# Biasing BJTs - part 1

#### Chuck D. Bones

##### Circuit Wizard
There's a lot to cover, so I'll break this up into parts.
This discussion will focus on the DC operating point of the transistor. First, some basics. Many of you will already know this stuff. Vbe is the base-emitter junction voltage. It's around 0.6V with silicon and around 0.15V with germanium. It varies with base current, temperature and the physical size of the device, but those numbers are pretty close to what we'll see in pedal circuits. Vce is the collector-emitter voltage. When the transistor is in the linear region, Vce will be greater than Vbe. When the transistor is saturated, Vce will be less than Vbe. Unless specifically stated otherwise in this discussion, the transistor is operating in the linear region. Ic is collector current, Ib is base current and hFE is the collector current divided by the base current.

There are many ways to bias a BJT, they all show up in pedal circuits and some of them are more stable than others. This first one is one stage of the Death by Audio Interstellar Overdriver. The transistor is silicon. The base current is established by the 9V power supply voltage (Vcc), the transistor's Vbe and the 910K resistor.
Ib = (Vcc - Vbe) / Rb = (9V - 0.6V) / 910K = 9.2uA.
Let's assume the transistor has hFE = 100. Ic would be 9.2uA * 100 = 920uA. That's not possible because the 180K resistor limits Ic to less than 50uA. This stage would be saturated. Let's change Rc to 4.7K. Now Vce = Vcc - Rc * Ic = 9V - 4.7K * 0.92mA = 4.77V. The transistor is biased into the linear region. What happens if hFE is larger? The base current is constant, so if hFE increases then Ic increases and Vce decreases. If hFE gets large enough, just over 200 in this example, then the transistor saturates. For this circuit to operate in the linear region, the transistor or the base resistor need to be hand selected. DbA intended this stage to operate in the saturated region, so selection of components is probably not necessary for them. This is the least stable biasing circuit because changes in power supply voltage or hFE will have a significant effect on the bias point. Here's another DbA circuit, from the Fuzz War v1. Although this schematic shows the transistor installed right-side-up, it is in fact installed upside-down in the Fuzz War. WTF? Turns out that BJTs work both ways, but the hFE is considerably lower when operated in reverse. R12 helps stabilize the bias point by providing some negative feedback. Let's assume the reverse hFE is 30. Vb is now Vbe + Ve and Ve = Ie * R12. What's Ie? Ie = Ib + Ic = Ib * (hFE + 1). A little more algebra gets us to
Ib = (Vcc - Vbe) / (Rb + (hFE +1) * Re). Plugging in the values shown gives us Ib = 19uA. Ic = 19uA * 30 = 570uA. This transistor is saturated because R11 and R12 limit Ic to just under 50uA. Again, this is what DbA intended. Suppose we change R11 to 10K. Now Vc = 3.3V and Ve = 0.23V. Vce = Vc - Ve = 3.07V, therefore the transistor is in the linear region. Still pretty sensitive to changes in Vcc and hFE, but better than the 1st circuit. Let's look at one more. This is the last stage in a BMP. Notice that the emitter resistor is much larger than the previous example. The larger the emitter resistor, the more stable the bias. It comes at a cost: the voltage drop across the emitter resistor reduces headroom. The voltage divider made up of the 390K and 100K resistors reduces the sensitivity to changes in Vcc. For easy math, we'll assume that the base current is small compared to the current in the 100K resistor. Vb = Vc * 100K / (100K + 390K) = 1.84V. Ve = Vb - Vbe = 1.24V. We'll also assume that hFE is >100 and therefore Ic is approximately equal to Ie. Ic = Ie = Ve / 2.7K = 460uA. Vc = Vcc - Ic * 12K = 3.50V. Vce = Vc - Ve = 2.26V. In reality, Ib will cause some additional voltage drop in Vb, reducing Ic a little and increasing Vce. This bias circuit is much more stable that the previous ones and shows up in many pedals. That's enough for now, stay tuned for part 2.

#### fig

##### Village Idiot
Why not boost the signal after this stage (BMP example) to compensate for the loss of headroom? • Chuck D. Bones

#### Mcknib

##### Well-known member
Chuck make part 2 in about 6 months I should have read and partially understood at least 3 paragraphs by then

Excellent work as always

• fig and Chuck D. Bones

#### Chuck D. Bones

##### Circuit Wizard
Part 2 gonna be short, probably write that one tomorrow.

Part 3 will be Fuzz Face and derivatives.

• Mcknib and fig

#### fig

##### Village Idiot
Part 3 will be Fuzz Face and derivatives.
Outstanding. I have 2 Sunflowers on the bench, and was just dialing in the trimmers this morning on the first.

• Chuck D. Bones

#### cooder

##### Well-known member
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