How to calculate the input resistance for the following?

Discussion in 'Homework Help' started by thenzengineer, Apr 12, 2008.

1. thenzengineer Thread Starter New Member

Aug 28, 2006
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An amplifier outputs 3.6Vpp into 470 Ohms load when connected to a 420mVpp input signal. When the same signal is feed via a 220 kOhms resistor to the input the output voltage is 2.8Vpp. Calculate the input resistance of the amplifier?

I have done a similar practical exercise in finding the input resistance of a DUT by connecting a variable resistor in between the signal source and the DUT and halved the output using the variable resistor. Whatever value of variable resistance halved the output voltage was my input resistance.

2. hgmjr Retired Moderator

Jan 28, 2005
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219
Can you take a stab at describing how you think it should be tackled?

hgmjr

3. thenzengineer Thread Starter New Member

Aug 28, 2006
4
0
Taking a stab I calculate the value as 500kOhms approx. As that would make the output voltage to drop from 3.6Vpp to 1.8Vpp. I did it using ratios since I know adding a 220k dropped it by 0.8V and by adding a 480k it would drop by 1.6V and I want 1.8V which is half of 3.6Vpp

4. thenzengineer Thread Starter New Member

Aug 28, 2006
4
0
The other way I could think about is by forgetting the practical method and by using simple concepts -

Gain of the amp is 3.6/0.420 = 8.57

So when output is 2.8, input is 2.8/8.57 = 0.327 volts

I'm going to assume the 470 ohm resistor is small compared to the input impedance.

so the drop across the 220k is 420-327mv or 93mV
I = 93mV/220k = 0.423µA
Input R = 327mV/0.423µA = 773kohm

But this does not seem to add up with the practical method I was taught.

5. hgmjr Retired Moderator

Jan 28, 2005
9,029
219
You have the right idea about using ratios however you need to take another look at your calculations.

The amplifier's input resistance together with the series input resistance of 220K form a voltage divider. The amount of attenuation of the input signal introduced by this voltage divider is related to the ratio of the attenuation on the output with and without the series input resistor.

hgmjr

6. hgmjr Retired Moderator

Jan 28, 2005
9,029
219
The value of 773Kohms you have obtain, while not exactly the same as mine, is quite close to the one I got. The method you have used is a reasonable one.

hgmjr