Transistor's Reverse Bias protective diode effect on small signal output

[masebee]

Hi all,

I'm wondering what the effect of the reverse bias protective diode is on the transistor's signal output. Does it introduce some clipping/feedback when the input signal swings below 0.7V? I understand that it is there to protect the transistor from its emitter/base breakdown voltage, but I would not expect the BE junction to reach ~ -6V or whatever with a 9V DC supply and a guitar input. Additionally, the amplifier is preceded by a common collector buffer.

For reference this is the circuit:


I built it last night, but I plan on tweaking it some. Thanks!

[R.G.]

Which transistor, which diode?

[masebee]

Which transistor, which diode?

The diode connected between the base and emitter of Q2.

[MaxPower]

I'm going to take a guess: It's clipping one half of the audio signal (one half of the negative feedback signal) and it's not serving any purpose as far as DC is concerned.

Just a guess though. Best wait for RG or one of the other gurus  for a proper answer.

[merlinb]

Without that diode, the coupling caps would charge up during loud passages and possibly cause brief cut-off of the transistor when the signal drops to a low level again (blocking distortion). But the diode allows the caps to discharge just as fast as they charge -it's called a 'catching diode'. As you observe, it is not intended to protect the transistor junction since the voltage can never reach damaging levels in this circuit anyway.
The diode itself has no direct effect on clipping itself, since the transistor is already cut-off (i.e. clipping) by the time the diode ever starts conducting, but the diode does affect the overall dynamic clipping behaviour of the transistor.

[masebee]

Without that diode, the coupling caps would charge up during loud passages and possibly cause brief cut-off of the transistor when the signal drops to a low level again (blocking distortion). But the diode allows the caps to discharge just as fast as they charge -it's called a 'catching diode'. As you observe, it is not intended to protect the transistor junction since the voltage can never reach damaging levels in this circuit anyway.
The diode itself has no direct effect on clipping itself, since the transistor is already cut-off (i.e. clipping) by the time the diode ever starts conducting, but the diode does affect the overall dynamic clipping behaviour of the transistor.

I'm new to the game, but this is the only circuit I've seen that configuration. Is it due to the amplifier being preceded by a buffer with high current gain that causes the caps to charge quickly? Also why is the transistor in cutoff when the signal voltage swings negative? Doesn't the DC bias provide enough collector current and Vce drop to keep it in the active region?

[masebee]

Without that diode, the coupling caps would charge up during loud passages and possibly cause brief cut-off of the transistor when the signal drops to a low level again (blocking distortion). But the diode allows the caps to discharge just as fast as they charge -it's called a 'catching diode'. As you observe, it is not intended to protect the transistor junction since the voltage can never reach damaging levels in this circuit anyway.
The diode itself has no direct effect on clipping itself, since the transistor is already cut-off (i.e. clipping) by the time the diode ever starts conducting, but the diode does affect the overall dynamic clipping behaviour of the transistor.

I'm new to the game, but this is the only circuit I've seen that configuration. Is it due to the amplifier being preceded by a buffer with high current gain that causes the caps to charge quickly? Also why is the transistor in cutoff when the signal voltage swings negative? Doesn't the DC bias provide enough collector current and Vce drop to keep it in the active region?

Bah, nevermind. Just to clear things up for my own understanding: The diode only conducts when Vbe = -0.7V (approximately). The transistor is clearly in cutoff region here and ic=0 therefore there is no drop across the 4.7k resistor and the transistor's output is clipped to 9V (?).

I guess I would have to actually view the output on a scope to see the diode's subtle effect on the clipping. Would it soften the hard clipping normally seen in transistors?

Thanks!

[masebee]

Without that diode, the coupling caps would charge up during loud passages and possibly cause brief cut-off of the transistor when the signal drops to a low level again (blocking distortion). But the diode allows the caps to discharge just as fast as they charge -it's called a 'catching diode'. As you observe, it is not intended to protect the transistor junction since the voltage can never reach damaging levels in this circuit anyway.
The diode itself has no direct effect on clipping itself, since the transistor is already cut-off (i.e. clipping) by the time the diode ever starts conducting, but the diode does affect the overall dynamic clipping behaviour of the transistor.

Do you have any resources that describe the diode being used in this way? I don't see how leaving it out would cause the capacitors to charge too quickly (also, which capacitors?).