Non inverting amplifier

[PRIYADHARSHINI PALANISAMY]

Hi all,

this is the basic concept..i have designed the circuit to amplify the 0.002v to 5 v..
according to the formula for non inverting amplifier the Rf and Rin values are 1.5M ohms
and 1K..to produce 5v as output..
i have simulated this circuit in Ltspice.
but i get the waveforms like this.what is the reason for this?
can u suggest me where i did the mistake?


thanks in advance

 

[chuckey]

This IC has a gain/bandwidth product of 80 MHZ, so at a gain of 2500, it will have a bandwidth of 80 KHZ/2.5 ~ 32 KHZ. Reduce your time base speed on the CRO display to get a meaningful result. use two in series each with a gain of 50, so overall gain is the same but the bandwidth will be 1.3 MHZ of each, so the overall bandwidth will be more like, 500 KHZ.
Frank

 

[LvW]

1.) Your feedback resistor has a value of 1.5 milliohms. (Use 1E6 instead)
2.) What is the purpose of V2?

 

[PRIYADHARSHINI PALANISAMY]

1.) Your feedback resistor has a value of 1.5 milliohms. (Use 1E6 instead)
2.) What is the purpose of V2?

thanks for ur reply
v2 is the voltage which is produced by the sensor.
i have changed the value of R2 as 1.5meg
but it produces 3v at the output side

 

[amayilsamy]

Make your amplifier gain is 1000 to 1500, you can get output voltage by 2 to 3 V.

After amplifying the sensor signal compare or feed into the micro controllers above case enough.

 

[LvW]

thanks for ur reply
v2 is the voltage which is produced by the sensor.
i have changed the value of R2 as 1.5meg
but it produces 3v at the output side

And what is the problem?
Vout=0.002*gain=0.002*(1+1.5*1E3)=3V.
Everything OK.

 

[PRIYADHARSHINI PALANISAMY]

And what is the problem?
Vout=0.002*gain=0.002*(1+1.5*1E3)=3V.
Everything OK.

oh sorry.
its my blender mistake...its correct
thanks for ur reply

- - - Updated - - -

i kept the gain as 2500 to get the 5v as output
v out=0.002*(1+2.49*10^(3))=5v output
thanks for ur replies

 

[amayilsamy]

i kept the gain as 2500 to get the 5v as output
v out=0.002*(1+2.49*10^(3))=5v output
thanks for ur replies

You can't get 5V output from amplifier because op amp supply voltage is 5V. You can get maximum output voltage 3.75 V to 4V

 

[Audioguru]

You can't get 5V output from amplifier because op amp supply voltage is 5V. You can get maximum output voltage 3.75 V to 4V

No. Please look at the datasheet for the LT1800 rail-to-rail opamp.
With a 5.0V supply and no load its output can swing from 0.02V to 4.98V. Its negative feedback resistor is its load which is such a high value that it will not reduce the maximum output voltage.

 

[PRIYADHARSHINI PALANISAMY]

No. Please look at the datasheet for the LT1800 rail-to-rail opamp.
With a 5.0V supply and no load its output can swing from 0.02V to 4.98V. Its negative feedback resistor is its load which is such a high value that it will not reduce the maximum output voltage.

what he said is right..its a rail to rail opamp..so it will give the output 4.976v

 

[amayilsamy]

No. Please look at the datasheet for the LT1800 rail-to-rail opamp.
With a 5.0V supply and no load its output can swing from 0.02V to 4.98V. Its negative feedback resistor is its load which is such a high value that it will not reduce the maximum output voltage.

Yes Agree with Mr. Audioguru...