Home Stereo Amplifier Overload Indicator

[Jon Snell]

NOTE Vebo ... Vebo means the maximum permissible reverse base emitter voltage before damage.
All silicon transistors have a Vbe of around 0.6volts at which point they switch on.

[bobloblaws]

NOTE Vebo ... Vebo means the maximum permissible reverse base emitter voltage before damage.
All silicon transistors have a Vbe of around 0.6volts at which point they switch on.

Oh, ok. Thanks, good to know. I see now that I was looking at the "maximum ratings: column.

So if the turn on voltage is 6V as opposed to 5V it would give me a little more wiggle room, assuming the circuit stays stable. But that might be wishful thinking given that I've seen some odd changes. The way it stands currently the only thing different from when I first started troubleshooting it is that R005 is replaced, and only because I had snipped one of the legs at one point. I'm wondering if one or more solder joints might be wonky. Since I'm planning to pull the PCB and clean everything up anyway I was thinking I'd go ahead and hit those surrounding components with a fresh bead.

[Jon Snell]

The junction point is between 0.5 and 0.6volts, not 5 or 6volts.

[Helmholtz]

All DC voltages have a polarity (a sign) and the sign matters, as if often determines if a component is forward or reverse biased.
A pnp transistor (Q3) turns on when its base is - 0.5V to - 0.6V wrt to its emitter. To measure correct polarity connect black meter lead to emitter and red to base.

[bobloblaws]

All DC voltages have a polarity (a sign) and the sign matters, as if often determines if a component is forward or reverse biased.
A pnp transistor (Q3) turns on when its base is - 0.5V to - 0.6V wrt to its emitter. To measure correct polarity connect black meter lead to emitter and red to base.

Darn, I am plenty confused.

I bread boarded up a circuit similar to the one on question. Wrtg I have 12.92 volts on the emitter and 12.83 on the base. But if I connect my black lead to the emitter and and the red one to the base my meter shows 0V. It's an autorange and I tried the other range settings as well, still 0V.

Also, I'm seeing a negative voltage on the collector, whereas in the actual amp circuit on the side (L) that is working properly the collector shows 0V which we established means it is turned off. So if this one is turned off as I would have expected it to be 0V also.

For the turn on voltage, do you mean it turns on at roughly -0.5V to -0.6V or that that is the extent of the range in which it will turn on? On the 2N3906 datasheet it shows Vbe min and max values of -0.65 and -0.85, respectively. Does that mean it turns off again around -0.8V to -0.9V?

[The Dude]

It's a silicon junction. Its maximum forward biased voltage drop is fixed. Let's say it's .6V B-E (it will vary slightly from transistor to transistor). If you hook up a 9V battery through a current limiting resistor to the junction (forward biased), it will still drop .6V. So no, it doesn't turn off again at .9V because it's not ever going to have .9V B-E unless it's a bad transistor. No disrespect intended, but it would probably be a good idea for you to read up on transistor theory and operation. It's too much to discuss in a single post. Please don't take that wrong. I'm happy to help if I can.

[bobloblaws]

It's a silicon junction. Its maximum forward biased voltage drop is fixed. Let's say it's .6V B-E (it will vary slightly from transistor to transistor). If you hook up a 9V battery through a current limiting resistor to the junction (forward biased), it will still drop .6V. So no, it doesn't turn off again at .9V because it's not ever going to have .9V B-E unless it's a bad transistor. No disrespect intended, but it would probably be a good idea for you to read up on transistor theory and operation. It's too much to discuss in a single post. Please don't take that wrong. I'm happy to help if I can.

No offence taken, you're absolutely right. I have read up on it before but it never sank in. I wish I had a better brain!

The problem is that a lot of the theory has little meaning unless I can see it work in practice.

[bobloblaws]

It's a silicon junction. Its maximum forward biased voltage drop is fixed. Let's say it's .6V B-E (it will vary slightly from transistor to transistor). If you hook up a 9V battery through a current limiting resistor to the junction (forward biased), it will still drop .6V. So no, it doesn't turn off again at .9V because it's not ever going to have .9V B-E unless it's a bad transistor. No disrespect intended, but it would probably be a good idea for you to read up on transistor theory and operation. It's too much to discuss in a single post. Please don't take that wrong. I'm happy to help if I can.

So I went and bread boarded the simple circuit shown here https://www.dummies.com/programming/...r-as-a-switch/. I can observe the voltage on the base is 0.724V (which is the emitter/base voltage since in this case the emitter is grounded) when the LED turns on. This happens when the input voltage is 2.34V. If I steadily increase the input voltage the Vbe increases as well until it maxes out at 0.845V (input voltage 11V). I take it that means the EB junction is saturated?

So that's all well and good but the application in the amp continues to perplex me. For starters the LED is back on so I'm back to square one. On the left side (the good side) the Q004 base has 15.4V, the emitter has 15.7V and the collector has 0V. Vbe is 0V. On the bad side the Q003 emitter is also 15.7V (obviously, they are connected) while the base is 15.6V and the collector is 14.84V (compared to 5V last night). The Vbe is -.023.

So apart from the collector voltage being higher than before, I don't understand why Q003 is on and Q004 is not given that the voltages on the respective bases and emitters are nearly identical. And why would Q003 be on if Vbe is less negative than -0.6V

I do think I understand a little better now about the Vbe. I had been thinking of it in terms of measuring voltage drop on a resistor for example where I could get the same result by measuring across the resistor or calculating the difference in voltage wrt ground between both ends. Mu understanding now would be that there is no voltage drop if the 0.6V threshold is not met since there is no current flow in that state.

[Helmholtz]

For the turn on voltage, do you mean it turns on at roughly -0.5V to -0.6V or that that is the extent of the range in which it will turn on? On the 2N3906 datasheet it shows Vbe min and max values of -0.65 and -0.85, respectively.

Maybe its more helpful to say that if Vbe (correct polarity) is below 0.5V, the transistor is most probably off or shorted.
As has been said there is some variation and dependance on operating conditions and temperature. You typically find Vbe to be around 0.65V +/- 0.1V in amplifier circuits.

Does that mean it turns off again around -0.8V to -0.9V?

No. As long as there is base current, there will be collector current. But the transistor will saturate and lose its controllability. Also steady state current limits may be exceeded.
I never saw a Vbe of 0.8V or more in an amp. Things are different in pulsed mode operation, though.

[g1]

Time to look at Q123 and associated circuitry. It is driving Q003.

[The Dude]

It would be Q223 for the right channel. Also, he unhooked D204 (right channel's D104) and the LED remained lit, so that should eliminate the amp section from the equation.

[g1]

he unhooked D204 (right channel's D104) and the LED remained lit, so that should eliminate the amp section from the equation.

I missed that part. Thanks.

I'm a little confused about why the reading differ for B & E with respect to ground as compared to measured directly from B to E for Q003. One way must be incorrect but not sure which.
Edit: that was a breadboarded circuit (post #20) so maybe not applicable.

[The Dude]

There are only a half dozen parts in the circuit. If they all check good, I would suspect there are solder/connection issues.

[bobloblaws]

I missed that part. Thanks.

I'm a little confused about why the reading differ for B & E with respect to ground as compared to measured directly from B to E for Q003. One way must be incorrect but not sure which.
Edit: that was a breadboarded circuit (post #20) so maybe not applicable.

I was confused about that myself but I was hoping I had come up with the answer in my last paragraph in post #23. But maybe not?

[bobloblaws]

There are only a half dozen parts in the circuit. If they all check good, I would suspect there are solder/connection issues.

Yes, I'm planning on pulling the board out tonight to fix everything up and re-solder the connections while I'm at it.