0
\$\begingroup\$

I've started learning about transistors in my class and am having a hard time understanding how to analyze circuits with them. I'm not sure how to calculate current changes around them or how they amplify current. I'm trying to find V_out in the following but don't know where to begin: enter image description here

V_in in this circuit is equal to 20 V DC

\$\endgroup\$
  • 1
    \$\begingroup\$ You need to know the transistor gain. \$\endgroup\$ – Leon Heller Oct 21 '15 at 2:01
  • \$\begingroup\$ Not really - it will only have second order effect. \$\endgroup\$ – Kevin White Oct 21 '15 at 2:13
  • 2
    \$\begingroup\$ First step is to recognize that this is an emitter follower, or common collector configuration. From there if you can assume the transistor isn't stupidly chosen, you can make a decent guess at the solution without knowing exactly the gain or writing any equations. \$\endgroup\$ – The Photon Oct 21 '15 at 2:14
5
\$\begingroup\$

Here is a simplified approach that hopefully give you the concepts.

Assume that the base emitter voltage of the transistor is 0.7V and the gain is large enough that the base current is negligible.

First of all calculate the voltage at the junction of the 10k and 6.8k resistors as if the transistor was not present.

Vbase = 20*6.8/(10+6.8) = 8.1 volts.

AS Photon points out the transistor is in emitter follower configuration so the emitter will follow the base but be lower by the base to emitter voltage.

In that case the voltage at the emitter (Vout) will be 8.1 - 0.7 = 7.4V. This will be the output voltage Vout.

Now lets how much the base current will affect things if we have a more normal transistor.

The current through the 1K resistor is Vout/1K = 7.4/1 = 7.4mA. If we assume the Hfe of the transistor is 100 (This isa reasonable gain although many transistors these days are better than that). The base current will be 7.4/100 = 74uA.

How much will that affect the voltage at the base?

The effective impedance at the base is equals to the Thevenin equivalent resistance which is 6.8K in parallel with 10k.

This is (R1*R2)/(R1 + R2) (10*6.8)/(10+6.8) = 4.04k.

The voltage drop due to the base current will be Ib * 4.04k = 0.074*4.04 = 299mV.

So the output with this correction will be about 300mV less than our original assumption when using a very high gain transistor. i.e. 7.1V.

This approach is not 100% accurate but will be very close and is more straightforward than the full analytical method, especially when some of the parameters such as the gain of the transistor are not known and will vary significantly from unit to unit.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.