On a long but educational process repairing a vintage German stereo. I have identified a failed (shorted) PNP transistor in the power amp section. (Schematic with highlight is below.) The part on the actual board is a BC158B; the schematic specifies only "BC158". The only difference between the BC158 and the B variant is the hFE gain value (higher value in the B). Here's a datasheet showing the BC158 series.

I have asked before here and gotten good advice on how to think about transistor equivalence in audio when going from vintage euro parts to current US replacements.

It appears to me that a standard general purpose 2N3906 would seem appropriate relative to the base BC158 (has same or higher max voltage, higher max current, and similar hFE.

I'm not knowledgeable enough to look at the circuit and see why the schematic shows the BC158 and the actual part used was a BC158B-- they appear to be likely different in amplification function, but maybe (presumably?) it doesn't matter here.

  1. Can anyone offer a theory for why the original part used could be a higher-gain model without messing anything up?

  2. Do I need to seek out a higher-gain replacement part? I'll use a 2N3906 if it would seem to work. If not, there are some random parts I groused up on Mouser based on the filters that seem like they might be appropriate, but I'm for those I'm casting around in the dark a bit.


Schematic-- T601 is the one that needs replacing:

enter image description here

  • 1
    \$\begingroup\$ A BC158B can be used anywhere the design says BC158, but not necessarily the other way round. They test the devices after manufacture and then put them in "bins" according to what they measure and mark with them with the letter, so if you need the higher gain you can specify the relevant part number. \$\endgroup\$
    – Finbarr
    May 28, 2020 at 19:16
  • \$\begingroup\$ ...and probably pay a bit more, obviously. \$\endgroup\$
    – Finbarr
    May 28, 2020 at 19:26
  • \$\begingroup\$ @Finbarr ahhh that's super useful and I didn't understand that. This implies that my (admittedly small and uninformed) intuition around what hFE is and how it's used/matched is way off. Got some studying to do. Thanks \$\endgroup\$
    – Ben Zotto
    May 28, 2020 at 19:47

2 Answers 2


This transistor is biased DC around 1-2 mA collector current.

  • Suppose HFE is at the low-end: 125...then DC base current is about 9.7 uA.
  • Suppose HFE is at the high-end: 500...then DC base current is about 2.4 uA.

This base current flows through base bias equivalent resistance of about 85k ohms. So at one HFE extreme compared to the other HFE extreme, DC bias would change only 0.625 V at T601. That's trivial compared to the 19V at transistor's emitter.
Since open-loop gain is very much higher than closed-loop gain, almost all transistor non-linearities are removed. So gain variations are removed too, because overall amplifier gain is set by resistor ratio of R608/R607.

T601 is not pushed anywhere near its operating envelope limits. 2N3906 is a decent substitute in this circuit.

  • \$\begingroup\$ Thanks. There's a lot of great context in here that is currently confusing to me but it creates a very helpful bunch of leads to Google around with. Thanks for the final assessment, too. \$\endgroup\$
    – Ben Zotto
    May 28, 2020 at 19:44

the T601 is where the input signal ( from Preamp ) is subtracted from the feedback signal to the loudspeaker.

The difference of Vbase and Vspeaker becomes a current across the 820 ohm between speaker and emitter. That (error) current exits the collector.

That current becomes a voltage across the 560 ohm between base of T604 and the thick black Ground buss.

Notice the 47pf on base_collector of T604; Miller Muitiplication boosts the cap by the voltage gain of T604, and sets the frequency rolloff.

There will be a dominate pole generated by the 560 ohm and the input capacitance [ 47pf * ( 1 + voltage_gain_T604) ].

The open_loop gain resistor --- R605 --- has 16 volts across it. Divide that voltage by the thermal_voltage of 0.026 volts (from the diode equation) and you find the low-frequency gain is 640X.

That makes the C_Miller be 47pf * 640X_gain === 30,000 pf; the time constant of 560 ohm and 30,000 pf === 15 microseconds, or 60,000 radian/second, making the pole be 2 * PI lower at 10,000 Hz.

With the dominant open_loop pole at 10,000 Hz, this amplifier likely has superb transient performance.



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