[SOLVED] Test Fixture blowing 741 Op amp from a Short on a Device under test PCB product

Then the collector-emitter has 13V of reverse voltage it was not designed to have.

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How is it reverse voltage, I see it has forward voltage +15vdc + 13 volts = +28 volts

Its emitter-base junction also has reverse voltage and will have avalanche breakdown at a very high current that destroys it.

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Where is the reverse voltage coming from?

danny davis said:

How is it reverse voltage, I see it has forward voltage +15vdc + 13 volts = +28 volts
Where is the reverse voltage coming from?

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Why should I answer when you do not know about the basics of transistors?
The collector of the NPN output transistor is at +15V. The emitter is supposed to be at +15V or LESS than +15V, not more than +15V.
When the emitter voltage of an NPN transistor is more positive than its collector voltage then the collector to emitter voltage is reversed (backwards) and the emitter to base voltage is also reversed.

When the emitter voltage of an NPN transistor is more positive than its collector voltage then the collector to emitter voltage is reversed (backwards) and the emitter to base voltage is also reversed.

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That's only if the output is on the collector and it has a LOAD , then the transistor is an inverter

The output you said was on the emitter , so its in phase with the input base signal or DC voltage

You're saying that's its out of phase and inverted

danny davis said:

That's only if the output is on the collector and it has a LOAD , then the transistor is an inverter

The output you said was on the emitter , so its in phase with the input base signal or DC voltage

You're saying that's its out of phase and inverted

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This has nothing to do with the phase of signal. Take a look at **broken link removed** for the internal schematic of a 741. You can see that the output pin is connected through a very small resistor to the collector of a transistor that goes to V-, and it is also connected to the emitter of another transistor. The emitter has a little arrow on it to remind you of the direction of the diode that is the emitter-base junction. If the output pin is very positive, then that positive voltage would be acting in the opposite direction of the arrow, so it is, as Audioguru said, reversed biased. That means the junction is going to try to oppose the flow of current. But that junction is a very delicate junction. It can only oppose the flow of current when the reversed bias voltage is very small. When that reverse bias voltage gets as large as it would be with +28 volts on the output, then it will certainly break down and burn up. Not only that diode, but the one just above it too - the one that is connected to V+. So the +28 volts goes roaring through those two transistors to get to the +15 volts. Current tries to flow from a more positive voltage to a less positive voltage.

If the output pin is very positive, then that positive voltage would be acting in the opposite direction of the arrow, so it is, as Audioguru said, reversed biased. That means the junction is going to try to oppose the flow of current. But that junction is a very delicate junction.

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I don't get it how it's reversed biasing when it's positive voltage, the output of the IC 741 is outputing a positive voltage, so how can adding a positive voltage of +28 volts be reversed biasing and flowing into the 741 output pin in the opposite direction into the output transistor to damage it?

I can only think of it as it's forward biasing it because it's in the same direction , the 741 output is positive voltage and so is the +28 volt short , so it should add together, not Oppose

danny davis said:

That's only if the output is on the collector and it has a LOAD , then the transistor is an inverter
The output you said was on the emitter, so its in phase with the input base signal or DC voltage
You're saying that's its out of phase and inverted

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Again you are showing us that you know NOTHING about electronics.
1) I said "the collector to emitter voltage was backwards (reversed) and the emitter to base voltage is also reversed". I did not say its load is on the collector. I did not say anything was an inverter.
2) I did not talk about phase.

When you connect the extremely small IC transistor with backwards polarities then it becomes destroyed very quickly.

When you connect the extremely small IC transistor with backwards polarities then it becomes destroyed very quickly.

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How is it the transistor connected backwards polarites?

Maybe there is a potential difference , the the potential differences from the output positive voltage and the SHORTED +28 volts, both potentials differences are POSITIVE

You're saying the Difference of the Potential differences are backwards polarites, even tho they are both positive voltages?

danny davis said:

I don't get it how it's reversed biasing when it's positive voltage, the output of the IC 741 is outputing a positive voltage, so how can adding a positive voltage of +28 volts be reversed biasing and flowing into the 741 output pin in the opposite direction into the output transistor to damage it?

I can only think of it as it's forward biasing it because it's in the same direction , the 741 output is positive voltage and so is the +28 volt short , so it should add together, not Oppose

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The emitter of an NPN transistor MUST NEVER be more positive than its collector. The collector was +15V and the emitter was +28V. Backwards.
When an NPN transistor turns off then its collector voltage is NORMALLY more positive than its emitter.
When an NPN transistor is partially turned on then the collector voltage is NORMALLY more positive than its emitter.
When an NPN transistor is turned on hard then its collector voltage is NORMALLY a tiny amount more positive than its emitter.

The collector was +15V and the emitter was +28V. Backwards.

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Right, it's two Positive voltages , that have a potential difference, but they are both positive ,

+28 - +15 = -13 volts

danny davis said:

How is it the transistor connected backwards polarites?

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The collector of the NPN output transistor is supposed to be more positive than its emitter. It was not, it had reversed polarity.
The collector is at +15V and the emitter was shorted to +28V so the polarity was reversed.
If the collector is at 0V and the emitter is at +13V then the polarity is still reversed.
If the collector is at -15V and the emitter is at -2V then the polarity is still reversed.
If the collector is at +115V and the emitter is at +128V then the polarity is still reversed.

The collector is at +15V and the emitter was shorted to +28V so the polarity was reversed.

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How it the polarity reversed? they are both positive potentials?

Do you mean the potential differences between +15v and +28v?

I don't see the polarities being reversed

danny davis said:

How it the polarity reversed? they are both positive potentials?

Do you mean the potential differences between +15v and +28v?

I don't see the polarities being reversed

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Yes, all that counts is the difference in potentials. The only way you can say the supply is +15v by using a voltmeter is the connect the black lead to ground and the red lead to the supply. The voltmeter measures the difference between the red lead voltage and the black lead voltage. All voltages are relative so some other point. So even though you say this point has +15 volts, what you really mean is that this point has +15 volts with respect to some assumed ground. What do you do when there is no ground? Or when there are several grounds that are not at the same voltage? That can happen. Then you will have to be very explicit when you say some point has a certain voltage. You will always have to mention the reference point for the voltage measurement. Same thing with this emitter-base junction. The only thing the junction in the transistor "feels" is the difference voltage between the two leads. If point A is +28v and point B is +15v, that feels exactly the same to the junction as if point A was +13v and point B was 0v. The current wants to travel from point A to point B. The only thing pushing that current is the difference in voltage.

It is sometimes useful to think of water pressure as an analog for voltage, and water flow as an analogy for electrical current. If you have a river that falls 20 feet through a power-generating dam, the power generated and the direction of the water flow are dependent only on the difference in water levels. If the dam is at 2000 feet above sea level and it falls to 1880 feet above sea level, it does the same kind of work as if the dam were at 1000 feet and fell to 880 feet. The absolute height does not matter. Only the difference matters.

If the dam is at 2000 feet above sea level and it falls to 1880 feet above sea level, it does the same kind of work as if the dam were at 1000 feet and fell to 880 feet. The absolute height does not matter. Only the difference matters.

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I guess the +28 volts , water level is Higher than +15 volts , which doesn't fall to a different level , instead it OVERRIDES and Overflows past the different sea levels , so the water goes Above

Reverse Biasing , it water flow that travels up the Fall from 880feet up to 1000feet , it's not a fall, its travelling up