Operational Amplifier circuit

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Hello,

I'm currently working on an op-amp circuit.
According to my calculations, the circuit I built should give me a gain around 1000 (30dB). However, it seems that the input voltage is dropping a lot when I check the op amp's output.

I've tried the 741 and the LM358N and none worked for me. I have also tried changing the resistors values but I don't see a considerable change. What am I doing wrong?

Here are the schematics:


http://img708.imageshack.us/img708/3939/opamp.jpg

Thanks in advance!

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Quote:
it seems that the input voltage is dropping a lot
What is the output impedance of the generator? You only have about 1K input impedance for the opamp. (if it worked)

A voltage ratio of 1000 is 60dB. Also it's unnerving seeing the op amp upside down. How is the chip biased if you have the +input connected to ground?

John Samperi

Ampertronics Pty. Ltd.

www.ampertronics.com.au

* Electronic Design * Custom Products * Contract Assembly

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It looks like the 2nd oscilloscope probe's shield is connected to the measurement signal. It this right? Hopefully it is isolated, and not the reference for that probe.

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I saw that but it looks like it should go to the ground dot of input A.

Of course there is also no input cap. An output cap would also be nice.

John Samperi

Ampertronics Pty. Ltd.

www.ampertronics.com.au

* Electronic Design * Custom Products * Contract Assembly

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It looks like a single supply (if Vss is also grounded) arrangement, so with the + input at ground, disappointment will be in the forecast 8-)

Tom Pappano
Tulsa, Oklahoma

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Apart from the +ve input being grounded (it should be half of the single sided supply voltage), have you actually read the datasheets if the opamps work with 5V single sided supply? LM358 should work with single supply 5V, but I think TL741 not.

Also what kind of signal you are amplifying? As the 741 has an open loop gain of about 80dB at 10V supply, making it amplify 60dB in single go is a lot.

Both opamps have bandwidth of about 1MHz, which means you can amplify 60dB only for up to 1kHz signals and 10kHz signals can be amplified by 40dB etc..

And yes the resistor values for 1000x gain are correct in theory, but in practice 1kohm on input may be too small and 1Mohm on feedback may be too high.

I suggest reading an opamp book (almost any analog electronics book will do) or at least reading the wikipedia article on opamps.

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lcruz007 wrote:
Hello,
I've tried the 741 and the LM358N and none worked for me. I have also tried changing the resistors values but I don't see a considerable change. What am I doing wrong?

Your inverting amplifier, as its name states, tries to invert and amplify the presumed positive signal you feed in. Inverting, in this case, means a negative voltage at output, but you do not have any negative power supply, so your output will stay to lowest voltage supplied - ground.

A non-inverting amplifier will work with this power supply.

Dor

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Input signal and non-inverting OP pin must have 2.5V offset.

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js wrote:
A voltage ratio of 1000 is 60dB.

Oh right. why is it db = 10log (x^2)? why squared if it's just a ratio between two quantities?

tlucas wrote:
It looks like the 2nd oscilloscope probe's shield is connected to the measurement signal. It this right? Hopefully it is isolated, and not the reference for that probe.

That was a mistake in my schematics. I don't have it like that in the real world. Channel B is connected to ground, and is only used to compare the original signal and the output.

Jepael wrote:

And yes the resistor values for 1000x gain are correct in theory, but in practice 1kohm on input may be too small and 1Mohm on feedback may be too high.

Hmm I've tried different values but I don't see a considerable change, what values can you suggest me? :)

Thanks.

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lcruz007 wrote:
why is it db = 10log (x^2)? why squared if it's just a ratio between two quantities?
It's a power reference. The definition of dB, in this case, is
10*log(Pout/Pin)

Power is P = VI = V^2/R. Assuming an equal load, it turns out to be

10*log((Vout^2/R)/(Vin^2/R)) = 10*log(Vout^2/Vin^2) = 20*log(Vout/Vin)
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Until you address the power supply-signal level incompatibility, not much else matters 8-)

Tom Pappano
Tulsa, Oklahoma

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You cant use such a hi gain with a regular old opamp.... its the gain-bandwidth product... you wont have any bandwidth at hi gain... better to have 2 stages of 40db and 20db for example. Keep the hi gain in the 1st stage. 2nd stage amplifies the noise of the 1st stage.

Imagecraft compiler user

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lcruz007 wrote:
js wrote:
A voltage ratio of 1000 is 60dB.

Oh right. why is it db = 10log (x^2)? why squared if it's just a ratio between two quantities?

gain in dB = 20 * log (Vout/Vin)

Jepael wrote:

And yes the resistor values for 1000x gain are correct in theory, but in practice 1kohm on input may be too small and 1Mohm on feedback may be too high.

Hmm I've tried different values but I don't see a considerable change, what values can you suggest me? :)

Thanks.

As suggested above, that gain is almost impossible with your opamp. Use two gain stages of about 30dB (32*32=1024) each as suggested. That should also reduce your resistor values to more sensible ones, i.e. leave you with high input impedance and more stiff feedback.

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Actually the gain of an inverted amplifier is negative.
In this case -1000,its too large for a single stage amplifier.And however the circuit needs +vcc,0,-vee power supply.
If a non inverted amplified with single power supply used the non inverted input must dc voltage biased.
A commonly made error is that the non inverted input must biased in vcc/2.Thats not correct since is a dc amplifier and the output voltage will be (vcc/2)*gain.
The bias voltage should be equal to(VCC/2)/gain,so the output will be in VCC/2.

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ideal op-amps have no limitations, real ones do! :)

Charles Darwin, Lord Kelvin & Murphy are always lurking about!
Lee -.-
Riddle me this...How did the serpent move around before the fall?