$$$V_{\text{C}} = 6.1142\text{V}\text{ - 67.9% of }V_{\text{CC}}$$

With a transistor ~\beta~ of 450:

$$V_{\text{C}} = 9 - (3.07\mu * 450 * 4700)$$$$V_{\text{C}} = 2.50695\text{V}\text{ - 27.9% of }V_{\text{CC}}$$

There is roughly a 3.6V range of bias voltages we can expect from ~\beta~ variation alone. Though obviously these get further from our ideal bias voltage, this kind of range is typical for this common emitter amp topology. You might wonder how we can minimize this variation for a set range of ~\beta~ values. Without a change in circuit topology, there is not much that can be done. It may appear at first glance that a different ~R_{\text{C}}~ value would result in a different bias range, but because ~R_{\text{C}}~ is inversely proportional to ~I_{\text{C}}~, to keep the bias range centered around ~0.5 * V_{\text{cc}}~, any changes to ~V_{\text{C}}~ should be offset by changes to ~I_{\text{B}}~.

To demonstrate, let’s look at an example with ~R_{\text{C}} = 10\text{k}~ instead of ~R_{\text{C}} = 4.7\text{k}~.

With a 10k resistor, 449μA needs to drop across ~R_{\text{C}}~ for ~V_{\text{C}}~ to bias to ~0.5V_{\text{CC}} = 4.5\text{V}~

Using our typical ~\beta~ value, we determine that the base current needs to be

$$I_{\text{B}} = \frac{449}{300}$$$$I_{\text{B}} = 1.497\text{ μA}$$

We calculate our base resistor with

$$R_{\text{B1}} = \frac{V_{\text{CC}} - V_{\text{BE}}}{I_{\text{B}}}$$$$R_{\text{B1}} = \frac{9 - 0.7}{1.497\mu}$$$$R_{\text{B1}} = \frac{8.3}{1.497\mu}$$$$R_{\text{B1}} = 5.54\text{ MΩ}$$

which we will round to 5.6MΩ. We then calculate the real base current with this resistor:

$$I_{\text{B}} = \frac{V_{\text{CC}} - V_{\text{BE}}}{R_{\text{B1}}}$$$$I_{\text{B}} = \frac{9 - 0.7}{5.6\text{ MΩ}}$$$$I_{\text{B}} = \frac{8.3}{5.6\text{ MΩ}}$$$$I_{\text{B}} = 1.482\text{ μA}$$

Calculating ~V_{\text{C}}~ with the new base resistor and ~\beta~ of 200:

$$V_{\text{C}} = V_{\text{CC}} - V_{\text{RC}}$$$$V_{\text{C}} = V_{\text{CC}} - (I_{\text{C}} * R_{\text{C}})$$$$V_{\text{C}} = V_{\text{CC}} - (I_{\text{B}} * \beta * R_{\text{C}})$$$$V_{\text{C}} = 9 - (1.482\mu * 200 * 10\text{k})$$$$V_{\text{C}} = 6.036\text{V}\text{ - 67.1% of }V_{\text{CC}}$$

Calculating ~V_{\text{C}}~ with the new base resistor and ~\beta~ of 450:

$$V_{\text{C}} = V_{\text{CC}} - V_{\text{RC}}$$$$V_{\text{C}} = V_{\text{CC}} - (I_{\text{C}} * R_{\text{C}})$$$$V_{\text{C}} = V_{\text{CC}} - (I_{\text{B}} * \beta * R_{\text{C}})$$$$V_{\text{C}} = 9 - (1.482\mu * 450 * 10\text{k})$$$$V_{\text{C}} = 2.331\text{V}\text{ - 25.9% of }V_{\text{CC}}$$

Even with ~R_{\text{C}}~ more than doubled, the range of bias voltages effectively stays the same.

Note that ~R_{\text{C}}~ has other effects on the circuit. You can see in the above equations that it has an effect on the current draw. With a ~\beta~ of 300 and a 10kΩ ~R_{\text{C}}~ instead of a 4.7kΩ ~R_{\text{C}}~, the current draw is

$$I_{\text{C}} + I_{\text{B}} = 449\text{μA} + 1.48\text{μA} = 450\text{μA}$$

With the 4.7kΩ ~R_{\text{C}}~, the current draw was

$$I_{\text{C}} + I_{\text{B}} = 922\text{μA} + 3.074\text{μA} = 925\text{μA}$$

Again, when designing a common-emitter amp, there will be some variation due to varying component specs and tolerance, but with the same target bias voltage, the current draw can be considered to be inversely proportional to collector resistance.

$$I_{\text{TOTAL}} = I_{\text{E}} = I_{\text{C}} + I_{\text{B}}$$

where

$$I_{\text{B}} = \frac{I_{\text{C}}}{\beta}$$

so

$$I_{\text{TOTAL}} = I_{\text{C}} + \frac{I_{\text{C}}}{\beta}$$$$I_{\text{TOTAL}} = I_{\text{C}} * (1 + \frac{1}{\beta})$$$$I_{\text{TOTAL}} = \frac{V_{\text{C}}}{R_{\text{C}}} * (1 + \frac{1}{\beta})$$

~\frac{1}{\beta}~ is very small compared to 1, so ~\frac{V_{\text{C}}}{R_{\text{C}}}~ is an approximation of the total current draw, where ~V_{\text{C}}~ is the voltage on the collector (our bias voltage - usually ~~0.5 * V_{\text{cc}}~).

~R_{\text{C}}~ also controls the voltage gain of the circuit. The voltage gain of a common emitter amplifier is given by the formula:

$$A = -\frac{R_{\text{C}}}{R_{\text{E}}}$$

The value is negative because a common emitter amplifier is an inverting amplifier. The above equation may look like we’re trying to divide by zero given that we have no emitter resistor, but the effective internal resistance of the emitter plays a part here. The total emitter resistance ~R_{\text{E}}~ can be considered to be ~R_{\text{e}} + r_{\text{e’}}~ where ~R_{\text{e}}~ is the emitter resistor and ~r_{\text{e’}}~ is the effective internal resistance of the transistor’s emitter. Here, ~R_{\text{e}} = 0~. The formula for internal emitter resistance of a small signal BJT is given by:

$$r_{\text{e’}} = \frac{25\text{mV}}{I_{\text{E}}}$$

This is an approximation of the effective emitter resistance at room temperature (68°F). ~I_{\text{E}}~ is ~I_{\text{B}} + I_{\text{C}}~, which is effectively the current draw of the entire circuit. In other words, ~r_{\text{e’}}~ is inversely proportional to the current draw. However, remember that current draw is itself inversely proportional to collector resistance, so ~r_{\text{e’}}~ is roughly proportional to ~R_{\text{C}}~. This means that as ~R_{\text{C}}~ increases, so does ~r_{\text{e’}}~, resulting in a voltage gain that stays roughly the same with changing ~R_{\text{C}}~ given the formula:

$$A = -\frac{R_{\text{C}}}{R_{\text{E}}}$$

With ~I_{\text{E}} = 925\text{ μA}~:

$$r_{\text{e’}} = \frac{25\text{mV}}{I_{\text{E}}}$$$$r_{\text{e’}} = \frac{25\text{mV}}{925\text{μA}}$$$$r_{\text{e’}} = 27Ω$$

The voltage gain of this amp is:

$$A = -\frac{R_{\text{C}}}{R_{\text{E}}}$$$$A = -\frac{4700}{27}$$$$A = -174.07$$$$A = 44.8\text{dB}$$

There will be some minor differences in voltage gain depending on the actual collector voltage, transistor specs, temperature, and the small amount of base current, but with no emitter resistor, the voltage gain will stay roughly the same for changing ~R_{\text{C}}~ values if the collector is biased similarly. When ~R_{\text{e}}~ is non-zero, the voltage gain formula is:

$$A = -\frac{R_{\text{C}}}{R_{\text{e}} + r_{\text{e’}}}$$

Increasing ~R_{\text{e}}~ will decrease gain, so we maximize the voltage gain by leaving out the emitter resistor (~R_{\text{e}} = 0~).

This common emitter amp is far from perfect, but it is perfectly usable. It may have a larger range of bias voltages due to transistor ~\beta~ (beta) variations than we’d like, but even in the worst case of beta and supply voltage that we are likely to see, the bias voltage range is near enough to ~0.5 * V_{\text{cc}}~ and far enough from the rails that it will be capable of amplifying a guitar signal relatively cleanly up to a certain point. With transistor betas that are on the outer edge of the 200-450 range, the signal will clip on one half of the waveform earlier than the other half. With a small amount of clipping, this can result in some subtle dirt being added to the sound.

Clipping is often meant to be avoided at all costs in amplifier circuits, but not in the guitar world. You may find circuits that are intentionally biased in a way that results in asymmetrical clipping. Target bias voltages for the Dallas Rangemaster Treble Booster, for example, are often closer to the -9V rail (around -7V) than to 0V. Note however that having a target bias voltage that is not centered between the rails is quite a bit different from targeting ~0.5 * V_{\text{cc}}~ but allowing for a widely varying range of biases. Aiming for an offset bias voltage is usually done to achieve a certain asymmetrical clipping sound, whereas varying bias is going to have unpredictable results on the sound if the signal is hot enough to clip.

Unpredictable bias voltages are undesirable across the board, but with the common emitter amp topology seen in Figure 1, there is little we can do to minimize the bias variance. In part 3 of the Silicon and Germanium Transistor Biasing series, we’ll look at a very effective way of limiting the bias variations in a common emitter amplifier with just a few simple tweaks to the circuit.

Note that the information presented in this article is for reference purposes only. CE Distribution makes no claims, promises, or guarantees about the accuracy, completeness, or adequacy of the contents of this article, and expressly disclaims liability for errors or omissions on the part of the author. No warranty of any kind, implied, expressed, or statutory, including but not limited to the warranties of non-infringement of third party rights, title, merchantability, or fitness for a particular purpose, is given with respect to the contents of this article or its links to other resources.