Practice problem

[huub8]

I'm trying to solve a practice problem, but I can't seem to get the calculation right.

I'm suppossed to calculate Vo and Vce
the current gain is 100, and the Vbe is 0.7 V

The answer is: 2.876 V and 1.984 V



I can only think of this:

KVL for the input loop gives:
-5 + 10 000 * Ib + 0.7 - 200 * (Ib - Ic) = 0

KVL for the output loop gives:
12 - 200 * (Ic - Ib) - Vce - 500 * Ic = 0

Ic = 100 Ib


But this gives me a wrong answer, what am I doing wrong?

 

[WimRFP]

In the first loop you meet the positive terminal of Vout first, so it should be
-5 + 10 000 * Ib + 0.7 + 200 * (Ib + Ic) = 0

the emitter current is the sum of Ib and Ic, given your current definitions.

the second equation is fine when the + of the collector resistor is at the 12V supply (so the minus is at the collector).

 

[huub8]

Thank you for your response, but I get a different answer if when put your formula in wolframalpha, do you know why?

 

[WimRFP]

Both base and collector current (from your circuit) go towards the transistor terminals, so Ib+Ic have to leave the emitter.

When I calculate IB and Ic based on the answers of #1, the first equation is fine.

 

[huub8]

If put this into wolframalpha:

-5 + 10 000 * b + 0.7 + 200 * (b + c) = 0, c = 100 b

i get: b = 0.000212871, c = 0.0212871

b + c is the collector current trought the 200 ohm resistor right? So the voltage acros it should be:

(0.000212871 + 0.0212871) * 200 = 4.3 V, not 2.876

 

[WimRFP]

When I put your values for b an c into the first equation (-5 + 10 000 * b + 0.7 + 200 * (b + c) = 0), the equation is not valid. The results from #1 give a valid equation.

 

[huub8]

Thank you very much, I just now found out that wolramalpha made 10 000 into 10 * 0

- - - Updated - - -

One more question, in this circuit, how can you calculate Vo and Io?



What I did is:

1 = 120 000 * Ib + 0.7
Ib = 2.5 * 10^-6

Ic =2.5 * 10^-6 * 80 = 2 * 10^-4

But I cant figure out how to aply KVL or KCL to the 10k and the transistor which to me seem te be in parallel?

Because: 20 = 10000 * (2.5 * 10^-6 + Io) + 10000 * Io
Doesn't work

 

[WimRFP]

Are you allowed to use Thevenin? You can convert the two 10k Ohms resistors into 5 kOhms with a 10V source.

This is a bit tricky circuit. As you are assuming an ideal transitor, Ib controls Ic, as long as the transistor doesn't saturate. So when Vce stays above say 0.3 V,
Ic = 100*Ib is valid, hence your calculated 0.2 mA=Ic (assuming now hfe=80).

for the last equation I would say: all current leaving from the vout +terminal should be zero.

So: Ic + Vce/10k + (Vce-20)/10k = 0.

Note that once VCE < 0.3V, you can no longer assume hfe=80 as the transistor will saturate.

 

[huub8]

b (beta) = 80 and Vbe = 0.7

So Ic = 80 * Ib right?

I believe that I'm supposed to assume that the transitor is in active mode.

The given answer is: 12V and 600uA

If calculate Io using your formula (with Ic = 2 * 10^-4) I find 9A

 

[WimRFP]

When I am using

Ic + Vce/10k + (Vce-20)/10k = 0 and enter Ic = 0.2 mA,

Vce = 9V, so the transistor is in its active region.

I get the same value when using Thevenin (10V source with 5 kOhm resistor).

So your Iout = 9/10 k = 900 uA.

Your answer is contradicting. 12V= VCE into 10 Kohm gives 1.2 mA.

 

[huub8]

So your saying that the given answer doesnt add up?

 

[WimRFP]

you mentioned Vout = Vce = 12V. The load (that carries Iout) = 10 kOhm, so Iout = 1.2 mA, I can't change it for you.

Did you use Thevenin also?

 

[erikl]

If VBE is assumed to be constant 0.7V , it's so simple:

IC*(1.01*200) + (IC/100)*10000 = 5 - 0.7

IC*(202+100) = 4.3

IC = 4.3/302 = 0.014238

See here the result:

 

[huub8]

Thats another practice problem erikl, my last questions are about a different one.

The exact question that came with practice problem 3.13 is:

"The transistor circuit in fig. 3.45 has B = 80, Vbe = 0.7V. Find Vo and Io."


And the exact answer is:

"Answer: 12 V, 600 uA


I did not use Thevenin

 

[godfreyl]

And the exact answer is:

"Answer: 12 V, 600 uA

That's wrong. I agree with WimRFP's calculation: Io = 900uA, V0 = 9V.

Today's lesson: Sometimes books contain mistakes.

 

[huub8]

Just one more question, this time I do need to use Thenevin equivalence:



I figured that I was dealing with a supermesh, since loop one and two share the 1.5Ix current supply.

Using mesh analysis I got this for the supermesh:

-6 + 5 * Ia + 7 * Ib – 4 *Ic = 0

And for node 1 I got this:

Ia + 1.5 * Ib = Ib

For loop 3:

1 + 4c – 4b = 0

Rth = 1 / Io, and Io = - Ic

But I get a different answer using these calculations.

 

[godfreyl]

This time I agree with the answer given, but my method of calculation probably wouldn't earn you any points in an exam.

 

[huub8]

I don't do this for a course (just in my free time), so if you could just point out where my fault lies that would be very helpfull.

 

[LvW]

For my opinion, it is a simple superposition problem (with Ib=2.5µA and Ic=80*2.5=200µA):
a) Setting the current source Ic=0, we have Io1=20V/20k=1mA=1000µA
b) Setting the voltage source to zero, the current Ic is devided between two equal resistors: Io2=-200uA/2=-100uA.

Hence, Io=Io1+Io2=900µA.