# Voltage gain of common emitter amplifiers with negative feedback

Im trying to make some fuzz pedals for guitar using a single transistor. The one on the right is as far as i understand a common emitter amplifier with negative feedback, I somehow can't find anything about how to calculate the voltage gain of this configuration, but its very common in fuzz pedal circuits.

The one on the left has a diode in place of the resistor and as far as i understand this isn't negative feedback. The diode is supposed to limit the voltage swing from the positive side. I made this circuit in LTspice, and i noticed that the DC voltage at the base and collector is the same, but I assumed that it would be 0.7V higher at the collector because of the voltage drop across a diode, that wouldn't allow the diode to limit the signal though, so I'm not sure how this exactly works. Why is the DC voltage the same at the base and the collector, and how exactly is the voltage gain calculated, because I can't find that online anywhere.

Thanks

• Firstly, the type of transistor you have drawn (BJT) is a current amplifier, not a voltage amplifier. There's lots to read about these but a good place to start is here: electronics-tutorials.ws/transistor/tran_1.html – MIL-SPEC Jul 23 '18 at 12:35
• MIL-SPEC I understand the transistor is a current amplifier, but the circuit is a voltage amplifier – TwoheadedFetus Jul 23 '18 at 12:36
• What is the purpose of the diode? In any case, such a feedback element must provide a proper positive DC biasing for the base terminal. It is quite normal to use a resistor (second circuit) for such a purpose. This resistor provides negative feedback (and bias stability) for DC only! In case you want also signal feedback, there is another resistor required left from the coupling capacitor. – LvW Jul 23 '18 at 13:11
• Both circuits are nasty enough to be relatively ill defined. CCt1 SHOULD stabilise, as you say, with Vc 0.6ish V above Vb. Has your SPICE diode got correct Vf behaviour or is it modelled as an ideal diode. CCt 2 will stabilise with ~~~ (Vc - Vb)/Rf x Beta = (9V-Vc)/Rc | Gain in all cases is 38.4 x (Vcc-Vc). | CCt Vb ~~= 0/6, Vc ~= say 1.1. Gain = 38.4 x (9-1.1) =~ 300. Clipping will happen almost instantly. | CCt 2 gain will be very slightly higher. – Russell McMahon Jul 23 '18 at 13:34

I made this circuit in LTspice, and i noticed that the DC voltage at the base and collector is the same, but I assumed that it would be 0.7V higher at the collector because of the voltage drop across a diode

We tend to say that a diode drops about 0.7 volts but this is a little simplistic. For instance, if you look at a typical signal diode like the 1N4148 it has a forward characteristic like this: -

As you can see it is only dropping about 0.5 volts with 100 uA forward current. If you manually extended/extrapolated the graph like this: -

You can now estimate what current is needed to drop 0.1 volts - it looks like about 60 nA to me. So when you said the collector and base were at the same voltage maybe you meant that there was a significantly-smaller-than 0.7 volts and more like 0.1 volts or 0.2 volts?

If it were 0.2 volts then there would be a base current of about 300 nA (because of the blocking capacitor) and this would (or could or might) result in a collector current of 20 uA to 500 uA (Beta dependent). If your collector resistor is quite high in value then that current can cause a significant volt drop across it and the resulting collector voltage will be maybe 0.1 to 0.2 volts higher than the base voltage.

If you saw 0 volts between base and collector then you are doing something wrong.

how exactly is the voltage gain calculated

It's like an inverting op-amp (-Rf/Rin). Rf is the resistor from collector to base and Rin is the resistance/impedance of your signal source. It's not as precise as an op-amp at DC because the open-loop gain of an op-amp at DC isusually 100,000 or so and the BJT might be only having 50 but it's near enough to make an estimate. More current through the collector means a higher "open-loop" voltage gain usually.

The one on the left has a diode in place of the resistor and as far as i understand this isn't negative feedback.

It is negative feedback but not a linear negative feedback.

I somehow can't find anything about how to calculate the voltage gain of this configuration

Assuming you're working in the passband, you can approximate the capacitors by short-circuits in AC. You can calculate the gain then by using the following small-signal model:

simulate this circuit – Schematic created using CircuitLab

I used the Extra-Element Theorem (EET) to calculate the gain here, but you might just as well use the KCL equation in the output node.

$$H_{g_m\to 0} = \frac{R_c}{R_f+R_c}$$ $$g_n = -\frac{1}{R_f}$$ $$g_d = \infty$$

\begin{align} H &= H_{g_m\to 0}\frac{1 + \frac{g_m}{g_n}}{1 + \frac{g_m}{g_d}} \\ &= \frac{R_c}{R_f + R_c}(1 - g_mR_f) \end{align}

If the input source has a significant output resistance, consider using the approximation Andy stated. If you really need the full expression, you can use the EET again to find

$$H = H_{R_s\to 0}\frac{1 + \frac{R_s}{R_n}}{1 + \frac{R_s}{R_d}}$$

Where $H_{R_s\to 0}$ is just the gain we found earlier. $R_n = \infty$ and

$$R_d = r_\pi || \frac{R_f}{1 - H_{R_s\to 0}}$$

or

$$H = H_{R_s\to 0}\frac{1}{1 + \frac{R_s}{r_\pi || (R_f/(1-H_{R_s\to 0}))}}$$

Or just solve the KCL equations again for the equivalent small-signal model including a series resistance.