Determine hFE of PNP-Transistor

ericgibbs

Joined Jan 29, 2010
18,766
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.??
 

Jony130

Joined Feb 17, 2009
5,487
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.

 

Thread Starter

S.Sphereson

Joined Apr 17, 2014
15
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?
 

Jony130

Joined Feb 17, 2009
5,487
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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ericgibbs

Joined Jan 29, 2010
18,766
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?
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.
 

ericgibbs

Joined Jan 29, 2010
18,766
So Ic = Ic - Ib because Ib provides negative feedback Collector -> Base?
Your equation should read Irc = Ic+ Ib.:)

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.
 

Thread Starter

S.Sphereson

Joined Apr 17, 2014
15
Your equation should read Irc = Ic+ Ib.:)

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.
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.
 

ericgibbs

Joined Jan 29, 2010
18,766
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.
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 .!
It should read, Ic= Irc-Ib
 

Thread Starter

S.Sphereson

Joined Apr 17, 2014
15
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.
 

Jony130

Joined Feb 17, 2009
5,487
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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