# Transistor Output Voltage Formula

Discussion in 'General Electronics Discussion' started by slamb, Aug 31, 2011.

1. ### slamb

3
0
Aug 31, 2011
Hi,

Given the data in the diagram below, is there a formula I can use to calculate what should be the output voltage?

Thanks.

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2. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,497
2,839
Jan 21, 2010
It depends on the characteristics of the transistor. Ignoring for a moment that they can vary significantly,

You would first calculate the base current (4.3v across 47k less 0.7v across 47k).

Then you would take a reasonable value for the gain (say the minimum) and estimate the potential collector current.

If the potential collector current is more that 5v - VCEsat across 1k, then the output voltage will be VCEsat (for that current). If not, use the current to calculate the voltage across the resistor.

Naturally, the difference in possible gain figures (hfe) for a transistor may vary greatly, so you may wish to repeat the calculation for another value (say, the max) and see if that makes a difference. In this case you may find you have a reliable figure, or some expected range.

3. ### slamb

3
0
Aug 31, 2011
[FONT=&quot]Hi Steve,[/FONT]

[FONT=&quot]Because I'm a rookie in electronics, trying to learn and understand it while I'm repairing, for the second time, my own LCD TV. The first time it was quite easy; 4 bulged e-caps on the power supply board. I replaced them and BAM! Problem solved. But, this time, it's a bit more tricky. So, I've done some voltage tests to try to find the problem's source. I've found a point where the voltage seems too low. Guess where, at the output of a transistor. Hence, my initial question to find out if the transistor can be defective or not and if I should remove it for further investigation. So, because I'm a rookie, I simply do not understand the terms "potential collector current" and "VCEsat". So, while I was searching the meaning of those terms, on the Internet, the following calculation caught my attention.[/FONT]

[FONT=&quot]V(base) = V(input) * ((R2 / (R1 + R2)) = 5v * ((47k / (47k + 47k)) = 2.5v[/FONT]

[FONT=&quot]V(emitter) = V(base) / 0.7v = 2.5v – 0.7v = 1.8v[/FONT]

[FONT=&quot]I(emitter) = V(emitter) / R(emitter) = 1.8v / 47k = 0.000038[/FONT]

[FONT=&quot]I(base) = V(source at collector) / (R1 + R2) = 5v / (47k + 47k) = 0.000053[/FONT]

[FONT=&quot]I(collector) = I(emitter) + I(base) = 0.000038 + 0.000053 = 0.000091[/FONT]

[FONT=&quot]V(drop at collector) = I(collector) * R(collector) = 0.000091 * 1k = 0.091v[/FONT]

[FONT=&quot]V(output) = V(source at collector) – V(drop at collector) = 5v – 0.91v = 4.9v[/FONT]

[FONT=&quot]Is it what you tried to explain, I mean, is it the way to calculate a transistor's output voltage?[/FONT]

4. ### Merlin3189

250
69
Aug 4, 2011
Transistor collector Voltage

Looking at the circuit you show, I would say the output Voltage is not critical. This appears to be a buffer inverter stage for a switching Voltage that is intended to be either on or off. With the 5V input you should get a low output, a few tenths of a Volt. If the input were low, say less than about 1V, then the output should be high, about 4.5V.
For intermediate values of input, the output would depend on the gain of the transistor.
This is not normal practice. If you want the output to be accurately related to the input over the intermediate range of Voltages, then the gain of the stage has to be set by the resistors so that it is relatively independent of the transistor's gain (ie. negative feedback.)
This is because the gain of a transistor is not set precisely by the manufacturing process and, even if the transistor were selected to have a specific gain, that gain would vary with temperature.

So I think your transistor is not at fault. When the input is high, the output should be low. The precise Voltage is unimportant so long as it is below about 0.7V.
There would not be much point in having this stage if the input were always 5V. So I presume sometimes it should be low and then the output should go high. If it stayed low when the input goes low, then the transistor would be faulty.

5. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,497
2,839
Jan 21, 2010
Knowing the reason for the question helps a lot.

Merlin's answer is a practical one which should lead you in the right direction.

Get the data sheet for this device and look up VCEsat. This is the voltage across the collector/emitter when the transistor is saturated. It it typically specified at a (or a number of) collector currents.

The "potential collector current" is the collector current that could potentially be flowing. Determine the current through the collector resistor when it has about 5v across it. This will give you a ballpark current to use when you're looking for the VCEsat.

The VCEsat will be the "few tenths of a volt" that Merlin mentions.

6. ### rogerk8

176
0
Jul 28, 2011
It is seldom that VCEsat has any importance whatsoever. I don't say this to be a besserwisser I just tell you the fact. I did some rough calculations and came to the conclusion that the transistor in mention should be well beyond saturation (i.e 0V) in the stated configuration. KR/Roger

7. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,497
2,839
Jan 21, 2010
That's a very bold statement.

I cannot see how the output voltage can fall any lower than VCEsat (without external influence).

the voltage across the CE (the VCE part) of the transistor isn't going to fall further if you give it more base current. (that's the "sat" part).

When the transistor is off (input grounded or open circuit) the output (unloaded) will be Vcc. When the transistor is on (input tied to Vcc) the output will be in the order of a few tenths of a volt, not 0.

If we look at a concrete example, say a BC548, the VCEsat for 10mA Ic, 0.5mA Ib is between 0.09 and 0.25V. We can see from the base resistor that Ib will actually be lower than this (under 0.1mA) but the Ic will also be lower (around 5mA) so it is likely that you'd get a similar voltage output given that the transistor won't be so far into saturation.

8. ### rogerk8

176
0
Jul 28, 2011
That may be true but those very accurate things you talk about is still seldom of practical importance. One of the only time I can think of is switched supply purposes. But BJT's are seldom used in those cases, as you probably know. KR/Roger

9. ### slamb

3
0
Aug 31, 2011
[FONT=&quot]Hi,[/FONT]

[FONT=&quot]Sorry for this long delay for reply but I was very busy with my day job.[/FONT]

[FONT=&quot]Ok, as you suggested, I got the transistor datasheet (see attachment). On page 2, at the ELECTRICAL CHARACTERISTICS section, it is written that the TYPICAL output voltage of the transistor is 0.1v. When I measure the voltage of this transistor (on board), here's what I have:[/FONT]

[FONT=&quot]Base: 5v.[/FONT]
[FONT=&quot]Load (before the 1k resistor): 5v.[/FONT]
[FONT=&quot]Output (after the 1k resistor and before the 10k resistor): 0.1v.[/FONT]

[FONT=&quot]So, based on what you said and with the voltage readings and the TYPICAL output voltage of the datasheet, the transistor seems to be fine.[/FONT]