0
\$\begingroup\$

A class AB audio amplifier has both positive and negative +/-40v rails. There is a differential amplifier feeding a trans-impedance transistor. The transistor is also a current source for an amplified diode. The other end of the diode goes to the negative rail through a couple of resistors. The diode provides adjustable + /- biasing voltages for both driver transistors, around +/- .5v, respectively.

Drivers and output transistors had shorted due to accidental overload and have been removed. Should bias voltages still be in range with these transistors out? If not, then how can I be certain there is no problems with bias before replacing the transistors? Are there any precautions I need to follow before powering up?

Here's a schematic that should help. A simpler way to ask this question is: with Q806. Q807, Q809 and Q810 removed, should the bias voltages at the collector and emitter of Q804 still be around + and - .5v, respectively?

enter image description here

Out of curiosity I powered this up again and measured +38v on the collector of Q804 and +36.5v on its emitter. Predriver Q803 has +38.1v on both emitter and collector and +37.4v on its base. So I guess now there is either a problem with predriver Q803 and/or the differential amplifier Q801, Q802. Anyone have anything to add?

Here is corrected schematic.

enter image description here


FOLLOWUP: Interestingly there is still over 35v bias. Here are the readings:

Q801 E= -.710  B= -.127  C=37.4
Q802 E= -.710  B= -.127  C=38.2
Q803 E= 38.1   B= 37.4   C=38.0
Q804 E= 36.2   B= 36.8   C=38.0 

R811,R812 junction -17.4

+40 actually is  38.2
-40 actually is -38.5

Q801. Q802 and Q803 have been replaced. Over current protection transistors, Q805 and Q808, are still installed. Any idea why the bias is still so high?

Thanks in advance.

Thanks Dave for the explanation for the high common mode voltage and clarification between bias and common mode voltage. I tried shorting Q804's emitter to P809 as you suggested and have 2V on Q804's collector and .125V on its emitter. Does this seem reasonable to you?

And I tried shorting Q804's collector to P809 and got .120V on it and -1.8V on its emitter. Does this mean I can now safely reinstall the driver and output transistors? Are there any precautions I need to follow before powering up?

Thanks Dave. So no other steps need be taken before installing the other transistors? I thought it might be a good idea to put a 75W lamp in series with the AC line or a 330 ohm resistor in series with each output transistor's collector in case the transistors saturate. Just want to avoid any unforseen disaster. Or don't you think any of this is necessary?

\$\endgroup\$
  • 3
    \$\begingroup\$ Not sure what an "amplified diode" is or even exactly how all this hooks up. If you could post a schematic you would have a much better chance of getting a good answer. \$\endgroup\$ – John D Feb 1 '16 at 23:53
  • 1
    \$\begingroup\$ The amplified diode is made from a transistor and has been popular ever since transistors got cheap .The transistor has two resistors around it .One is BE the other is BC .These two resistors form a potential divider which effectively amplifies the VBE of the transistor allowing good flexability in setting up bias volts and hence idle current .Most importantly it amplifies the well documented negative temp co of VBE .This helps prevent thermal runaway and keeps idle current reasonably constant when supply volts varies.Its best to mount this transistor on the main heatsink for good tracking. \$\endgroup\$ – Autistic Feb 2 '16 at 9:59
  • \$\begingroup\$ @Autistic Thanks, I always called that a VBE multiplier, never heard it called an "amplified diode" but I can kind of see where the name comes from. \$\endgroup\$ – John D Feb 2 '16 at 15:44
  • \$\begingroup\$ Q 809 is a terrible mistake in your schematic .Reverse CE . \$\endgroup\$ – Autistic Feb 2 '16 at 20:05
  • \$\begingroup\$ Autistic, good catch on the schematic error! \$\endgroup\$ – zarko Feb 3 '16 at 17:02
1
\$\begingroup\$

Without a specific schematic, one can only speak in general terms. But yes, in a robust design, the actual bias current of the output devices should be a relatively small fraction of the total current through the bias network, in order to reduce the effects of the manufacturing tolerances of specific devices. This means that the voltages you measure should be close to the values you would get with good devices in place.


Responding to the followup information you added, I can see that there is some confusion.

The bias is the voltage drop across Q804, which amounts to about 1.8V, based on your measurements. This seems like a reasonable value.

The fact that the common mode voltage is on the order of 37V just reflects the fact that the feedback path is currently open, because the output transistors are not in place. The differential amplifier is trying to increase the output voltage slightly, but since it can't tell that it's happening, it ends up driving to the positive rail.

If you temporarily connect the emitter of Q804 to P809, you'll create an alternate feedback path, and you should see the common mode voltage drop to close to 0V. Specifically, you should see Q804 emitter at 0V and its collector at +1.8V. If you instead connect its collector to P809, you should then see the collector at 0V and the emitter at -1.8V. These checks would verify that Q801-Q804 are all working correctly.

\$\endgroup\$
  • \$\begingroup\$ Thanks for the feedback Dave. Sorry for the late reply, I got sidetracked on something else. \$\endgroup\$ – zarko Feb 16 '16 at 20:00
  • \$\begingroup\$ Hello DaveTweed, @Zarko. Zarko, you added some data for Dave Tweed to look at. But you editted it into his answer and so it was rejected. Please edit the info into your question, then everyone can see it! Thanks. \$\endgroup\$ – bitsmack Feb 16 '16 at 21:36
  • \$\begingroup\$ @bitsmack: Thanks! I moved the new material to the question and then responded to it by editing my answer. \$\endgroup\$ – Dave Tweed Feb 16 '16 at 23:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.