Behavior of a common emitter BJT with diode from collector to base

Question

This is the schematic for a popular fuzz pedal. I'm struggling to comprehend the mechanism. For the DC signal V1, the forward drop of the diode V_F and V_BE are both overcome, but when the positive part of V_in is applied, isn't V_BE brought high enough to damage the transistor? (I believe guitar pickups can give voltages as high as 1V, so 1.7 VAC; max rating for this BJT is 0.7VDC).

Furthermore, how does the diode create distortion? It seems as though any current that feed backs to the base is miniscule.

^{simulate this circuit – Schematic created using CircuitLab}

Answer 1

The diode is a voltage limiter so you get 0.7 to 0.8 Vpp with no load and then 50% if equal load or 1% in your case.

If the Vol/Io=Rce is about 5 ohms when saturated the Rbe resistance will be around 10% of this. I expect the guitar coil pickup to be around a few kilohms so this attenuates the signal exponentially.

I put some design corrections for you on my simulation.

Feel free to ask any more questions.

You can use anything from 2V to 12V and just scale the collector Rc. Normally saturation is rated at Ic/Ib= 10 but in this case if you consider a load pot greater than Rc, you can use a lower ratio like 2 or 3.

Otherwise you get a dip in the middle of the +ve square output.

• note how massive that dip is in @transistor 's sim using your incorrect values.

If I used your RC values, it would give skinny spikes for each edge.

Answer 2

^{simulate this circuit – Schematic created using CircuitLab}

Figure 1. OP's schematic with addition of Vb and Vc test points for simulation.

The simulation results are interesting.

Figure 2. V_in, V_b and V_c.

... but when the positive part of Vin is applied, isn't VBE brought high enough to damage the transistor?

The base-emitter junction protects itself! Remember that the b-e junction is like a diode and will clamp V_b at about 0.7 V. This doesn't cause any damage because the current drive is limited by C1.

Furthermore, how does the diode create distortion? It seems as though any current that feed backs to the base is miniscule.

• When V_in goes negative the base is pulled negative and, interestingly, V_c goes negative because D1 pulls it down.
• When V_in goes positive V_c follows it until V_b gets high enough to turn on Q1.
• Q1 then saturates and the collector is pulled to ground until the input signal drops below the Q1 turn-on voltage again.

The effect gives a crude frequency doubling (and a load of harmonics) on positive half-cycles while leaving negative half-cycles relatively unscathed. The effect is dependent on the signal amplitude so I would expect it to be noticeable on the initial attack of a note at high volume and to die out as the note fades. At low volume settings there will be little distortion.

Answer 3

but when the positive part of Vin is applied, isn't VBE brought high enough to damage the transistor? (I believe guitar pickups can give voltages as high as 1V, so 1.7 VAC; max rating for this BJT is 0.7VDC).

That is true but, the impedance of the pick-up is thousands of ohms so, it inherently current limits.

Furthermore, how does the diode create distortion? It seems as though any current that feed backs to the base is miniscule.

It's a classic asymmetrical clipping circuit - the diode doesn't provide much feedback when the input signal is small (the odd mV p-p) but, as the signal grows current from the collector does pass through the diode and produces significant negative feedback that clips the voltage seen on the collector. This produces the fuzz effect. As the input signal gets bigger the asymmetrical distortion leads into full top and bottom waveform clipping.