# Determine hFE of PNP-Transistor

Discussion in 'Homework Help' started by S.Sphereson, Apr 20, 2014.

1. ### S.Sphereson Thread Starter New Member

Apr 17, 2014
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The resistor over Ucb is giving me a hard time...

I need to find hFE for given transistor circuit but get stuck at evaluating Ib and Ic.

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2. ### ericgibbs Moderator

Jan 29, 2010
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hi,
The question gives you the Collector voltage and the Collector resistance, so it should be easy.

What do you calculate the Collector current value.??

3. ### S.Sphereson Thread Starter New Member

Apr 17, 2014
15
0
10mA.... Not correct though, I know. What am I missing?

4. ### ericgibbs Moderator

Jan 29, 2010
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Ohms law,,, Ic = Vc/Rc , so why do you think 10mA is wrong.?

Where Vc= 2.3v and Rc = 230R

5. ### Jony130 AAC Fanatic!

Feb 17, 2009
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As you know current that is flow through Rc resistor is equal to
IRc = Vc/Rc = 10mA But IRc current is not equal to Ic current. Try to think about base current and 20K resistor.

6. ### S.Sphereson Thread Starter New Member

Apr 17, 2014
15
0
OK so I have Ic, it seems. I suspected I had it... now I know I had it and hence Ib is my problem.

In a previous problem Ib, as was Ic, went to ground through a resistor Rb. Now base and collector are connected. How do I approach something like that?

7. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Look at this diagram

Where in red you have the IB current path. And in blue you have Ic current path. Also notice that you know the voltage across 20K resistor. So if you know voltage across resistor, you can easily find current that is flow through this resistor.

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8. ### ericgibbs Moderator

Jan 29, 2010
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The Collector to Base resistor provides negative feedback, you must assume the voltages given in the question are steady state.

Your working calculations so far are on the right track.
Consider what Jony is pointing out so that you can complete the calculation.

9. ### S.Sphereson Thread Starter New Member

Apr 17, 2014
15
0
Ubc = 2V

So is the current running through the 20k resistor I20k = 2V/20kohm = .1mA ??

If so, how would that help me??

10. ### ericgibbs Moderator

Jan 29, 2010
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So what do you consider is the actual Collector current Ic value.?

Given Irc=10mA and Ib=0.1mA

EDIT:
hFE = Ic/Ib

Last edited: Apr 20, 2014
11. ### S.Sphereson Thread Starter New Member

Apr 17, 2014
15
0
I'm thinking Collector current Ic value is the sum of Ib and Irc...

When I do that I get hFE = Ic/Ib = 0.99

12. ### S.Sphereson Thread Starter New Member

Apr 17, 2014
15
0
Is that correct?

I get 0.99 => I must be on to something..?

13. ### Jony130 AAC Fanatic!

Feb 17, 2009
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And apply KCL for IRc current.

14. ### ericgibbs Moderator

Jan 29, 2010
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Look at this expanded diagram of your circuit, showing the individual transistor currents.
E

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15. ### S.Sphereson Thread Starter New Member

Apr 17, 2014
15
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$h_{FE}=\frac{0.01A-0.0001A}{0.0001A}=99$

So Ic = Ic - Ib because Ib provides negative feedback Collector -> Base?

16. ### ericgibbs Moderator

Jan 29, 2010
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Both the Ic and Ib currents 'flow' thru the 230R

Your hFE looks OK to me.

You will cover negative feedback in your future studies, for the time being consider this method of collector to base of biasing sets the DC operating conditions of the circuit.

17. ### S.Sphereson Thread Starter New Member

Apr 17, 2014
15
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OK. If hFE = Ic/Ib... is hFE = Irc/Irb ok? If I do that I get hFE = 101.

I want hFE = 99

Quite confused at moment.

18. ### ericgibbs Moderator

Jan 29, 2010
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I did not say that...

hFE is not Irc/Irb

hFE= Ic/Ib = 99, where Ic = Irc-Ib.

Its Collector Current/ Base Current

E

You said earlier Ic = Ic - Ib .!

19. ### S.Sphereson Thread Starter New Member

Apr 17, 2014
15
0
OK. Not getting it right at all.

Can you give me the collector and base currents so that I can compare to my calculations?

For some reason it is not working out.

20. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Ib = (Vb - Vc)/Rb = (4.3V - 2.3V)/20kΩ = 100μA
IRc = Vc/Rc = 2.3V/230Ω = 10mA

And from KCL we have

IRc = Ib + Ic and Hfe = Ic/Ib

Ic = IRc - Ib = 10mA - 0.1mA = 9.9mA and Hfe = 9.9mA/0.1mA = 99

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