# Current to voltage converter based on op-amp

Discussion in 'The Projects Forum' started by goodbyegti, Jun 18, 2008.

1. ### goodbyegti Thread Starter Well-Known Member

Apr 28, 2004
59
1
Here's my problem:

I'm planning to build the current to voltage converter below for measuring very low currents over a wide range (~1pA to ~1uA). The idea of the circuit is that initially Uout1 is used until ~3.5V develops across the limiter at which point resistor nR is bypassed and the second output Uout2 is then used for the higher current measurements.

The back to back zener diodes in the limiter have the following problems:

> High, non linear leakage current
> High capacitances
> Hard to switch off once conducting

I need to come up with a new diode network to remove resistor nR from the circuit with a threshold voltage of ~3.5V, low leakage current and capacitance.

Unfortunately I think this requires some very in-depth knowledge (which i don't have but am trying to learn!) and i'd be very grateful if any of you enthusiasts could point me in the right direction!

Thanks!

2. ### John Luciani Active Member

Apr 3, 2007
477
0
Measuring currents over six decades with a single circuit could be
difficult. I think you are better off with two circuits one for 1pA-1nA, another for
1nA to 1uA. Use a relay to switch in one or the other. For low leakage relays
take a look at Coto.

For the pA-nA I would use a capacitor to integrate the current
over a fixed time period (I = C dv/dt). For the nA-uA I would use
a resistor.

(* jcl *)

3. ### SgtWookie Expert

Jul 17, 2007
22,183
1,729
Instead of trying to use two op amps and switch between them based upon the 1st op amps' output, why not use a single op amp, and change the value of the feedback resistor using a window comparator that selects the next higher value feedback resistor when the Vout high threshold is exceeded, or the next lower value feedback resistor when the Vout falls below the low threshold?

4. ### goodbyegti Thread Starter Well-Known Member

Apr 28, 2004
59
1

The problem with using a relay to switch between two separate circuits is that during the switchover (i'm guessing this will be ~ms) the current will be undefined which in this particular case isn't acceptable. The same is true for switching the feedback resistor.

I suppose the other option is a logarithmic amplifier but then it's very hard to achieve an accurate calibration over the full range.

5. ### John Luciani Active Member

Apr 3, 2007
477
0
Place the two feedback resistors in series with the relay in parallel with
the larger resistor. You could also place a resistor and capacitor in series
with the relay across the capacitor. If the current reading is low open

The large values of resistors can be problematic. If possible I would use
a resistor and a capacitor.

(* jcl *)

6. ### goodbyegti Thread Starter Well-Known Member

Apr 28, 2004
59
1
Thanks John, I had a look at those Coto relays and they seem well suited, the only thing that worries me is the size as the circuit has to be as small as possible.

I understand your point about high value resistors, for nR i need ~0.1GOhm. The problem is the sampling frequency needs to be fairly high, i need a rate of about 5000Hz so I'm worried about introducing additional capacitance into the circuit.

I suppose the next problem is how to switch the relay when ~4V develops across nR without adding extra capacitance, i will have a think about this.

7. ### John Luciani Active Member

Apr 3, 2007
477
0
I believe that Coto makes some SMD relays (I am not sure of the leakage).

Coto relays were used in all of the analog switch and mux test fixtures when
I was at Harris Semiconductor (now Intersil).

At 100MOhm 1pA is going to give you 100uV. Measuring 100uV accurately
in 200uS could also be difficult. For the pA switch and mux leakage measurements
we would use a capacitor and integrate over a period of 10-100mS.

I have not looked at the state-of-the-art in analog switches but I wonder if the
relay could be replaced with a solid state switch. If the temperature of the
measurement circuit is stable you may be able to calibrate out the leakage.

(* jcl *)

8. ### goodbyegti Thread Starter Well-Known Member

Apr 28, 2004
59
1
Thanks again John, i'll follow up those relay ideas tomorrow. In worst case i can always reduce the sampling frequency to a few hundred Hz.

I've now simulated the above circuit in multiSIM and it's working as expected (save for trouble with the resistor switching!). I've also had an idea about modifying the existing zener diode setup which came from looking at state of the art ESD suppression devices with very low capacitance (for use with high speed signals etc.).

I thought i could use GaAs LED (extremely low reverse leakage), a zener diode and an ultra low capacitance diode (maybe a couple) to neutralise the relatively large capacitances of the other two diodes. A rough sketch is below. I would make connections at 5 and 6, hoping the the circuit would become conductive at some threshold around +/- few volts.

Does this sound insane?!

9. ### John Luciani Active Member

Apr 3, 2007
477
0
Your voltage limiter is in parallel with nR which is a large value. The leakage
may change the parallel resistance and affect your OUT1 measurements.
If your measurement circuit is at a constant temperature maybe you could
calibrate the error out.

Your measurements are slow enough that capacitance doesn't seem like
it would be an issue.

Your zener diode is shown with a Vz of 5V. The LED will add a few more volts.
Isn't this above your threshold design goal?

(* jcl *)

10. ### goodbyegti Thread Starter Well-Known Member

Apr 28, 2004
59
1
Yes, the biggest problem i have now is the leakage current from the limiter reducing the effective value of nR. With an input current of 10nA i get -1.4V from the first channel which is almost a factor of ten lower than expected.

As long as the limiter switches on before the first channel saturates (~12v) i don't think it's too critical exactly when it happens, but as you correctly point out there'll be a relatively large voltage drop across the LED. I should probably replace the zener with another LED.

I tried putting a parallel capacitor across the limiter in multiSIM. Using an AC input current of 1nA at 1kHz I get a peak to peak voltage output of 27mV using 10pF, 98mV using 1pF, 135mV using 0.1pF and 140mV without the capacitor. At lower frequencies it's not such a problem.

For interest I should mention the motivation for building the amplifer, it will be used for measuring the conductance of individual molecules, an entirely different can of worms .

11. ### John Luciani Active Member

Apr 3, 2007
477
0
Are you sure you need the voltage limiter?

If the output of OA1 is at the rails the only issue I see would be recovery time.
Your system is slow so it may not matter.

(* jcl *)

12. ### Andrew.Thompson New Member

Jun 27, 2008
1
0
Hi, I've only just found the forum today but as I was having a look about I found this thread and as it's something I've been working on for the last few years I though I'd chip in my two cents worth.

From practical experience of building these circuits I can say the two bigest problems you will encounter are related to noise and bandwidth in your measuring stage.

One of the design variants I've been working with is a simple transimpedance amplifier (TIA, it's too long to keep writing that) with 500M as it's feedback resistor. In a best case senario, once this is laid down on a PCB, I end up with a single stage which has a bandwidth of around 250Hz and an RMS noise fugure of around 125fA (peak to peak noise of 1.25pA).

Now, as the noise output of a TIA is proportional to one over the square root of the feedback resistor, this would mean, for a system with the same 250Hz bandwith as above but with a 100M feedback resistor your noise levels will be around 280fA RMS (or 2.8pA peak to peak). Furthermore, as the noise is very similar in structure to white noise, if you were to take as much of an increase in bandwith as you can get (assuming similar parrasitic capacitances slowing things down) you would end up with a system which has a bandwidth of around 560Hz and an RMS noise level of approximately 400fA (4pA peak to peak).

There are ways to reduce the noise levels on the measurement if these figures sound a little on the high side, the easiest way is to add more capacitance accross your feedback resistor to limit the bandwidth of your measurement circuit and hence the bandwidth of the noise that is passes. The second, and slightly more difficult method would be to move to some sort of cryogenic setup which you could use to reduce the thermal noise in the system.

However, if you look at the fact that with 1pA of current you would only be looking at measuring around 6,000 electroncs per sample (assuming a 1000Hz sample clock, related to the above circuits bandwidth limitations) which will likely mean that you are going to a have a lot of shott noise in your measurement in any case.

One further problem you may encounter when using switched feedback elements is that the charge injections you will have occuring each time you use the switches (relays would probably be worse than the better semiconductor switches, but even they are quite poor) will likely saturate you measuring system's lower range for several measurement cycles. So, unless you can accept large settling times each time you change the characteristics of the circuit you may have trouble here as well.

Some questions which would be useful to know the answers to if I can offer any more help are:

Do you need to be able to use the full range and full speed of the measuring system all of the time? (ie would it be possible to set up one slow, low noise amplifier for the lower ranges and then physicaly move to a different amplifier for the situations where you need the higher input current and can accept a higher noise floor to get your measurement speed back)

How will you be coupling the input signal to your amplifier? (a co-ax, Tri-ax or some other cable for example)

Do you have a rough idea of the likely source characteristics that you will be recieving the signal from (output impedance, capacitance, etc)

How much of a budget do you have for this? (there are a number of off the shelf products which are reasonably good from a company called femto if I rember correctly)

Anyway, sorry for the length of this post (and probably the spelling/typing) and I hope what I've said is reasonably clear, if not just ask and I should be able to explain things in a little more deatail if you'd like.

13. ### goodbyegti Thread Starter Well-Known Member

Apr 28, 2004
59
1
Hi, thanks for all the input!

The device is based on a paper titled:

"Current measurements in a wide dynamic range—applications in electrochemical nanotechnology"

According to the authors not using the limiter will result in n times greater noise on the second output compared to the first.

The results they obtained using this setup are really good, they're resolving features in the pico amp regime sampling at ~ 1KHz with the ability to make seamless measurements as high as ~150uA.

Using my current pre-amp I have no trouble performing the low current measurements it's just i would really like to increase my current range.

The input to the current op amp is almost directly connected to a 0.25mm diameter tube, ~10mm long into which a STM tip is inserted. The tube is encased in teflon, i presume to avoid leakage currents. There is also a guard ring round the input on the PCB for similar reasons.

Since the pre-amp must fit on a small PCB (10mm x 20mm) which is mounted onto a pizeo scanner i'm not sure it would be easy to couple the system to an external device.

Maximum accuracy is required when performing the low current measurements, the high current channel is not quite as important. The sampling frequency ideally should be ~1000Hz on the low current channel so capacitance needs to be low.

The source current arises from applying a bias across a molecule or quantum mechanical tunneling between the tip and substrate.

In anycase i hope the above paper may help you if you haven't seen it already (let me know if you can't get hold of it).

I haven't had a chance to work on this for the past few days but i'll update the thread if and when i make any progress!

14. ### John Luciani Active Member

Apr 3, 2007
477
0
Interesting article. Thanks.

That's not the way that I read that. When the limiter is off (low currents) the input
impedance is greater (N*R) and so is the noise. You always get the noise with the
high resistance. You get the high resistance when the limiter if off or removed.

If the recovery time of your op-amp is fast enough you should not need a limiter. Try testing the recovery time of your op-amp. It will be slower than the diode limiter
but may be OK. See if it is specified by the manufacturer.

Are the currents positive and negative? What is the power supply range of your opamp?
For example if the power supply is +-15V and the currents where only negative you could bias the power supply at +25V and -5V. You would at least double your range.

The teflon is used for its insulation resistance. If the guard ring on the PCB is driven
to the same voltage as the input then it is for leakage otherwise it is for radiated noise.

Since you are now doing the low current measurements it will not be if but when ;-)

(* jcl *)

15. ### goodbyegti Thread Starter Well-Known Member

Apr 28, 2004
59
1
Thanks again John, i'm glad you found the article interesting. I agree with your interpretation on the noise, sorry for the confusion.

I currently don't have any real components, all the circuits have been built in the free version of national instruments multisim (Analogue Devices edition). Fortunately Analogue Devices supply a large number of suitable op amps and a lot of the models for these are included with multisim.

It should be possible to undertake measurements using only negative currents but this puts sigificant constraints on the experiment which i would like to try and avoid.

I tested your idea of not using the limiter and it seems to work perfectly until the the first output saturates. At this point i suppose the negative input on the first op amp is no longer a virtual ground.

I found a neat page here:

http://www.ecircuitcenter.com/Circuits/op_limiter1/op_limiter1.htm

About isolating leakage using diodes (see circuit below).

It seems to have helped the situation but i get a lot of ringing at certain stages so i've been reading about stabilising the circuit by placing a capacitor in parallel across the feedback resistors. Unfortunately the capacitor starts to reduce the output voltages at higher frequencies so there is a tradeoff here.

Changing the capacitance by as much as ~0.5pF makes a significant difference to the voltage outputs so i can only look forward to laying this out on a PCB!

Last edited: Jul 11, 2008
16. ### John Luciani Active Member

Apr 3, 2007
477
0
I should have thought about that a little more. It can't be a virtual ground.

Here is a current amplifier idea for you. The downside is that you need to do
a floating measurement across RL. This may be difficult to implement.
The equation is for the HI Range.

17. ### ciavardello New Member

Dec 1, 2010
1
0
Hi,

I see this thread is rather old, just I did not find this forum before. So if you are still interested, just post a "yes" or send me a message and I will give the circuit solution, and also some hints. Also I see there are couple of misunderstandings on circuit noise...