VBE multiplier question on Farfisa amp repair

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Hi,

I'm working on this broken farfisa.

As i can't find the schematic i redraw it. (only the power amp part...)





On the power amp board i found a faulty darlington PNP transistor (Q2). I changed it and the keyboard is now alive.


But i've got some questions:

1/ i've replaced the SE9400 with a BDW94C wich seems to be a direct remplacement.
Do i need to adjust something in the circuit according to this? What do i need to check to make a good replacement? Is it better to change the working SE9300 with a new BDW93C?

2/ the part with transistor Q, R12, R10 and R11 seems to be a biasing circuit for the darlington pair.
If i'm correct this configuration is called a "Vbe multiplier".
I've read that the good way to bias is to  set current on R7 and R8 (0,22 ohms resistors) between 20mA and 200mA , is that right?

So i need something like between 4mV and 40mV across R7 and R8.

Today I've got 300mV overthere so maybe this is why the darlington get really hot?



Thanks

Morgan

Nice re-drawing.

> between 20mA and 200mA , is that right?
> something like between 4mV and 40mV across R7 and R8.

Yes. I would aim for more like 30mA-100mA (7mV-22mV), but 20mA to 200mA is acceptable.

> I've got 300mV overthere

YOW!! 1.36 Amps!!

> so maybe this is why the darlington get really hot?

Transistor voltage is roughly 30V total. 30V at 1.36A is 40 Watts. About 20 Watts per transistor, heat in the heatsink.

Maximum audio output is like 15W in 8 Ohms or 30W in 4 Ohms. So the heat in the heatsink "should be" about 8 or 16 Watts (66% efficiency). It is not supposed to have *40* watts of heat in the heatsink.

R11 is the trimmer for this bias. *Carefully* trim it while watching the bias current.

If, with the new (improved?) transistor, you can not get down below 100mA (20mV), then change R12 (1,5K) to 1K or maybe 820 Ohms. (or tack 3.3K or 2.2K across the existing 1.5K.)

Thanks a lot for your answer PRR!

I'm gonna check that.

Best

Morgan

I would target closer to 25mA for class A current, and if you need to alter the Vbe multiplier I would offer a little advice. In amp design of Vbe multipliers a question to always ask, is what happens if the trimpot opens up? A common failure mode. The worst case is for the Vbe multiplier to go max voltage after fault, so the safer topology is for the trimpot to be used as a shunt from base to emitter, so if it opens up the Vbe goes to minimum voltage and minimum class A current.

Little things like this can make the difference between expensive and modest repairs.

JR

JohnRoberts said:

I would target closer to 25mA for class A current,
JR

Click to expand...

I'm seeing a class B amplifier.  Do you mean quiescent current ?

gridcurrent said:

JohnRoberts said:

I would target closer to 25mA for class A current,
JR

Click to expand...

I'm seeing a class B amplifier.  Do you mean quiescent current ?

Click to expand...

Look again, it is a class AB amplifier. Class A current in the same thing as quiescent current in the output stage with no AC signal. The Vbe multiplier establishes high enough voltage that both output devices are slightly turned on with around 25mA of class A current, the output current higher than 25mA is class B .

JR

Randy Sloan and Douglas Self are excellent resources explaing the details of Class B verses Class AB.  We want the Vbe multiplier to provide just enough bias to overcome the Vbe drops in the output transistors.  Any more bias introduces gm doubling.

> gm doubling.

Jeez, it's a Farfisa!!!

We want more than Zero current but MUCH LESS than the _1.3_AMPS_ he has now.

10mA, 100mA, even 5mA or 500mA.....

Leave the Self/Sloan/etal "optimizations" for the golden-ear gang.

I hope the measured voltages posted are wrong. For example the left side of R12 should not be .2V higher than the right side. The current is flowing right to left in that divider.

A more reasonable quiescent current would result in more like 10mV across the two emitter resistors than the 600mV indicated.

JR