# Amplifier (non inverting) question

1. Nov 11, 2007

### peanutbutter

1. The problem statement, all variables and given/known data
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

2. Relevant equations
Gain of a non inverting amplifier = (R1+R2)/2

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

2. Nov 12, 2007

### peanutbutter

Can anyone help?

3. Nov 12, 2007

### peanutbutter

1. The problem statement, all variables and given/known data
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

2. Relevant 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

3. 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?
I've shown my working.