danny davis
Banned
How is it reverse voltage, I see it has forward voltage +15vdc + 13 volts = +28 volts
Where is the reverse voltage coming from?
Then the collector-emitter has 13V of reverse voltage it was not designed to have.
Its emitter-base junction also has reverse voltage and will have avalanche breakdown at a very high current that destroys it.
Why should I answer when you do not know about the basics of transistors?How is it reverse voltage, I see it has forward voltage +15vdc + 13 volts = +28 volts
Where is the reverse voltage coming from?
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.
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.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
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.
Again you are showing us that you know NOTHING about electronics.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
When you connect the extremely small IC transistor with backwards polarities then it becomes destroyed very quickly.
The emitter of an NPN transistor MUST NEVER be more positive than its collector. The collector was +15V and the emitter was +28V. Backwards.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
The collector was +15V and the emitter was +28V. Backwards.
The collector of the NPN output transistor is supposed to be more positive than its emitter. It was not, it had reversed polarity.How is it the transistor connected backwards polarites?
The collector is at +15V and the emitter was shorted to +28V so the polarity was reversed.
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.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
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.
We use cookies and similar technologies for the following purposes:
Do you accept cookies and these technologies?
We use cookies and similar technologies for the following purposes:
Do you accept cookies and these technologies?