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Output Inverting Amplifier not as formulated.

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Member level 1
Jan 2, 2015
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I try to designing my inverting amplifier circuit using LT1360 and simulate using Multisim before plan to implement it for real circuit.
Let say my sinewave input is 500mV with frequency 2 MHz and I want my output around 10Vp-p.
So, by using the standard formula of inverting amplifier my gain should be around -20.
Thus, I decided to use resistor that feed through inverting input, R1=1k ohm and feedback resistor, Rf=20K to get the gain,A=-20.

However, the result shown was not exactly 10 Vp-p even in simulation and it looks like has a lit bit shifting.

So, I tried play around with the input signal. I noticed that, if the input signal is just in Volt (i.e 5Vp-p) then by using gain 2 (R1=1k ohm and Rf=2K) i got exactly 10Vp-p without shifthing.

Does it because of this ic cant catter for small input like mV or higher gain?How can I know it?Or is there any other factors effects my simulation result?

Also, I have modified the circuit to become 2 stage amplifier with the first stage is amplify to 2.5Vp-p and 2nd stage then I amplify to get around 10Vp-p, It still has a little shifting but much better if compared to the single stage amplifier.

Please help.thank you


The LT1360 has 50MHz gain bandwidth. Like the phrase states it is a product of gain x frequency (bandwidth).
So with a frequency of 2MHz the internal gain is only 25.
This is too close to your desired 20. (I should have recogniced this in the other thread).
It needs some headroom for regualtion. Therefore the reduced output voltage and the phase shift.

A two stage is an improvement. Gain5 and gain4 to get 20 is OK.

Or you use a much higher gain bandwidth OPAMP.


Hi klaus,
Thank you for the explanation.
This is what im thinking too.
But, im dont know wherether it was correct or not.
So, im glad that you explained it.
Thanks again?

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