how to calculate the collector-emitter voltage?

Question

friends please teach me how do u calculate the collector-emitter voltage of a transistor in a circuit,ex:-
" http://www.pan-tex.net/usr/r/receivers/5109sigav.gif " please tell me the result and please explain me,it would be the greatest help,please.thank u

Answer 1

Billrussel was nearly right, except he didn't notice the collector is not at 12V, it is powered through the 100R resistor.



That makes the analysis more complcated. One way is to set up the network equations and solve them. The best way to do that is use a computer circuit simulator, but that doesn't teach you anything much so here is a different way to do it by hand.



Ignore the base current of the transistor, and assume the DC resistance of the inductors is zero.



First assume the Vc is 12V (even though we know it can't possibly be that value). As Bill said, that gives

Vb = 12 x 0.47/1.47 = 3.8V

Ve = 3.8 - 0.7 = 3.1V

Ie = 3.1 / 0.056 = 55 mA

Current through the base divider is 12 / 1.47 = 8mA



So the current through the 100R resistor is 55+8 = 63mA. The voltage drop across it is 63 x 0.1 = 6.3V so Vc would be at 12- 6.3 = 5.7V, not 12V.



So try again with the different guess for Vc, say Vc = 6V

Vb = 6 x 0.47 / 1.47 = 1.9V

Ve = 1.9 - 0.7 = 1.2V

Ie = 1.2 / 0.056 = 21 mA

Current in base divider = 6 / 1.47 = 4mA

Total current = 21+4 = 25mA

Voltage drop across 100R = 2.5V

Vc = 12 - 2.5 = 9.5V not 6V.



The circuit is linear so we can do a linear interpolation between the two guesses, to find the actual value of Vc

If we guess Vc = 12V we get Vc = 5.7V

If we guess Vc = 6V we get Vc = 9.5V

So, if we guess Vc = 12 + x(12-6)

we will get

Vc = 5.7 + x(5.7-9.5)



Solve for x

12 + 6 x = 5.7 - 3.8x

9.8x = -6.3

x = -0.64

Vc = 12 - 0.64 x 6 = 8.2V



If you don't like the algebra, you can do it graphically. Just plot straight line through the points (12, 5.7) and (6, 9.5) and read off the point where the value on both axes is the same which will be (8.2, 8.2)



So now we can do the "proper" analysis with Vc = 8.2



Vb = 8.2 x 0.47 / 1.47 = 2.6V

Ve = 2.6 - 0.7 = 1.9V

Ie = 1.9 / 0.056 = 34 mA

Current in base divider = 8.2 / 1.47 = 6mA

Total current = 34+6 = 40mA

Voltage drop across 100R = 4V

Which is gives Vc = 12 - 4 = 8.0V which is near enough 8.2 for a practical answer.

So Vce = 8.2 - 1.9 = 6.3V



As Bill said. you can't do an AC analysis without knowing the input frequency and the inductor parameters.

Answer 2

The DC voltage is calculated as follows: (assuming the IB is low)

The voltage divider on the base sets the base voltage at 12*(470/1470) = 3.8 volts. the emitter voltage is 0.7 volts lower, so Ve = 3.1 volts. Collector is at 12 volts, so Vce = 12–3.1 = 8.9 volts.



As a check, the Ie = 3.1/56 = 55 mA. I can't find the datasheet, but assuming HFE = 50, then Ib = 55/50 = 1.1 mA. Current in the divider string is 12/1.5k = 8mA. so there is some error, 1.1 is not very small compared to 8.



AC voltage on the collector is more complicated. It depends on the input voltage and frequency and on the choke values, but if the transistor remains within it's linear region, then the max collector voltage is 2x12 = 24 volts. Subtract Ve of 3 volts for 21 volts Vce. But the transistor could be overdriven into class C operation which would cause the choke in the collector to operate as a flyback, so the voltage could be much higher than that.



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Answer 3

The collector voltage is 12 volts minus the IR drop through the 100 ohm resistor. The collector voltage is about 8.1 Volts, the emitter voltage is around 2 Volts, and base voltage is about 2.6 Volts. The collector emitter voltage is about 6.1 volts. You assume that the collector current is equal to the current leaving the emitter. The base voltage is about 0.6 volts above the emitter voltage.

Start by assuming the emitter voltage is 1 volt, the collector voltage would be around 10 volts. The base voltage is about 0.6 volts above the emitter voltage or 1.6 volts. The 470 ohm base resistor would have a voltage of 1.6 volts, the 1 kohm resistor would have a voltage drop of 3.4 volts, the combined voltage of the base bias resistors is 5 volts. The voltage across the base bias resistors doesn't equal the collector voltage. So the emitter voltage we picked is too low.

Try a larger emitter voltage like 2.0 Volts, the collector voltage is 8.1 volts. The base voltage is 2.6 volts, the 470 ohm resistor has a voltage drop of 2.6 volts, the 1k ohm resistor has a 5.5 volt drop. The total voltage across the base bias resistors is 8.1 volts which matches the collector voltage.

Answer 4

Ohms law