# Interesting transistor problem

#### CrktMan

Joined Nov 29, 2005
34
Please see the circuit diagram as attached.
Given values are: β = 100, VBE= 0.7V, VCE= 0.3V and VA= infinite

Notice: base and collector are grounded!
We have to find Ve and Ic
I have come up with the following solution
I applied Thievenin's theorem and found: Vth = 5/(1+1) = 2.5v and Rth = 1||1 = 0.5 KΩ

I consider the pnp transistor is operating in the active region and base is forward biased and collector is reverse biased.

So, Ie = (2.5-0.7)/0.5 = 3.4 mA and Ve = 0.7 volt
Therefore, Ic = αIe = β/(1+ β)*Ie = 0.99*3.4 = 3.36 mA.

Anybody wants to correct me if I'm doing anything wrong here?

Thanks

#### hgmjr

Joined Jan 28, 2005
9,029
Originally posted by CrktMan@Dec 19 2005, 01:28 PM
Please see the circuit diagram as attached.
Given values are:  β = 100, VBE= 0.7V, VCE= 0.3V and VA= infinite

Notice: base and collector are grounded!
We have to find Ve and Ic
I have come up with the following solution
I applied Thievenin's theorem and found: Vth = 5/(1+1) = 2.5v and Rth = 1||1 = 0.5 KΩ

I consider the pnp transistor is operating in the active region and base is forward biased and collector is reverse biased.

So, Ie = (2.5-0.7)/0.5 = 3.4 mA and Ve = 0.7 volt
Therefore, Ic =  αIe =  β/(1+ β)*Ie = 0.99*3.4 = 3.36 mA.

Anybody wants to correct me if I'm doing anything wrong here?

Thanks
[post=12568]Quoted post[/post]​
It looks like you have obtained the correct values as far as I can tell.

The current flowing in the base should be the difference between your value for Ic and Ie.

hgmjr

#### JoeJester

Joined Apr 26, 2005
4,360
Don't see anything wrong ... it tracks well with the attached ...

#### n9352527

Joined Oct 14, 2005
1,198
Originally posted by CrktMan@Dec 19 2005, 07:28 PM
Please see the circuit diagram as attached.
Given values are: β = 100, VBE= 0.7V, VCE= 0.3V and VA= infinite

Notice: base and collector are grounded!
We have to find Ve and Ic
I have come up with the following solution
I applied Thievenin's theorem and found: Vth = 5/(1+1) = 2.5v and Rth = 1||1 = 0.5 KΩ

I consider the pnp transistor is operating in the active region and base is forward biased and collector is reverse biased.

So, Ie = (2.5-0.7)/0.5 = 3.4 mA and Ve = 0.7 volt
Therefore, Ic = αIe = β/(1+ β)*Ie = 0.99*3.4 = 3.36 mA.

Anybody wants to correct me if I'm doing anything wrong here?

Thanks
[post=12568]Quoted post[/post]​
Shouldn't you consider that Ve is fixed at 0.7V in your Thevenin reduction? If you do it'd give a slightly higher Ie.

#### CrktMan

Joined Nov 29, 2005
34
Originally posted by n9352527@Dec 19 2005, 07:22 PM
Shouldn't you consider that Ve is fixed at 0.7V in your Thevenin reduction? If you do it'd give a slightly higher Ie.
[post=12573]Quoted post[/post]​

Do you think I need to consider Ve = 0.7V to get Thevnin's equivalent circuit? I'm not sure!

#### hgmjr

Joined Jan 28, 2005
9,029
cktman,

Since the 0.7 volts present at the pnp's emitter is not an independent voltage source, I believe that the thevenin calculation that you made is correct.

The simulation results that joejester has provided seems to be in good agreement with your calculations.

hgmjr

#### n9352527

Joined Oct 14, 2005
1,198
Originally posted by CrktMan@Dec 20 2005, 12:35 AM

Do you think I need to consider Ve = 0.7V to get Thevnin's equivalent circuit? I'm not sure!
[post=12576]Quoted post[/post]​
Well... bear with me, if you consider Ve=0.7V, the current through the emitter resistor would be 4.3/1k=4.3mA. This current is the sum of the currents flowing through the resistor across the emitter to ground and the Ie. The current through the emitter-ground resistor is 0.7/1k=0.7mA, that would make the Ie=4.3-0.7=3.36mA. Ic can be obtained from the formula you have used Ic=3.36*100/101=3.33mA. Ib=3.33/100=33.30uA.

It does give a slightly lower Ie and Ic.

Opss, I've just noticed I wrote higher in my previous post, my mistake.

#### CrktMan

Joined Nov 29, 2005
34
Originally posted by JoeJester@Dec 19 2005, 06:12 PM
Don't see anything wrong ... it tracks well with the attached ...
[post=12571]Quoted post[/post]​
Hi Joe, when you do circuit simulation do you actually ask the computer to do the calculations and computer does the analysis and draws the circuit diagram with results?

#### JoeJester

Joined Apr 26, 2005
4,360
I construct the circuit and then select whatever analysis I need. In this example, I added the current meters and used a DC analysis. I then pasted the circuits to adobe photoshop and created the png file from there.

I followed your math and there were no obvious errors in your thinking. I went that extra step to see how a 3905 transistor would perform in your circuit. Your thinking and the simulation were congruent.

On a side note:

To me the thevnin model was reducing the voltage divider connected to the emitter and not including the b-e junction. If one reduced the whole circuit to a thevnin model, but that defeats the purpose of seeing if the student understood the interaction happening in the circuit. Yes, one could win the arguement with the reduction, but one must realize what the professor or instructor was seeking in the problem.

#### JoeJester

Joined Apr 26, 2005
4,360
The current through the emitter-ground resistor is 0.7/1k=0.7mA, that would make the Ie=4.3-0.7=3.36mA.
N93 my friend ... you might want to recheck that Ie.

#### CrktMan

Joined Nov 29, 2005
34
Originally posted by n9352527@Dec 19 2005, 07:54 PM
Well... bear with me, if you consider Ve=0.7V, the current through the emitter resistor would be 4.3/1k=4.3mA. This current is the sum of the currents flowing through the resistor across the emitter to ground and the Ie. The current through the emitter-ground resistor is 0.7/1k=0.7mA, that would make the Ie=4.3-0.7=3.36mA. Ic can be obtained from the formula you have used Ic=3.36*100/101=3.33mA. Ib=3.33/100=33.30uA.

It does give a slightly lower Ie and Ic.

Opss, I've just noticed I wrote higher in my previous post, my mistake.
[post=12579]Quoted post[/post]​
It appears that n9352527 you've got a very argumentive point here. In fact, in order to apply Thevenin theorem we would have to consider only INDEPENDENT voltage sources, as required by the theorem. So the Vbe should be out of the Thevenin analysis here; that's what I'm thinking here.

Joined Jan 19, 2004
220
i guess for thevenins theorem u can hav both independent and dependent sources as along as there is no coupling between the load circuit and source circuit..by coupling i mean that the dependent voltage source variable must be in the same network(load or source as the case may be)

#### n9352527

Joined Oct 14, 2005
1,198
Originally posted by JoeJester+Dec 20 2005, 03:34 AM--><div class='quotetop'>QUOTE(JoeJester @ Dec 20 2005, 03:34 AM)</div><div class='quotemain'>N93 my friend ... you might want to recheck that Ie.
[post=12585]Quoted post[/post]​
[/b]

Ah.. right, I've made an error in the calculation. My apology.

Changing the numbers gives the correct results:

... Ve=0.7V, the current through the emitter resistor would be 4.3/1k=4.3mA. This current is the sum of the currents flowing through the resistor across the emitter to ground and the Ie. The current through the emitter-ground resistor is 0.7/1k=0.7mA, that would make the Ie=4.3-0.7=3.6mA. Ic can be obtained from the formula you have used Ic=3.6*100/101=3.56mA. Ib=3.56/100=35.60uA.

But the numbers are still not in agreement with the initial CrktMan numbers. Checking the math again, I think CrktMan also made a calculation error, which made me doubt his Thevenin reduction in the first place.

<!--QuoteBegin-CrktMan
@Dec 19 2005, 07:28 PM
So, Ie = (2.5-0.7)/0.5 = 3.4 mA and Ve = 0.7 volt
Therefore, Ic = αIe = β/(1+ β)*Ie = 0.99*3.4 = 3.36 mA.
[post=12585]Quoted post[/post]​
[/quote]

Should be Ie=(2.5-0.7)/0.5=3.6mA and Ic=3.56mA

My apology once again and thanks for pointing the error Joe.

#### CrktMan

Joined Nov 29, 2005
34
Originally posted by n9352527@Dec 20 2005, 06:06 AM
Ah.. right, I've made an error in the calculation. My apology.

Changing the numbers gives the correct results:

... Ve=0.7V, the current through the emitter resistor would be 4.3/1k=4.3mA. This current is the sum of the currents flowing through the resistor across the emitter to ground and the Ie. The current through the emitter-ground resistor is 0.7/1k=0.7mA, that would make the Ie=4.3-0.7=3.6mA. Ic can be obtained from the formula you have used Ic=3.6*100/101=3.56mA. Ib=3.56/100=35.60uA.

But the numbers are still not in agreement with the initial CrktMan numbers. Checking the math again, I think CrktMan also made a calculation error, which made me doubt his Thevenin reduction in the first place.
Should be Ie=(2.5-0.7)/0.5=3.6mA and Ic=3.56mA

My apology once again and thanks for pointing the error Joe.
[post=12590]Quoted post[/post]​
Yes you are right, I had errors in the calculations. After all it all looks to be in the good agreement!

#### JoeJester

Joined Apr 26, 2005
4,360
Everyone makes a basic aw $hit now and then. You recognize it when the first thought in your mind is ... aw$hit.