Breaking "rule of thumb" for common emitter?

Question

I'm trying to understand this common emitter amplifier that is part of the of a guitar pedal. The image is take from the ElectroSmash analysis from the Big Muff Pi .

The values for R6 and R4 are clear for me. 10KOhms is a common value in guitar pedals and 3.3KOhms for R5 makes sense since the desired gain is moreless G=4.5, and the gain of this amplifier is R=R6/R5. However, i don't understand the value for R3. As far as i know the "rule of thumb" in a common emitter amplifier is that I_r3 > 10*Ib. I have calculated the quiescent current of the schematic to be around 0.375mA for the collector, so it should be around 3.75uA for the base current. The tension for the base in the Q point is supossed to be around 1.3V. So the current in R3 will be I_r3 = Vb/R3 = 13uA which is not bigger than 10x3.75uA. Am i missing something here? Thanks!!

Answer 1

While your actual calculations are slightly off, your general conclusion is correct, that this illustrated stage is not following the 10x Ib rule of thumb. Rules of thumb are just suggestions. If you know what you're doing, and why, you can reject their advice.

Back_of_napkin_1

Voltage at base bias point = 9V x 100k / (100k+430k) = 1.7 V
Voltage at emitter = 1 V
So emitter current = 300 uA
So base current (assuming beta = 100) = 3 uA

Back_of_napkin_2

Base bias Thevenin impedance = 100k // 430k = 81k
So base bias drop at 3 uA = 0.24 V
So new base voltage = 1.46 V
So new emitter current = 230 uA
New base current = 2.3 uA
100k current = 14.6 uA
430k current = 16.9 uA

I make that a factor of 7x, which for an engineer is roughly 10 ;-)

Do we need the full x10 factor?

Repeat the calculations above with beta = 50 and beta = 200, and see whether the change of bias point is significant in your application. Repeat with values other than the 0.7 V I've assumed for VBE (temperature!) and see whether that's significant.

As it happens, I did have a similar situation early in my career where I was asked to debug an RF amplifier whose frequency response was sensitive to temperature. It turned out that the emitter voltage was too small, so the emitter current was sensitive to temperature, varying the collector voltage, and with it the collector-base capacitance. Not a problem you'd see in an audio amplifier.

Answer 2

Typical base current will be around 1uA (hFE = 400) with maximum (worst-case hFE = 120) less than 2.5uA. With a BC239C it will be lower (hFE 380 to 800 with 500 typical)

The current through the divider is about 17uA ignoring the base current.

So it's roughly inside the rule-of-thumb territory. It's a rule of thumb, not a strict rule. The penalty for going a bit less is that variations of the bias will be more. 'Will it be big enough worst-case to cause saturation with the expected signal values' is the question, and the answer in this case is negative. Thermal variations could be larger, depending on the temperature range the product is expected to work over (outdoor concerts in Calgary winter? summer in Dubai?).

The designer also may have decided to aim for the minimum acceptable divider current in order to get slightly more life out of the 9V battery.