Yamaha G100B-212 help with troubleshooting and repair.

[Helmholtz]

Forward voltage of a silicon diode depends on current and diode type.
At 0.1mA a 1N4007 measures around 0.45V and a 1N4148 measures 0.5V.
At 1mA a 1N4007 drops around 0.55V and a 1N4148 around 0.6V.
At higher current forward voltage will be higher, just check diode datasheets.

[x-pro]

And if you put germanium diodes, distortion will start even at low input signal. However, in high signal mode it will be hell!

[JAelec]

A silicon diode doesn't noticeably conduct below a voltage of around 0.45 to 0.6V depending on diode type.

There's a 11:1 voltage divider before C7, so voltage across the diodes should be around 0.55Vpeak judging from your scope pic.
This is assumig that the inner knob of the channel sensitivity selector switch is in its calibrated (arrested) position.

Interesting... trying to understand, how are you calculating that? When the distortion is maxed (last picture) the voltage is about 0.05V peak, right? And before that it is higher then? Where is the voltage divider? Is it the two resistors of 1K and 10K between the diodes and C5?

[x-pro]

He knows that the diode opens at a voltage of about 0.55-0.65 V. So, for the sine of the signal to be trimmed (limited), this signal must have a peak greater than 0.55-0.65V, for example, 1V.
A simple voltage divider on 10k/1k resistors is used for calculation: 0.55V/10=0.05V.

[x-pro]

Further, to make calculations, you need to know the parameters and understand the purpose of the parts.

[Helmholtz]

Yes, the voltage dividers is composed of the 1k and 10k resistors.
Just google the term and do some forward and backward calculations.
A voltage divider is probably the most essential basic circuit of electronics.

To verify just scope the voltage across the diodes.

[JAelec]

Thanks,

I think I understand the circuit a bit better now...

is this right, then?

1. The signal gets split into a clean path and DT path
2. The DT signal has to get amplified to above 0.55-0.60 V so that it can be clipped by the diodes, this is done by TR3 and TR4 in the DT-circuit
3. Now the DT signal is too loud compared to the clean signal so a voltage divider is used to bring the level down.

So the voltage at the minus side of C7 will be 0.05 V when the diode begins to clip, and increase from there.

Vout = 10k / (10k+1k) x 0.55 = 10/11 x 0.55 = 0.05 V


I do not have acces to the amp anymore so cannot do any measurements....

[Helmholtz]

Correct.
Only, when the diodes start to clip, the signal won't increase much anymore, because clipping means limiting.

[JAelec]

I see, so now I understand why it was so easy for you to see the exact clipping level of the diode, because you know that the clip point/peak voltage will be the same even when increasing the DT-knob, which I did not realise.

Calculating backwards will then be easy knowing the formula for voltage dividers:

Vout = 50mVp = 0.05 Vp
Vout = R2 / (R1+R2) x Vin
0.05 = 1 / (10+1) x Vin
0.05 = 1 / 11 x Vin
Vin = 0.05 x 11 = 0.55 Vp

Sorry for the repetition, just want to sum up in case my future self wants to check this thread.

[x-pro]

Vout = R2 / (R1+R2) x Vin
0.05 = 1 / (10+1) x Vin
0.05 = 1 / 11 x Vin

why so complicated?
10k/1k = 10
The 0.55V limiting voltage is divided by the divider factor, which is 10. That's it!

[Helmholtz]

why so complicated?
10k/1k = 10
The 0.55V limiting voltage is divided by the divider factor, which is 10. That's it!

Divider factor is 11 (not 10!).
The OP is correct.
Look up voltage divider.

[x-pro]

• I agree