# Amplifiers

Discussion in 'Electronics Homework Help' started by Mbetts, Feb 24, 2013.

1. ### Mbetts

4
0
Feb 22, 2013
Hi all, In this question I had already received the answers but have no idea how to get to there in the first place, I'll take you through what I've decided on so far.

This is the layout I get from the question.
transducer input amplifier output
--- 200R --- 10KR --- [> hfe 40 --- 1K ---

Input resistance of 10kΩ
Output resistance of 1kΩ
Gain 40
Transfucer voltage 1mV/oC
Internal resistance of 200Ω
Room temp 20oC

To come out with the answer of 784mV.

If someone could guide me towards the answer I think I could do it but I don't know of which equation to use, couldn't grab my lecturer in the tutorial because he is always too busy so feeling a little lost.

Thanks

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2. ### duke37

5,364
771
Jan 9, 2011
1. What is the transducer output at room temperature?
2. What will the transducer out be when loaded with the amplifier. A 200 and 10k voltage divider.?
3. What will be the amplifier output when this is amplified?

Do the same maths for the second amplifier.

3. ### Mbetts

4
0
Feb 22, 2013
1. 20 x 1mv = 20mV

2. From what I understand VDR VR1= R1/ (R1+R2)*Vs.
So you're saying 200/(10k+200)*20mV the maths doesn't add up?

3. I guessing this is the answer to 2 by 40 (the gain)?

All help is massively appreciated!

4. ### Harald KappModeratorModerator

11,444
2,628
Nov 17, 2011
It does add up. That is the effective input voltage to the amplifier. Now multiply by the gain and get the output voltage for the single amplifier case.

As Duke said:
Consider the output impedance of the first amplifier and the input impedance of the second amplifier. They too form a voltage divider. This gives the effective input voltage to the second amplifier. And so on.

To calculate the saturation effect, set the output of the second amplifier to 36V and do the whole math backwards.

5. ### duke37

5,364
771
Jan 9, 2011
1. OK
2. You have the divider upside down. The source resistance is lower than the load resistance so the voltage will only be dropped a little.
The equation becomes 10k/(10k+200)*20mV.
3. As you say.
4. As Harald says for the second amp etc.

It is coming along nicely.

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