# Calculate gain in BJT

Discussion in 'Homework Help' started by smrams28, Apr 29, 2013.

1. ### smrams28 Thread Starter New Member

Mar 29, 2013
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0
This is a pic of the BJT.
http://i44.tinypic.com/zkg9qv.jpg[/IMG]

For the amplifier circuit above, find a) RB and b) the gain (∆Vout/∆Vin) such that Vout = 8.3 V . Assume β = 155, RC = 8.6 kOhm, RE = 0.5 kOhm, VCC = 10 V, VBB = 1.02 V and VBE = 0.7 V.

I need to find the value of resistor Rb and the gain. For Rb, I thought I could just find the currents through Re and Rc and then use KCL to find the current through Rb. Then I could use that to find Rb but no luck.

Can someone help?

Last edited: Apr 30, 2013
2. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,990
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Show as your work. The method you use to find RB resistance.

3. ### WBahn Moderator

Mar 31, 2012
18,064
4,904
Give me a value of Rb and I can give you a value of VBB that results in exactly the same conditions at the transistor base (and at all points to the right of that).

4. ### WBahn Moderator

Mar 31, 2012
18,064
4,904
I thought about it some more and need to ask a clarifying question. When you say that you want to calculate the gain, what gain are you talking about? Vout over what? VBB? Normally a voltage with a double subscript is a bias voltage and not a signal source. But if you are treating it as a signal source, then the gain and the value of Rb are related.

5. ### Ron H AAC Fanatic!

Apr 14, 2005
7,050
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Furthermore, the value of Rb can only be calculated if there are some known numerical values in the circuit. I think the OP has omitted some information.

6. ### WBahn Moderator

Mar 31, 2012
18,064
4,904
A specific, numerical value of Rb certainly requires this. But it can be solved for in terms of the other resistances and the desired gain (and the transistor beta unless that is assumed to be sufficiently large). I can't tell from the OP whether that is what was being referred to or what.

7. ### Ron H AAC Fanatic!

Apr 14, 2005
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Well, he specifically said "value of", as opposed to "expression for". Perhaps that's just sloppy writing on his part.

8. ### WBahn Moderator

Mar 31, 2012
18,064
4,904
That's what is unclear. When I see "value of" I generally think of a numerical value. But then he described finding the currents in the other two resistors even though they also have no values, which would imply either that he is doing it symbolically or that he has values for those resistors.

I also see (and occasionally use) the term "value" being used generically to include symbolic expressions.

Hopefully he will clarify things.

9. ### smrams28 Thread Starter New Member

Mar 29, 2013
13
0
Sorry, I forgot to add a portion.

For the amplifier circuit above, find a) RB and b) the gain (∆Vout/∆Vin) such that Vout = 8.3 V . Assume β = 155, RC = 8.6 kOhm, RE = 0.5 kOhm, VCC = 10 V, VBB = 1.02 V and VBE = 0.7 V.

10. ### WBahn Moderator

Mar 31, 2012
18,064
4,904
Thanks. Now you just need to remember the other portion you forgot. Namely.... your work!

1) We won't work your homework for you. It is YOUR homework!

2) We need to see your best attempt so that we can get a peek into your thinking and where you are going okay and where you are going astray.

The help we can provide you is MUCH more effective if you will show your best efforts to date. Don't worry if it's wrong -- that's why you're here!

11. ### smrams28 Thread Starter New Member

Mar 29, 2013
13
0
ok, so here was my thought process

I know that (Vbb - Vbe)/Rb = IB
However, we only know the voltages and have two unknowns. I was thinking that if we got the currents from the two other sections, I could use KCL and solve it.

So I did
IC = VCC/RC = 0.001163 A
IC = VBE/RE = 0.00204 A

IB = 0.00204 - 0.001163 = 0.000877 A

(1.02 - 0.7)/0.000877 = 364 ohms

12. ### WBahn Moderator

Mar 31, 2012
18,064
4,904
You are already going astray. So let's get you back on track.

Ohm's Law related the voltage ACROSS a resistor to the current through THAT resistor. The voltage ACROSS a resistor is the voltage on one side of the resistor minus the voltage on the other side.

Here, you have Vbb, which is the voltage on one side of the resistor, but Vbe is NOT the voltage on the other side of the resistor -- it is the voltage ACROSS the base-emitter junction, which is something else entirely.

Work from the ground up ... literally. Start with the ground node at the bottom of the emitter resistor and work your way back to Vbb writing the voltages across the resistors along the way in terms of the currents in them.

And start using units in your work. That means tracking them through your work from start to finish and not just tacking them on at the end. Most mistakes you make will mess up the units letting you detect that an error has occured -- but only if the units are there to get messed up and only if you check them as you go.

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13. ### smrams28 Thread Starter New Member

Mar 29, 2013
13
0
Ok guys. I got the resistor value. It was around 172kohms.

Since gain is Vo/Vin, would the gain in this instance not just be 8.3/1.02?

14. ### WBahn Moderator

Mar 31, 2012
18,064
4,904
That's not what I got, and I can't begin to tell you where you went wrong (or perhaps I went wrong, certainly possible) because you won't show your work.

You have to make up your mind about which gain you are talking about. Is it Vo/Vin, as stated above, or ΔVout/ΔVin as stated in a prior post? They are NOT the same thing.