EEVblog Electronics Community Forum
Electronics => Projects, Designs, and Technical Stuff => Topic started by: raff5184 on September 17, 2019, 12:19:04 pm
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Hi all,
I'm trying to design an amplification circuit using an OPA835.
My input signal is a 150kHz 0-5mV signal (or -5mV-0 depending on the configuration I use). I want a voltage gain of 600x (56dB), or the best I can have with single rail supply, powering the opamp with a 3.8V 500mAh battery.
Fig.17 of the datasheet tells me that at 150kHz my gain is around 40-45dB, at best. Let's assume that it is fine. I did a simple inverting configuration, and chose R_F=2kOhm as specified on page 31 of the datasheet and, consequently, R_G =3 Ohm.
But I don't see any amplification of AC signals on the output.
Is it ok to chose the two resistors the way I did?
Thank you
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Please post a schematic.
To use a single supply, the op-amp must be AC coupled with capacitors.
(https://www.eevblog.com/forum/beginners/tl072-microphone-preamp-(differential)/?action=dlattach;attach=388679;image)
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Why do you select an inverting amplifier? Since the frequency peaking is reduced when the Rf resistor is only 2k then the input resistor value must be very low for a high gain. Then the input source impedance must be extremely low.
Here is your amplifier:
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This is the circuit. I followed the TI manual http://www.ti.com/lit/an/sloa030a/sloa030a.pdf (http://www.ti.com/lit/an/sloa030a/sloa030a.pdf) Case 1 page 8.
The problem I have with this circuit is that it is already saturated and I measure a DC value of 3.7V
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Why do you select an inverting amplifier? Since the frequency peaking is reduced when the Rf resistor is only 2k
Simply because it is mostly recommended, but honestly I'm not sure when it is fine to use a non-inverting design. Is there a rule of thumb?
However, I used a non-inverting one.
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Why do you select an inverting amplifier? Since the frequency peaking is reduced when the Rf resistor is only 2k
Simply because it is mostly recommended, but honestly I'm not sure when it is fine to use a non-inverting design. Is there a rule of thumb?
However, I used a non-inverting one.
This is confusing; you first said you used an inverting configuration with 2k feedback and 3 ohm input resistor, but you now show a non-inverting configuration with 750k and 3 ohms. That gives a gain of 250001, which is clearly ridiculous.
Besides the gain problem, you have not correctly configured the op-amp for single rail operation as per Zero999's post.
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Yep, is the 750k value in your posted schematic what you used, or is it just a typo?
If you used that, well, obviously due to the input offset of the opamp, with such a gigantic gain, it can only saturate.
Another point: your biasing.
With the values shown on your schematic (given that your input signal is DC-coupled), the + input will "idle" (input voltage when your input signal is 0V) at 0.1V. With a gain of 600... you know what you'll get.
See above for correct way of biasing your opamp on a single supply. Adapt depending on whether you need your signal to be AC or DC-coupled.
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This is confusing; you first said you used an inverting configuration with 2k feedback and 3 ohm input resistor, but you now show a non-inverting configuration with 750k and 3 ohms. That gives a gain of 250001, which is clearly ridiculous.
Besides the gain problem, you have not correctly configured the op-amp for single rail operation as per Zero999's post.
Sorry for the confusion, I did a first INVERTING design with a R_F 2k and R_G 3. I thought this could be the issue, so I decided to use larger resistors.
Then in my second design, NON-INVERTING of which I attached the schematic, there is a typo, R_G is 1k, not 3.
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Ok, but this is still a biasing problem.
If you want your input signal to be DC-coupled, look at what Zero999 posted, and at Fig. 60 of the datasheet. If you do as in this Fig. 60, don't forget that your input voltage must have an offset equal to the reference (bias) voltage.
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Ok thanks,
I'm trying to fix the biasing.
Btw, I want the input to be AC-coupled. I don't need to amplify DC values, only AC
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For AC coupling, you will need a series capacitor and biasing the + input. (Note: the - path will also need proper biasing. Refer to Zero999''s examples again, or the datasheet example.)
Once the biasing is fixed, you will still have something else to consider: bandwidth. The bandwidth of an opamp is basically a function of the gain, more or less linear. This is the GBW product figure.
For such a high gain (x750), and a 150kHz sine input signal , you would need a GBW of at least 750x150e3 = 112.5 MHz! The OPA835 would not cut it.
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Split the total gain between two opamps. Then the gain of each opamp will be only 27.4 times. Then the high frequency response will be much higher.
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Ok thanks,
I'm trying to fix the biasing.
Btw, I want the input to be AC-coupled. I don't need to amplify DC values, only AC
The DC gain must be unity. Your circuit has a high gain at all frequencies, so the op-amp is trying to amplify Vref and saturating.
Add a capacitor between R_G and ground to ensure the gain is 1, at DC and R_G only enters the circuit, at the frequency of interest. R2 will keep the op-amp biased at half the supply and R1 should be replaced with a capacitor, to block DC.
As mentioned above, you need a op-amp with more bandwidth to get the gain you desire at the such a high frequency. Either use a faster op-amp or two stages in series.
What's the lowest frequency of interest?
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thank you everybody for the answers. It's working better, I can amplify AC signals
Add a capacitor between R_G and ground to ensure the gain is 1, at DC and R_G only enters the circuit, at the frequency of interest.
Done.
What's the lowest frequency of interest?
140 kHz
A couple of more questions:
R1 should be replaced with a capacitor, to block DC.
I only added the capacitor in series with R1 but didn't remove R1. Any thoughts on this?
Either use a faster op-amp or two stages in series.
I tried two stages in series, it works, but I have a problem, I see some noise in output at about 600kHz and 80mV and I have no idea where it is coming from. I have the same problem with 1 stage opamp. I tried to use an RC low pass filter in output to cut it but it's basically worse. Suggestions?
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The single stage circuit should look something like this, but as previously mentioned, it won't work up to 160kHz, without a much faster op-amp. R1 will form a potential divider with R2 and reduce the gain. If you don't need so much gain, then reduce the gain of the amplifier. Having a potential divider, to attenuate the signal before the amplifier makes no sense.
[attachimg=1]
As far as the two stage design is concerned: I suspect it's oscillating. Are the two separate amplifiers in series or have you tried to put two op-amps in the same loop? Please post a schematic.
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yes that's what the single stage looks like, thanks for confirming. Yes the gain problem is clear, I know I can't reach the 60dB. I just want to understand the circuit and the OpAmp now.
The two opamps are simply in series, not in the same loop.
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When you said in series, did you mean like this?
[attachimg=1]
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If the input signal is AC-coupled anyway, you may want to AC-couple the second stage as well, thus avoiding for it to amplify the output offset of the first stage (in case it could be significant).
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If the input signal is AC-coupled anyway, you may want to AC-couple the second stage as well, thus avoiding for it to amplify the output offset of the first stage (in case it could be significant).
Well no because:
1) You'd need an extra resistor between the non-inverting input and Vref, to bias the op-amp.
2) AC coupling the second stage is unnecessary, as the second stage has a unity gain as DC, so it won't amplify the offset voltage.
3) DC coupling the two stages gives the lowest part count.
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Well no because:
1) You'd need an extra resistor between the non-inverting input and Vref, to bias the op-amp.
Well, of course. You'd need to duplicate the first stage, biasing the input properly. Didn't mean to simply add a series capacitor.
2) AC coupling the second stage is unnecessary, as the second stage has a unity gain as DC, so it won't amplify the offset voltage.
Right, thanks to the capacitor between R_G and ground. I had another configuration in mind. So nevermind, with your schematic, it won't *amplify* the output offset of the first stage, as this is a gain of 1 at DC.
To nitpick though (even though again I actually had another topology in mind), there is still a unity gain at DC, and not zero gain. So the output offset of the first stage will still add up to the output offset of the second stage. Probably not a big deal here. But that's still not quite like AC-coupling between the first and second stage. :P
3) DC coupling the two stages gives the lowest part count.
That's for sure.
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When you said in series, did you mean like this?
(Attachment Link)
Yes this is the type of series. Of course the 2 Vcc's are the same. Vref is derived from Vcc throuugh a 2426 type of virtual ground chip.
However I tried the circuit with these values but it gives a saturated 3MHz output
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When you said in series, did you mean like this?
(Attachment Link)
Yes this is the type of series. Of course the 2 Vcc's are the same. Vref is derived from Vcc throuugh a 2426 type of virtual ground chip.
However I tried the circuit with these values but it gives a saturated 3MHz output
Do you mean that it's self-oscillating at 3MHz? If so, is the first stage oscillating, not saturating, and then the oscillation gets amplified by the second stage, which saturates?
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When you said in series, did you mean like this?
(Attachment Link)
Yes this is the type of series. Of course the 2 Vcc's are the same. Vref is derived from Vcc throuugh a 2426 type of virtual ground chip.
However I tried the circuit with these values but it gives a saturated 3MHz output
Since you are not showing the exact circuit, we can only guess!
For the moment, my guess is that, if you study a little more the datasheet, you will find what happens when using higher value resistors and what we can do to improve things (tip: put some capacitance in parallel with RF. Don't miss the reference to the loading of the amplifier (... these amplifiers are designed for a 2kΩ load...).
By the way, what is the input signal (shape) and how it is connected to the circuit?