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Op Amp - what am I missing

sirch

Dec 6, 2012
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Been at this for days and can't see what is wrong, have read everything I can find, even got out the Art of Electronics...

I have set up the simplest of op-amp circuits with 1/4 of a TL084, I am supplying it with a 40kHz signal from a piezo transducer. I have a 2.2k resistor on the input and a 39k feedback resistor, so was expecting a gain of about 18. The attached image (input on the -ve pin of the op amp in green, output in yellow) shows that the output is about half a volt from a 2 volt input.

In this image the output is clipped because I have not biased the input. I have tried biasing the input inbetween +ve and gnd. I have tried a non-inverting set-up. I have tried a different chip and different op amps on the same chip, I have tried different supply voltages - I always get a lower output than input.

What am I missing????
 

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poor mystic

Apr 8, 2011
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:)
Like every current source, the piezo has an internal equivalent impedance. What is this impedance for your transducer?
 

sirch

Dec 6, 2012
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Well, I believe it's about 2200pf so assuming that 1/(2 Pi f c) applies to these devices I make that 1.8k.

But it's not obvious to me how that affects the op-amp gain?
 

duke37

Jan 9, 2011
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The inverting op-amp has the - terminal at virtual earth. The reason you are seeing anything here is that the amplifier is clipping so not providing feedback. You should have a dual supply to get it to work properly.

The amplifier has to work less hard if you use it as a follower with gain (input to + terminal).

I have not looked up the frequency response of the amp, Have you tried this on a simulator using this particular amp?
 

sirch

Dec 6, 2012
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Ok, just tried it with a virtual ground - i.e. 9v across two 1k resistors to get +/- 4.5v. it's no better. One thing I did notice is that it makes little difference if the op amp has power or not.

If I am reading the data sheet correctly (tl084.pdf) the TL084 does not start to drop off until around 200kHz.

Not tried it on a simulator - don't have one.
 

duke37

Jan 9, 2011
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I have looked at an op-amp with gain-bandwidth product of 3MHz and gain of 200k.
Inverting, input 2k2, FB 39k, using 5spice.

The frequency response seems to be adequate.
 

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duke37

Jan 9, 2011
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The virtual ground is the - terminal.
You would be better to use two 9V batteries with the + connected to the junction.
 

4algeria

Feb 13, 2013
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What I would suggest, is if you have a function generator, try and connect it, without dc offset, and see what you'd get.
Also I noticed that the phase between the input and the output is a bit odd.
 

gorgon

Jun 6, 2011
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From the datasheet I can't see that you can use rail to rail input, and certain not output. If your input is centered around 0V you exceed the negative input and run the opamp into whatever domain. You should start with a dual supply and use 0V as ground, including the output. The opamp is not made to output to the rails. The amplification is also too great, and the output will distort anyway, if the input is in the magnitude shown.
 

sirch

Dec 6, 2012
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Thanks for the input. Turns out it was down to a combination of a dodgy resistor and poor breadboard connections.

Now on the next issue, as can be seen from the attached the gain is now about 50 when I was expecting 17, not that I am complaining because at least it now amplifies...
 

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woodchips

Feb 8, 2013
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Never used one, but isn't a piezo sensor basically a capacitor? You have some voltage generating crystal with metalisation either side so when the crystal is deflected then it produces a voltage.

If so, then surely the normal amplifier gain of feedback over input resistances won't apply? The output is producing a voltage which flows through the feedback resistor to the virtual earth point at the inverting input. But then where does the current flow? It can't go through the piezo crystal because it is an infinite resistance, it has to go into the opamp and upset all the input bias levels. If you have a high impedance DVM, even a 10 Mohm scope probe might work, look at the voltage, waveform, at the inverting input, should be a DC level of course.

Bob
 

sirch

Dec 6, 2012
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Since I have a working solution using an inverting op-amp configuration, I thought I try a non-inverting variant, i.e. basically the same circuit but with the piezo on the +ve input just for my own edification. The result is ... it just doesn't work. I have tried connecting the piezo from the +ve rail to the op-amp, from the virtual ground to the op-amp, I've tried various combinations including a resistor in parallel to reduce the input impedance (at least that's what I think would happen?).

The thing is that as I understand it the difference between inverting and non-inverting is that in the latter the input impedance is very high, and so I was expecting the output of the piezo to be higher (less drain on it) however it seems like the effect is to just ground it.
 

duke37

Jan 9, 2011
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The non-inverting op-amp does not have a vertical ground.
The + input must be biassed somewhere between the power supply inputs. Take some DC measurements to make sure the op-amp is biassed properly.

Show us the circuit.
 

woodchips

Feb 8, 2013
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I still think what has been forgotten is that the opamp has DC currents flowing around it. Whilst the inverting input is a virtual earth it does not also follow that there is no current flowing PAST the pin. When the output is at, say, +5V at a particular time then a current flows through the feedback resistor, PAST the inverting input, and then through the input resistor to be sunk in the source. The ratio of output voltage to input voltage is of course the voltage gain, and if 5V at the output then with a gain of 100 then it will be 50mV at the source. Note that the current flowing is the same in both the feedback and input resistors.

So, with a piezo sensor then no current is going to flow into the sensor, so where does it go? As I said before the only place it can go is into the inverting input of the opamp, upsetting the bias voltages. This will go someway to explain the strange voltage waveforms. The piezo sensor needs a place for this current to flow without upsetting the opamp. Could be a resistor, transformer, or something more sophisticated.

Bob
 

sirch

Dec 6, 2012
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I'll attempt to draw up the circuit this weekend.

Woodchips - you may well be right. As I said I did try putting a resistor across the piezo between +ve and virtual ground I guess I could try piezo to -ve?
 

sirch

Dec 6, 2012
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Here's a rough circuit diagram for the non-inverting version
 

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Laplace

Apr 4, 2010
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The datasheet for the TL084 shows the input bias current to be up to 200 pA and up to 10 times that over the full operating temperature range. So where does that DC current flow in the '+' input circuit? I doubt the DC current will flow through the piezo crystal.
 

KrisBlueNZ

Sadly passed away in 2015
Nov 28, 2011
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Your last circuit has no DC path for the non-inverting input.

The design in your first post should be fine if you fix the bias on the non-inverting (+) input; connect it to a voltage divider made from two equal resistors (e.g. 10k each) connected between +V and 0V. The op-amp's power connections should also be to +V and 0V.

There's no need for a JFET-input op-amp. I would go for something more modern with a better GBP.

Also, I would use a higher supply voltage than 5V.

Edit: This one looks good and works fairly well down to 3V supply:
STMicro TSH22IN
http://www.digikey.com/product-detail/en/TSH22IN/497-7686-5-ND/1040434
 
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sirch

Dec 6, 2012
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Again, thanks for the input.

Kris - I had got the inverting version going the way you suggested, I was looking at getting a non-inverting version going just for my own education. I have a handful of TL084s but I may well try the op-amp you suggested as a comparison.

Anyway I have now sort of got that going too, by using the circuit in this document which uses much higher value (mega ohm) resistors and as has been suggested a path between +ve and ground
 
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