Non-Inverting Amplifier (AC Coupled) : How to solve this circuit with lower and high frequency?

Thread Starter

Dragnovith

Joined Apr 16, 2021
16
I need to design this circuit with a gain of 26 for signals where the frequency varies from 800Hz to 8kHz and I also need to find out the current on the RL load. (VA = 100mV)
Will C1 behave like an open circuit in ac analysis?
I am getting to lower the frequency (800Hz) but I am not able to project to the frequency of 8kHz. Regarding the current I have a little doubt if it will be only 2.5 V divided by the RL.
circuitopamp.jpg
 

Papabravo

Joined Feb 24, 2006
16,773
It looks like you hooked up the power backwards &
The components don't seem to have values
For AC analysis use AC amplitude 1 so the output will be properly scaled in dB
 

Jony130

Joined Feb 17, 2009
5,243
All the capacitors you have now in your circuit will only influence the overall lower cut-off frequency. Because all the capacitors you have now form high pass filters only.
Thus the higher cut-off frequency will be determined by the opamp gain-bandwidth product (GBP) and the mid-frequency gain Av = (1 + R2/R1).
Thus the 3dB higher cut-off frequency is equal to Fh = GBP/Av
 

Thread Starter

Dragnovith

Joined Apr 16, 2021
16
All the capacitors you have now in your circuit will only influence the overall lower cut-off frequency. Because all the capacitors you have now form high pass filters only.
Thus the higher cut-off frequency will be determined by the opamp gain-bandwidth product (GBP) and the mid-frequency gain Av = (1 + R2/R1).
Thus the 3dB higher cut-off frequency is equal to Fh = GBP/Av
Okay, I was in doubt with the highest frequency. Thanks
 

crutschow

Joined Mar 14, 2008
27,697
You model has only a 208KHz gain-bandwidth so the -3dB rolloff is 208kHz divided by the circuit gain.
Also your amp power is still incorrectly connected, which you were already told about.
 
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