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Need some help please with my circuit

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sinusoidal signal with peak-to-peak amplitude 0.2V, frequency 5kHz and offset 0.05V, and two outputs:non inverting opamp-schematic.png

A sinusoidal signal at the same frequency as the input, peak-to-peak amplitude 2.2V and offset 0.55V.
A sinusoidal signal at the same frequency as the output, peak-to-peak amplitude 1.14V, and no offset

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Message 1 of 12
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Solution
Accepted by topic author adi80

Hello,

 

your supplies should be VDC sources instead of sine waves. As they are, the opamp is supplied with 5V, 5kHz sine waves. Replace V1 and V2 with 5V batteries.

 

Also, your non-inverting amp gain is (1+47/10)=5.7. With an 5V input signal this will saturate your opamp and you will not see the correct output. Try with 100mV input signal in V3 with some offset if you want.

Message 2 of 12
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Thank you very much. Could you guide on these VDC source. I am not an electronic student. I am just trying to learn some basic about circuits. 

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Message 3 of 12
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You can find the VDC source (battery) in the Sources category (DC Voltage VCC). This provides a continuous voltage as required by the amplifier supplies. The polarities you used are correct, the opamp has symmetrical supplies referenced to GND.

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Thank you very much once again. For simulation do i need to add values or it will do it automatically?

 

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Message 5 of 12
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Solution
Accepted by topic author adi80

Yes, you need to specify the voltages. For an ideal opamp model these can be chosen. For real opamps the supplies must comply to the device datasheet.

Message 6 of 12
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Could you please help me with the calculation of the following. Thanks

A sinusoidal signal at the same frequency as the input, peak-to-peak amplitude 2.2V and offset 0.55V.
A sinusoidal signal at the same frequency as the output, peak-to-peak amplitude 1.14V, and no offset

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Message 7 of 12
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Hello,

 

you have a non-inverting amplifier configuration in which the gain is (1+Rf/Rg). Rf is the feedback resistor (47k) and Rg is the gain setting resistor (10k). The third resistor is simply shorted and has no effect. Your gain is then (1+47k/10k)=5.7.

 

The circuit amplifies both the signal and the offset. Therefore:

 

1. For Vin=0.55V+2.2V/2*cos(2*pi*freq*t) you get Vout=0.55*5.7+2.2V/2*5.7*cos(2*pi*freq*t)=3.135V+6.27*cos(2*pi*freq*t). This means that your DC voltage at the amplifier output is 3.135V and you will have a +-6.27V variation around this DC value. Your output sine wave will change between -3.135V and 9.405V with the selected frequency. Beware that the instantaneous output voltage should be between the supplies at all times. Otherwise the amplifier output is simply clipped at the supply that is saturated.

 

2. For the no-offset version you simply remove the DC component and redo the calculation.

 

Also, you should choose your frequency to be sufficiently low so that you are within the closed loop bandwidth of the amplifier.

Message 8 of 12
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Thank you very much for your guidance. I think i missed the first line to mentioned before.

input a sinusoidal signal with peak-to-peak amplitude 0.2V, frequency 5kHz and offset 0.05V, and two outputs.

Could you please guide me further ? How can i design this circuit which will two Vout as well.

Kind Regards,

A

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Message 9 of 12
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Could you please also give me the calculation for Non-inverting with no offset please.

Thanks,

A

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Message 10 of 12
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