# non inverting amplifier

Feb 6, 2014
137
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?

#### Attachments

• NON INV.png
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• NON INV1.png
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#### Harald Kapp

##### Moderator
Moderator
Nov 17, 2011
12,884
R2=1.5m = 1.5milliΩ ! Therefore gain is ~1 and vout=vin (approx.)

LTSPICE requires R2=1.5Meg for 1.5MΩ

Feb 6, 2014
137
thanks for ur reply..but now it provides 3v at the output

#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,506
Several thinks to consider now:

1) What is the gain of the circuit?
2) Are there any limitations on the output of the circuit?

#### LvW

Apr 12, 2014
604
What do you expect for 2mV input and a gain of 1500?

Feb 6, 2014
137
no limitations..just i want to design the non inverting amplifier with the gain of 2.5 with 2mV input
but i did nt get in simulation.
i want to know the reason..that s it

#### Harald Kapp

##### Moderator
Moderator
Nov 17, 2011
12,884
How do you compute the gain for a non-inverting opamp configuration? Show us your calculation.
Certainly with the values you use you will not arrive at a gain of 2.5 but 1501.

Last edited:

Feb 6, 2014
137
How do you compute the gain for a non-inverting opamp configuration? Show us your calculation.
Certainly with the values you use you will not arrive at a gain of 2.5 but 1600.
sorry Harald kapp
it produces the correct output only
Vout=0.002*gain=0.002*(1+1.5*10^3)=3V.
thanks for ur replies

#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,506
And here was me thinking you wanted a 5V output and a non-inverting amplifier and perhaps a gain of 2.5 (or some combination thereof). Sorry I must have mis-read the original question.

#### Harald Kapp

##### Moderator
Moderator
Nov 17, 2011
12,884
Sorry I must have mis-read the original question.
Not completely, Steve, rest assured you're not insane. The op reads 5V (although at an input of 0.002V).

Feb 6, 2014
137
actually i need the 5v for 0.002v input
so i have to keep the gain as 2500
here i get the output in ltspice

#### Attachments

• n-i.png
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• n-i1.png
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#### Harald Kapp

##### Moderator
Moderator
Nov 17, 2011
12,884
You seem to work by trial and error. Which equations do you use? The standard equations for a non-inverting opamp will not give you 2.49Meg/1k for a gain of 2500.

Feb 6, 2014
137
Vout=0.002*gain=0.002*(1+2.49*10^3)=4.97V.

#### Harald Kapp

##### Moderator
Moderator
Nov 17, 2011
12,884
Right, but not what I meant.
What I'd like to know how you calculate gain from the resistors or how you need to calculate the resistors. Provided that gain is given as 2500 and assuming R1=1k, which value for R2 will set the gain at 2500?

Feb 6, 2014
137
i have calculate the gain like this..(1+2.5meg/1k)
is this not a right method?i don t know..if its wrong kindly explain me the right way to choose the resistors.

#### Harald Kapp

##### Moderator
Moderator
Nov 17, 2011
12,884
i have calculate the gain like this..(1+2.5meg/1k)
This is correct, so why do you use 2.49MΩ instead of  2.499MΩ in your simulation? No wonder gain is lower than expected.

Note that in a real circuit 2.5MΩ is unreasonably large for many (if not most) applications. Also a gain of 2500 will reduce the useable bandwidth of the opamp noticeably. In a real circuit it may be advantageous to split the gain using two separate gain stages (opamps) with a smaller gain for each gain such that gain1*gain2=gain_total=2500.

Last edited:

Feb 6, 2014
137
This is correct, so why do you use 2.49MΩ instead of  2.499MΩ in your simulation? No wonder gain is lower than expected.

Note that in a real circuit 2.5MΩ is unreasonably large for many (if not most) applications. Also a gain of 2500 will reduce the useable bandwidth of the opamp noticeably. In a real circuit it may be advantageous to split the gain using two separate gain stages (opamps) with a smaller gain for each gain such that gain1*gain2=gain_total=2500.
thanks for ur explanation
this means
may split the gain of 2 stages like this
(i.e)gain1=50
gain2=50
so two stages 50*50=2500
is this right?

#### Harald Kapp

##### Moderator
Moderator
Nov 17, 2011
12,884
My pocket calculator agrees

Feb 6, 2014
137
here have modified my circuit..please check this
and i know that GBWP=Av * f
this is theoretical explanation
practically how the gain affect the bandwidth of an opamp?
i have made the google search..let me know that there is any link to see this concept in CRO Display

#### Attachments

• 2 stage.png
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• 2 stage-1.png
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Last edited:

#### Harald Kapp

##### Moderator
Moderator
Nov 17, 2011
12,884
here have modified my circuit..please check this
Looks o.k.

OpAmp theory is well described here.
let me know that there is any link to see this concept in CRO Display
You need to perform a small signal analysis (.AC in SPICE) of the circuit to see the amplitude and phase vs. frequency.

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