Solving Amplifier Question: Find Vout for a,b,c

[peanutbutter]

Homework Statement

An amplifier has negative feedback applied to it such that feedback fraction is 0.5%. If the open loops gain of the amplifier is 10^5 what is the closed loop gain? I get this as 199.6

The amplifier is connected to a signal source of internal resistance 300kohms which provides a sinusoidal signal of 30mV amplitude, when not loaded. If the amplifier has an input resistance of 150kohms and an output resistance of 600ohms what is the amplitude of the voltage signal that would appear at the output terminals of the amplifier when it is
a) open circuit
b) connected to a load resistance of 8 ohms
c) connected to a load resistance of 600 ohms

Homework Equations

Gain of a non inverting amplifier = (R1+R2)/R2

Any help would be great. It's multiple choice and the options are parts a,b and c are;

0.003V, 0.03V, 0.3V, 1V, 2V

The Attempt at a Solution

a) open circuit, R1 = 0, R2 = infinity so gain is 1 ==> Vout = 0.03
b) Gain = (600+8)/8 = 76 => Vout is approximately 2
c) Gain = (600+600)/600 = 2 ==> Vout is 0.06 but there isn't an option for this. :S
Attempt
a) open circuit, R1 = 0, R2 = infinity so gain is 1 ==> Vout = 0.03
b) Gain = (600+8)/8 = 76 => Vout is approximately 2
c) Gain = (600+600)/600 = 2 ==> Vout is 0.06 but there isn't an option for this. :S

Could someone help?

 

[Jhero]

Hello,

Your calculations for parts a and b are correct. For part c, the gain should be (600 + 600)/600 = 2, but the output voltage should be 0.06V, not 0.6V. So the correct answer for part c would be 0.3V.

To understand why the output voltage is not 0.6V, we can use the voltage divider formula. The output voltage is equal to the input voltage multiplied by the gain. In this case, the input voltage is 30mV and the gain is 2, so the output voltage should be 60mV. However, when we connect the output to a load resistance of 600 ohms, this forms a voltage divider with the output resistance of 600 ohms. This means that the output voltage will be divided by 2, resulting in an output voltage of 30mV, which is 0.06V.

I hope this helps clarify the calculation for part c. Let me know if you have any other questions.