# Calculation for Transistor Circuit

Discussion in 'Homework Help' started by sdas86, Mar 18, 2015.

1. ### sdas86 Thread Starter New Member

Mar 18, 2015
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Hi all,
I have a circuit as attached. I am facing problem to calculate the voltage across the Resistors.
I am also not sure what is the state of the BJT transistors (Cutoff, Linear or Saturation).

Need help here. Thanks.

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2. ### DickCappels Moderator

Aug 21, 2008
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What can you say about the currents in the circuit so far? What is the voltage at the bast of Q3 and what would you expect the current through the 6.2 volt Zener to be with that base voltage. Everything else follows from that.

3. ### sdas86 Thread Starter New Member

Mar 18, 2015
26
0
Hi,
The voltage for the base of Q3 is 5.1/27.1 * 12 = 2.25V. I am not sure about the current flow.

4. ### DickCappels Moderator

Aug 21, 2008
3,918
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What would the voltage on the emitter of Q1, and therefor the cathode of the 6.2 volt zener be?

5. ### sdas86 Thread Starter New Member

Mar 18, 2015
26
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The Q1 Emitter voltage will be the reverse bias voltage of Zener Diode D33 which is +6.2V. What is the turn on voltage for NPN transistor,Q3? (Datasheet: http://www.farnell.com/datasheets/1856333.pdf)
But I do not quite understand the datasheet information as I cannot find what voltage is required to turn on Q3 on the datasheet.

I am not sure how to calculate the voltage across R53 and R54.

And for the PNP transistor, Q2, from what I know, PNP transistor is "normally ON" which means no current is required at the Base to switch it on and it is always On unless a current is applied to Base to turn it off.

What is the output current at the Q2 Collector? Is Q2 turn on or off?

6. ### DickCappels Moderator

Aug 21, 2008
3,918
1,084
In a working circuit the voltage at the cathode of the zener would be equal to the reverse breakdown voltage, but this is a "trick" circuit. The only part of the circuit that does anything is the voltage divider made by R51 and R52.

Because the zener has reverse leakage, in order for the zener to have voltage across it, there must be some current applied to it. The only place that current can come from is the emitter of Q3. We can see the parameter VBE which is typically 0.66 volts at a nominal collector current.

You found that the voltage at the base of Q3 is 2.25 volts. Taking 2.25 volts and subtracting the base-emitter drop, we get 2.25 - 0.66 = 1.59 volts. What does a 6.2 volt zener do when you put 1.59 volts reverse bias across it? Very little. The correct answer would be some number of nanoamps or microamps, but in this circuit, that wouldn't matter so it is safe to say that the emitter current for Q3 is 0 amps.

Now the voltage drop across R53 and R54 is easy to calculate: zero amps x 22k = 0 volts; also 0 amps x 15k = 0 volts.

Q2 is a PNP but just as with an NPN, you only get collector current (above leakage) when the base is forward biased, meaning that it is normally off. Since there is 0 volts across R53 there is 0 volts across the base and emitter of Q2, so Q2 must be off as well.

Now is it clear that only part of the circuit working is the voltage divider made up of R51 and R52 because there is no appreciable current flowing through the any other circuit components.

If R52 had a high enough resistance, enough to get the emitter of Q3 to the zener breakdown voltage the calculation would be more involved because there would be flowing through all of the parts.

In real life, there will be leakage currents flowing, but from looking at the circuit, they would so small as to not be important.

7. ### sdas86 Thread Starter New Member

Mar 18, 2015
26
0
Hi Thanks for the explanation.
I understand now. Thumbs up for the great help!