# Transistor Amplifier gain

1. Feb 8, 2014

### rms5643

1. The problem statement, all variables and given/known data
A typical transistor amplifier is shown in the Figure. Find the amplifier gain G.
Circuit diagram: http://i.imgur.com/Rzv4JR7.png

Vs = 280mV
R1 = 100Ω
R2 = 7kΩ
R3 = 100Ω
Vd = 4*10^5 Ib (Ib is current through R3)
R4 = 7kΩ
R5 = 900Ω
Vo = ? (Voltage drop of R5)

2. Relevant equations
Ohms Law, current division, voltage division

3. The attempt at a solution
1) To find the current through R3, I have to find the voltage drop first. So, I wanted to use current division from the start, but I don't know the current coming out of Vs, the 280mV source. To do so, I combined R2 & R3 which are in parallel: 1/(1/7000+1/100)=98.59Ω

2) Combine my Req from step 1 with the R1, which are in series: 98.59Ω + 100Ω = 198.59Ω

3) Now I have 1 resistor in a loop with my voltage source. Applying Ohm's law: .28/198.59 = 1.4mA, which is the current coming out of the source.

4) Now I can apply current division to find the current Ib in the figure: 7000/(7000+100)*0.0014=1.38mA

5) Since Vd is defined as 4*10^5*Ib(0.00138)= 555.6v (<-- My first indication that something went wrong)

6) Applying voltage division on R5, the 900Ω resistor: 900/(900+7000)*555.6 = Vo = 63.29v. After dividing the Vo and Vs, I get a 226.05 gain, which is incorrect. Where did I go wrong?

2. Feb 8, 2014

### CWatters

As a check...

To work out Ib you can ignore the 7k resistor because the 100R in parallel with it will dominate. So Ib is approximately 280/(100+100) = 1.4mA

The rest seems ok to me.

1.4 * 4*10^5 = 560

Vo = 560* 900/(900+7000) = 63 approx.

Gain = 63/0.28 = 225 approx.

What was the correct answer given?

3. Feb 8, 2014

### rms5643

Hello CWatters, thanks for your help! Since this is a homework question, the correct answer is not given. I put in 226.05 as my answer for the amplifier gain (The answer has a +/-2% tolerance), and the website told me that answer was incorrect.

I think my calculation for Ib is incorrect as it seems really strange that a dependent voltage source would be producing ~556V when the source voltage is something as small as 280mV.

4. Feb 10, 2014

### CWatters

Yes its unlikely but then a current gain of 4*10^5 or 400,000 is also unlikely unless it's a multi stage amplifier, something like an opamp in open loop mode (eg no feedback).

5. Feb 10, 2014

### Jony130

This exercise is very strange.
First of all the BJT transistor don't behavior as a current controlled voltage source (CCVS). BJT is not a transresistance amplifier. And this 4*10^5 has a unit of OHM Vou/Ib = 400kOhm.

From witch site or book you have this question ?

6. Feb 10, 2014

### Staff: Mentor

If you convert that Thevenin into a Norton, it's a CE amplifier with hfe = 57 and a collector resistor of 7k, driving an external load of 900 Ohms. Sounds okay.

7. Feb 11, 2014

### Staff: Mentor

There should be a negative sign associated with the gain, for this amplifier.

8. Feb 11, 2014

### CWatters

Well spotted that man.

9. Feb 11, 2014

### rude man

One of the delights of growing old is to watch how standard circuit symbols change all the time. I have been faked out by this many times already on this site.

The latest rule seems to be: if it has + and - signs, it's a voltage source. If it has a diamond shape it's a controlled source. I think if it's not diamond shaped then it's round. If it's an ac source there is a sine wave symbol inside the round symbol but unfortunately phasing (0 or 180) is no longer defined. (I guess, if you have to ask, you don't belong in the course ...). If it has an arrow in it it's a current source. If it has an arrow AND + and - signs then the arrow predominates. NO RECTANGULAR SHAPES ALLOWED! Got all that? :tongue:

In my day we had round shapes for voltage and square for current, and controlled sources were clearly annotated next to those two shapes, avoiding the need for exotic and confusing subsymbol terminology as exemplified here and on many other PF posts. And it's not the OPs' fault.

So the symbol annotated "4e5 Ib" means it's a current-controlled voltage source with value 4e5 x Ib volts.

Just another example of societal dumbing-down, like eliminating "rest mass" so now it's no longer E = mc2 but E = γmc2. Or writing the dot product of two vectors a and b as ab. Go figure.

10. Feb 11, 2014

### rude man

Yes, but you can combine the current-controlled votage source with the 7K resistor to make a current-controlled current source. Don't ask me why they presented the equivalent circuit in such an oddball way.

11. Feb 12, 2014

### CWatters

How does that quote go.... The nice thing about standards is that there are so many to choose from :-)