# BJT amplifier problem

Discussion in 'Homework Help' started by PsySc0rpi0n, Jul 12, 2015.

1. ### PsySc0rpi0n Thread Starter Well-Known Member

Mar 4, 2014
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Hello...

I'm trying to solve a new amplifier problem with a BJT.

The circuit is attached and also a screenshot.

I think I have already found:

$\displaystyle {R_{i}=\left [ \left ( R_{b1}\parallel R_{b2} \right ) + hie + \left ( hfe + 1 \right )\cdot \left ( R_{e} \parallel R_{Load}\right )\right ]}$

But now I'm struggling to find Vi. Can anyone give a hand?

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2. ### PsySc0rpi0n Thread Starter Well-Known Member

Mar 4, 2014
1,184
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Can I say that Vi = Ri*Ib knowing that Ib is the input current?

And consequently that
$\displaystyle {A_{v}=\frac{v_{o}}{v_{i}}=\frac{hfe\cdot \left ( R_{e} \parallel R_{Load}\right )}{\left [ \left ( R_{b1} \parallel R_{b2}\right )+hie + \left ( hfe+1 \right )\cdot \left ( R_{e} \parallel R_{Load}\right ) \right ]}}$

Last edited: Jul 12, 2015
3. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Your gain expression is wrong and I also do not like your expression for Ri . As for Vin or I should say Vb
Vin = Ib*hie + (β+1)*Ib*Re

Also your small-signal diagram are drawn horrible, non-intuitive at all.

Last edited: Jul 12, 2015
4. ### PsySc0rpi0n Thread Starter Well-Known Member

Mar 4, 2014
1,184
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Can you then explain or give a hint on how to find correct Ri and Vi?

What I did was to remember what I used to do with resistive circuits to find an equivalent circuit resistance, meaning that I have shorted all power supplies and replaced the current source by an open circuit. Then I wrote that expression for Ri based on what was left after the described procedure!

For Vi I just multiplied Ri by Ib.

The Small AC signal equivalent circuit, I just rotated BJT by 180º so that I would have an equivalent circuit that looked more like what I was used to see. The Hybrid model, was just follow the previous circuit. How could it be more intuitive? Maybe it helps me understanding your equation for Vi where you're just adding voltage drop at hie and the voltage drop at Re...

5. ### shteii01 AAC Fanatic!

Feb 19, 2010
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If I were you, I would get a textbook about transistors and start reading it.

6. ### PsySc0rpi0n Thread Starter Well-Known Member

Mar 4, 2014
1,184
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That was not of a big help. I have exam next friday and obviously I have no time to read an whole book about transistors or amplifiers!

Edited;

Jony130, for Vin, shouldn't we also consider R_Load? If not, then wouldn't the input impedance be always the same regardless the load we hook up there?

So, after re-analysing things I would say that Vi = hie*Ib + (β+1)*Ib*(Re//R_Load)

Last edited: Jul 12, 2015
7. ### Jony130 AAC Fanatic!

Feb 17, 2009
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The small-signal diagram should look like this

And from there we see that:
Vb = Vbe + Vo = Ib*hie + (Ib + hfe*Ib)*Re||RL = Ib*hie + Ib* (hfe +1)*Re||RL

As for Ri we see that RB1||RB2 is in parallel with hie+(hfe +1)*Re||RL not in series as in your expression.

Last edited: Jul 13, 2015
8. ### PsySc0rpi0n Thread Starter Well-Known Member

Mar 4, 2014
1,184
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Ah ok, then R_Load is now included in Vi. The previous expression of your's didn't.

And now your equations looks more like mine after re-analysing the circuit. At your Vi expression, I think there might be missing there a parenthesis to make Ib to multiply for the whole expression, right?

About Ri, I think you're right.

So, re-writing both expressions:

$\displaystyle {R_{i}=\left [ \left ( R_{b1} \parallel R_{b2} \right )\right \parallel \left ( hie + \left ( h_{FE} + 1 \right )\cdot \left ( R_{e}\parallel R_{Load} \right ) \right ) ]}$

$\displaystyle {V_{i} = I_{b}\cdot \left ( hie+\left ( h_{FE} + 1 \right )\cdot \left ( R_{e} \parallel R_{Load}\right ) \right )}$

$\displaystyle {V_{o} = I_{b}\cdot \left ( h_{FE} +1 \right )\cdot \left ( R_{e} \parallel R_{Load}\right )}$

$\displaystyle {A_{v}=\frac{\left ( h_{FE}+1 \right )\cdot \left ( R_{e} \parallel R_{Load}\right )}{hie+\left ( h_{FE}+1 \right )\cdot \left ( R_{e} \parallel R_{Load}\right )}}$

$
\displaystyle {R_{o}=\frac{v_{x}}{i_{x}}\left\{\begin{matrix}
\\
V_{sig} = 0\\
\end{matrix}\right.=\frac{i_{x}\cdot R_{e}}{i_{x}}=R_{e}}
$

Last edited: Jul 13, 2015
9. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,936
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Yes, I forget about Ib in the second part of a equation.
But notice that yours Vi is in fact Vb in my diagram. So Vo/Vi = Vo/Vb which is not equal to Vo/Vsig.

10. ### PsySc0rpi0n Thread Starter Well-Known Member

Mar 4, 2014
1,184
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I've edited my previous post.

I don't know if I got your point about Vo/Vi = Vo/Vb and not being the same as Vo/Vsig.

Can you please rephrase or explain differently?

What you called Vb I think it's my Vi and it's Vi (or your Vb) that I'm looking for for my voltage gain.

11. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,936
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Last edited: Jul 14, 2015
12. ### PsySc0rpi0n Thread Starter Well-Known Member

Mar 4, 2014
1,184
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Ok, if I think about that node at the top of Re, I can see Ix going in, Ib going out, hfe*Ib going out and Ie also going out.

Ix = hfe*Ib + Ib + Ie
Ix = Ib*(hfe + 1 + (hfe + 1)) = 2*Ib*(hfe + 1) ????? I don't like this at all!

Edited;

If I go on with this:

Ro = vx/ix
Ro = Ib*(hFE + 1)*Re/(2*Ib*(hFE + 1)) = Re/(hFE + 1)

Never seen this result before and don't like it!

Last edited: Jul 13, 2015
13. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,936
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First try simplify the diagram.
First notice that Re is in parallel with Vx and Rsig||(Rb1||Rb2) is in series with hie. Do you see this?
So we have this circuit

Ix = Ib + Ic = Ib + β*Ib = Ib * (β + 1)
And
Rx = hie +Rsig||(Rb1||Rb2)

14. ### PsySc0rpi0n Thread Starter Well-Known Member

Mar 4, 2014
1,184
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I can see what you said about the cirucit resistors in parallel and in series.
But then why did you omitted the rest of the circuit on the above diagram? Also that Ix = Ie and Rx = Re. I'm not getting the point of this!

I can't also see why is Rb1//Rb2 in parallel with hie and the rest, and not Rb1//Rb2 in series with hie and then, in parallel with Re//R_Load

Last edited: Jul 13, 2015
15. ### ericgibbs Senior Member

Jan 29, 2010
2,413
369
hi.
Jony has not omitted the rest of the circuit, he has shown the equivalent total resistive value of the circuit shown in this image as Rx , in order to simplify the circuit.

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16. ### PsySc0rpi0n Thread Starter Well-Known Member

Mar 4, 2014
1,184
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Ok, but what about Re? Re is not included in his equation for Rx!

And why is Rb1//Rb2 in parallel with hie and the rest, and not Rb1//Rb2 in series with hie and then, in parallel with Re//R_Load?

I mean, why is

$\displaystyle {R_{i}=\left ( R_{b1}\parallel R_{b2} \right )\parallel \left [ hie+\left ( hfe+1 \right )\cdot \left ( R_{e}\parallel R_{Load} \right ) \right ]}$

and not

$\displaystyle {R_{i}=\left [ \left ( R_{b1}\parallel R_{b2} \right )+hie\right ]\parallel \left [ \left ( hfe+1 \right )\cdot \left ( R_{e}\parallel R_{Load} \right ) \right ]}$

Last edited: Jul 13, 2015
17. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,936
1,088
Maybe my latest posts have been unclear for you. I'm sorry for that.

You want to find the output resistance for a CC amplifier. And this is why I give you this small-signal circuit.
View attachment 88448
So you can find output resistance Ro = Vx/Ix. But this circuit is hard to analysis by hand.
So I show you a simplified diagram

And I omitted Re because Re is in parallel with Vx so we can include Re later in our calculations.
And as I said earlier the Rx resistance is a equivalent resistance of this part of a circuit:

Rx = hie + Rsig||(Rb1||Rb2)

So now you can use this circuit:

View attachment 88460

And all you need to do know is to find Rt = Vx/Ix and after you find Rt we can finally find Ro by adding Re, Rout = Rt||Re. Can you do that ?? And we doing all of this to make the math easier
Please do not mix input resistance with a output resistance.

Input resistance is equal to

$\displaystyle {R_{i}=\left ( R_{b1}\parallel R_{b2} \right )\parallel \left [ hie+\left ( hfe+1 \right )\cdot \left ( R_{e}\parallel R_{Load} \right ) \right ]}$

Last edited: Jul 14, 2015
18. ### PsySc0rpi0n Thread Starter Well-Known Member

Mar 4, 2014
1,184
3
Ok. I think I can say that

$\displaystyle {I_{x}=I_{b} \cdot \left ( hfe + 1 \right )}$

$\displaystyle {R_{x}=hie + R_{sig}\parallel \left ( R_{b1}\parallel R_{b2} \right )}$

But isn't Vx = Ix*Rx???

So, if I divide Vx by Ix I will end up with Rx, right?

And about Ri, my question is why isn't Ri the equation I said, but the one you said. There is a difference where I say that Ri is the Rb1//Rb2 in series with hie, and this in parallel with Re//R_load and you say that it's Rb1//Rb2 in parallel with hie in series with Re//R_Load. Why is

Last edited: Jul 14, 2015
19. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,936
1,088
No, please do the math first. And remember that Ix = Ib*(β+1)

I don't understand your question. But If you talking about input resistance we clearly see that Is current is spiting into I_RB and Ib . The Ib current is also flowing through Re||RL plus Ic of course. And this is why Rb1||RB2 is in parallel with (hie + Re||RL). If you don't see it, please do the math.

20. ### PsySc0rpi0n Thread Starter Well-Known Member

Mar 4, 2014
1,184
3
Ok,

Ix=Ib*(β+1)

Rx = hie + Rsig||(Rb1||Rb2)

Vx - I'm not sure what to say about Vx.

But if Vx = Rx*Ix, then Rt = (Rx*Ix)/Ix = Rx