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Nano-amp amplifier - 1nA to 1V

by Wapochief
Tags: amplifier, nanoamp
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Wapochief
#1
Feb14-12, 03:10 PM
P: 28
Hi Guys and Gals,

I am trying to linearly amplify a 1nA-10nA current to 1V-10V. I need at least 10KHz bandwidth. I have tried using a 3 stage amplification (negative feedback) system using an lt1012 op-amp (1 nA to 1mv, then 100,10 gain), but the noise is very high (-30/40 db). I need at least -50 db.

I have built the circuit with surface mount components on PCB, with a ground plane and enclosed in a grounded metal box. I have voltage regulators, as well as capacitors to reduce supply voltage noise.

Another strange thing is that the gain increases (yes, increases!) with frequency, and I expected the gain to be flat and then roll off as frequency increased. The real world behavior of this circuit in no way matches any of the simulations I have done.

Is there any other way to amplify that small of a current with low noise? I have found transimpedance amplifiers, but they are for uA up.

Thanks,
Wapo
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yungman
#2
Feb14-12, 03:47 PM
P: 3,898
You have two problems. First, for 1nA to 10nA range, your amplifier has too much bias current at 100pA. Here is two that I find from a few minutes of checking. You might find much better. You want max of 1pA input bias current.

http://cds.linear.com/docs/Datasheet/1052fa.pdf

http://cds.linear.com/docs/Datasheet/146465f.pdf

Second You don't want to use too many stages as noise is dominated by the first stage. If you follow the first stage with a gain of 1000 in your case, you multiply the noise of the first stage by 1000. That's the reason you see a lot of noise. I designed a lot of transimpedance amp like this kind even much lower current. Go look up information in transimpedance amp. For 1V/nA, you need a transimpedance resistor of 1GΩ.

10KHz is not high frequency for this kind of amp. You can do the first stage using 100KΩ feedback resistor, then a gain of 10 on the next stage. For 10KHz, I personally would even try using a 1GΩ resistor and do it in one stage. Make sure you put a small cap across the said feedback resistor, do calculation so the C=1/[2∏Rf] where R is the feedback resistor and f is say a little above 10KHz.

This design will give 100mV/nA on the first stage. You'll find noise going to be a lot lower. If you need any more detail, post back. Don't be afraid to use big resistor. Only precaution is make sure you clean the board really well as a little dirt residue will have enough leakage to cause error. But reading 1nA is not pushing in any sense. If you need suggestion, I can even draw one or two circuit for you.
Wapochief
#3
Feb14-12, 06:32 PM
P: 28
Thanks for the reply. I agree with what you are saying, but I am worried that stray capacitance will limit the band width, especially with the 1Gohm resistor. Under simulation and according to your equation, a low value capacitor (a couple of pf) across the 1Gohm resistor will limit the band width to 1kHz. That is why I chose multiple stages with lower gain. Do you think that will be a problem?

yungman
#4
Feb14-12, 07:14 PM
P: 3,898
Nano-amp amplifier - 1nA to 1V

Quote Quote by Wapochief View Post
Thanks for the reply. I agree with what you are saying, but I am worried that stray capacitance will limit the band width, especially with the 1Gohm resistor. Under simulation and according to your equation, a low value capacitor (a couple of pf) across the 1Gohm resistor will limit the band width to 1kHz. That is why I chose multiple stages with lower gain. Do you think that will be a problem?
I don't know what is the input device. I did a lot of photo detector that has capacitance. This will cause a poll at the input and raise the noise and cause instability. You have to tell me what is the input to determine. Usually the small feedback cap is for lowering the noise. As I said, put a cap that give you a little over 10KHz. When making small cap, I use those wire wrap wires and solder on both side of the resistor, I then twist the two together to form a cap, twist tighter to get more cap and looser to get less!!!. Or you can get even smaller cap. Those day, I also use Johanson(spelling) adjustable cap that you can get sub pF adjustable caps.



BUT, I thought we are talking about 100K for first stage and gain of 10 at the second. This will already improve the noise drastically. My original suggestion is using two stages, but I put as much transimpedance gain as possible on the first stage. You don't need three stages, two stages are plenty. I only did two stages for measuring sub pA!!!!

Make sure you look for another amp, what you have don't cut it. the bias current is over 50 or 100pA, you introduce 5 to 10% error by default!!!
Wapochief
#5
Feb14-12, 09:39 PM
P: 28
Thanks again,
The input impedance is about 1Gohm/200pf. I need 1nA input, 1 Volt out. Thats a gain of 10^9. So I would need a 100M and a 10. (is that a typo above?)

Just ran some numbers and the lt1052/lt1464 wont cut it. The gain rolls off around 2KHz with a 1pf cap and 100Mohm. 0.5pf just gets 10Khz, but wont the board have more capacitance than that?

I can't seem to find a op-amp that has high enough bandwidth/gain and low input base current.
yungman
#6
Feb14-12, 11:14 PM
P: 3,898
Quote Quote by Wapochief View Post
Thanks again,
The input impedance is about 1Gohm/200pf. I need 1nA input, 1 Volt out. Thats a gain of 10^9. So I would need a 100M and a 10. (is that a typo above?)

Just ran some numbers and the lt1052/lt1464 wont cut it. The gain rolls off around 2KHz with a 1pf cap and 100Mohm. 0.5pf just gets 10Khz, but wont the board have more capacitance than that?
How do you get your numbers? I did the calculation for you below.
I can't seem to find a op-amp that has high enough bandwidth/gain and low input base current.
The amp has plenty of bandwidth. You are limited by the input capacitance. For two stage with first stage feedback resistor of 100K, the feedback cap that give zero frequency at 12KHz is C=1/[2πRf] = 132pF where R=`100KΩ and f =12KHz. So for 1GΩ feedback resistor, the cap is 1/10 which is 13pF. I don't know how you get 2pF.

The 1GΩ input resistance is not important. The 200pF input capacitance to the ground is the gating factor as when frequency going up, the reactance goes down and your closed loop gain of the first stage goes up as 100K divided by the reactance of the 200pF input capacitance. At 12KHz, the reactance of the 200pF is Z=1/[2πfC]=-j66.4KΩ. So your noise gain using a 132pF would be about the real number of (100-100j) divided by j66.4. It is not too bad. The total noise gain is 10 times as you have a gain of 10 following. That's the reason you try to avoid high gain in the following stage after the transimpedance amp.

If you use the 1GΩ feedback resistor and 13.2pF cap, the noise gain is 10 times less.

You should read up how to get the noise gain of transimpedance amp. I think you have plenty of BW using those amps. You might find a lower noise amp though as I only spent 2 minutes on DigiKey to look for it. The amp has never been a factor, it is the feedback cap that limit the BW. But you need the feedback cap to limit the noise gain as I explained in detail above. You might want to look for a lower voltage noise amp or even lower bias current op-amp. Your requirement is not high, you should be able to find something good.

It is a dance between BW vs acceptable noise. What I calculated for you is BW of 12KHz, it should be better than your requirement.
Wapochief
#7
Feb15-12, 07:34 AM
P: 28
1Gohm. = 10^9
100Kohm = 10^5
1Mohm = 10^6

I think you are mistaking Gohm for Mohm

Using 100Mohm I calculate a 0.132 pF cap at 12 KHz. I don't see how to control this small of a capacitance.

I have since read about how noise with amplifiers works - subsequent stages amplify the noise of the first stage, and the subsequent stages insert noise to some degree, but I still can't figure out how to get the gain and bandwidth I need with only 1 or 2 stages.

thanks for your help,
Wapo
yungman
#8
Feb15-12, 11:59 AM
P: 3,898
Quote Quote by Wapochief View Post
1Gohm. = 10^9
100Kohm = 10^5
1Mohm = 10^6

I think you are mistaking Gohm for Mohm
Excuse me, you are right.
For 100M follow by gain of 10, you use a 1.2pF and it will be close to my original calculation. The noise gain from calculation is a lot higher, but that's life. It is going to be a lot lower than your original 3 stages with post gain of 1000. With 100M tranimpedance, post gain is 10 so the noise is 100 times less. If you use 1G, you have to use 0.13pF. I think you might be able to find a Johanson tweek cap that go this low. Theory of the post before still stands, I just make the mistake from M to K!!! It's only 1:1000 error!!!
There's always the option of using wire wrap wire. It work like magic!!! I even use it on RF amplifiers!!!
Sorry.
Wapochief
#9
Feb15-12, 02:25 PM
P: 28
Still a decimal place off

Have not found 0.13 pf variable caps...so I guess the only option is to make a cap. About making the cap, does it look like this?
       Wire 1           Wire 2
        |                  |
        |                  |
--------+-----[====]-------+----
                R
Then twist the two wires together? Do you use 30 gauge wire?

Would that level of capacitance already be present on the board?
yungman
#10
Feb15-12, 03:07 PM
P: 3,898
Quote Quote by Wapochief View Post
Still a decimal place off

Have not found 0.13 pf variable caps...so I guess the only option is to make a cap. About making the cap, does it look like this?
       Wire 1           Wire 2
        |                  |
        |                  |
--------+-----[====]-------+----
                R
Then twist the two wires together? Do you use 30 gauge wire?

Would that level of capacitance already be present on the board?
Yes, again!!! I don't know what happened, maybe I was at the same time helping someone in solving grounding problem in a guitar amp and I was bouncing between here and there!!!! Hey compare to 1000:1, this is only 10:1, what's the big deal!!!!

Yes, that's the idea of the cap, I twist the two ends together, the tighter the turn, the more cap. Another way is to twist together, then use a cutter and cut off a little at a time to get the right response!!! I believe is 30 gauge common wire wrap wires. The capacitance on the board is not across the feedback resistor. You need the cap to control the noise gain and stability for the 200pF input cap.
Wapochief
#11
Feb15-12, 03:11 PM
P: 28
I got 0.13pf for 100Mohm.

What do you think of this op-amp?
http://www.ti.com/product/opa656

Low noise, typically 1-2 pA base current, better gain.
yungman
#12
Feb15-12, 03:21 PM
P: 3,898
Quote Quote by Wapochief View Post
I got 0.13pf for 100Mohm.

What do you think of this op-amp?
http://www.ti.com/product/opa656

Low noise, typically 1-2 pA base current, better gain.
Quick look and it look good. I mainly looking at the low frequency noise and is less than 100nV/√Hz @ 10Hz which is good, also 7nV/√Hz at higher frequency is very good. Be a little careful on the layout as this is 5MHz amp. 2pA is slightly high, but it's not the end of the world. I don't know the production number you are aiming, if it is small run, I would bend the -ve input up in the air and solder the stuff on it. This will prevent flux residue causing leakage. I don't think it's a good idea to use surface mount resistor for the 100M resistor, even 0805 is very short. Unless you can experiment with the surface mount resistor without the cap. Maybe the cap across the body is good enough as the feedback cap!!!! If you have to use surface mount resistor for this, cut a slot on the pcb right under the resistor to open the path of the potential leakage and is easier to clean the under side of the resistor. Flux residue is a big problem for us, we actually had an ultrasonic freon bath for cleaning!!!
Wapochief
#13
Feb15-12, 03:34 PM
P: 28
Thanks a bunch yungman. I will keep you posted.
yungman
#14
Feb15-12, 03:37 PM
P: 3,898
Oh!! I take back on what I said about the 100M first stage will have 100 times lower noise as yours. I based on the wrong calculation of the 100K to 100M and concluded the noise gain was low. But with the corrected noise gain. you are not going to see the noise lower by 100 times. It is going to be lower, but you need to do calculation to find out exactly the improvement. With my track record on numbers the last two days, you better do it than me!!!
Wapochief
#15
Feb15-12, 09:16 PM
P: 28
Can you point me to a good resource for learning exactly how to calculate the noise?
yungman
#16
Feb15-12, 11:21 PM
P: 3,898
Here are some articles that I pull up really quick:

http://www.analogzone.com/avt_1204.pdf

http://www.analog.com/static/importe...als/MT-050.pdf

The first one is more basic noise model of op-amp, the second is what I am talking about the noise gain due to the input capacitance that cause the noise gain to go up. Youer 200pF is causing the noise gain to go up. If you look back to my post when I made the mistake(yeh yeh yeh!!!) on the K to M, the noise gain calculation is quite low of like around 2, that's the reason I said the 100M first amp is going to be about 100 time better in noise than yours. Now that we know my mistake ( Again.....yeh yeh yeh!!!) the noise is very high. The advantage is going to be much less because in your case you first stage has a much larger feedback cap and the noise gain is very much reduced in you first stage. But the one with 100M first stage is going to be lower. But you have to really get down and calculate the numbers using the info from the articles as it get tedious.

As you can see, the voltage noise of the amplifier is very important and I think you choose one with very low noise this time so I even let it go on the 2pA input bias current. Also as the noise gain goes up, you need a higher speed op-amp because as you can see from the second article that the closed loop gain got bended up and intersect the open loop graph at much lower frequency. The opamp you chosen is quite fast.

Usually there are articles from op-amp manufacturers. I studied the old PMI application notes at the time.

Keep me inform on the result.
Wapochief
#17
Feb24-12, 01:48 PM
P: 28
Looks like the large feedback resistors will limit the noise floor...

The key noise equation for our system turns out to be the thermal noise of the resistor.

En_r = sqrt(4*k*temp*req*BW) = Vrms

V = 6 * V rms

Noise rms = (Rf/R1+1) * dominating noise source

k = boltzmans
temp = degrees in k
req = Rfeedback // Rsource
BW = 10kHz

I calculated 181uV rms of noise for the opa656 and a 100Mohm input and feedback resistor. This is about 1mV of noise or a -60dB (voltage) noise floor first stage, 10 mV or -40 dB after the second stage (gain of 10). I need to do better!!!
yungman
#18
Feb24-12, 02:03 PM
P: 3,898
Now you are using 100M and gain of 10. The only way to do better is 1GΩ my friend!!! Then you have to massage the BW and see what you can give. You are getting to the end of the road. The next is to do Nitrogen to lower the temperature as you find the thermal noise dominates!!! I don't know nitrogen, that's the end of the road for me too!!!

Sorry.


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