Forum: Analog Circuits Current Amplification (nA => uA)

von Enrique P. (flote21)

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My problem is that I dont know if there is any IC which can amplifies 
low current with a very low noise. I need to amplify the low current 
generated by an photovoltaic detector (around nA) to something bigger 
(around uA). And i can't use a transductor amplifier....

I was thinking about a mirror courrent source with the WIDLAR 
strucuture. But I don't know which trt't use in order to achive a 
properly working with nA current (See doc. attached)...Can you give me 
some advise?

Thank you very much for your help!


Enrique P.

: Edited by User
von ulrich (Guest)

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Der Stromspiegel wird nicht gut funktionieren. Da ist einfach das 
Rauschen und die Drift vergleichsweise hoch.

Was ist das Problem mit dem Transimpedanzverstärker ?
Da gibt es ggf. Varianten auch für spezielle Fälle.

Um welche Frequenzen geht es ?

I really don't think you really consider a transductor (magnetic 
amplifier) based amplifier. A agree that that would be a rather bad 
choice - though it has some interesting points.

von Enrique P. (flote21)

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first of all thank you so much for your answer.

Regarding the question. we want to take measurements in ms. So the 
maximum freq could be 300KHz.

At the output of the photodiode we have already a switched integrator 
transimpedance amplifier: IVC102. And we need to use this IC because it 
provides a reset circuit (which only works with current) for reseting 
the photo diode. Take a look to the picture attached.

The main problem of IVC102 is the low sensitivity (uA). The photodiode 
gives the information in nA and IVC102 would not detect so small 
variations of current.

So then my question is how to amplify the output current of the photo 
diode to be detected by the IVC102 IC?

thank you so much.

von M, nicht Q (Guest)

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Keithley has published a low-level measurement handbook where they 
discuss some measurement principles for picoampere measurement. Maybe 
you can get some ideas from that handbook. Available for free on the web 
or from Keithley.

von Erich (Guest)

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As I know, photodiodes also work with very low current output.
Maybe you get some information using GOOGLE
  "nA" "photodiode" current "amplifier"

von ulrich (Guest)

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The IVC102 is alleady quite good in working with very small currents. 
It's actually quite hard to make it better. So nA currents should not be 
a problem. With 10 pF capacitance 1 ms at 1 nA allready gives 0.1 V 
change at the output.

So the points would be to make the IVC102 work well: keep the 
temperature down and the supply voltage very clean und HF signals out. 
Also a very clean circuit is important, and avoid cables as much as 
possible. Maybe there are slighly better suited (or graded) versions 
available (e.g. only one capacitor, ceramik case).

One can get more current likely easier by using a larger detector or 
even better optical concentration (lense, mirror).

von Reinhard Kern (Guest)

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Enrique Perez wrote:
> And we need to use this IC because it
> provides a reset circuit (which only works with current) for reseting
> the photo diode.

I dont understand what the IC does, but whatever, it will not function 
backwards through the Current Amplifier. If it is true what you say you 
have to use the IVC102 as the first stage.

sy Reinhard

von Enrique P. (flote21)

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yes I need to use the IVC102 like the first stage. And reading carefully 
the datasheet of this IC. I figured out that it is possible to control 
the gain with an external cap. So I don't already need a previus 
amplification stage. The IVC102 is able to amplify low currents.

But now I need to compensate the dark current of the photodiode. it 
provides a offset current around 100 uA. And the max input current for 
the IVC102 is also 100uA. So the IVC102 could be damaged. I was thinking 
to remove this offset dark current using a digital potenciometer (see 
picture attached) But I am open for other low noise solutions...Any 


von Achim S. (Guest)

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So you have a signal in the nA-range, but a dark-current of 100µA, which 
is "just" 100000 times bigger and even exceeds your amplifier input 

Then I would recommend you to find a way to modulate your fotocurrent 
(i.e. a chopper in front of your detector). Amplify your signal with a 
standard transimpedance amplifier, which can handle the 100µA dark 
current and the modulation frequency. (the IVC102 is an integrating 
ampflifier and will probably be to slow to follow the modulation of the 
signal). And then let a lockin-amplifier take care about filtering the 
(modulated) signal from the background and noise.

von ulrich (Guest)

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A current of 100 µA won't damage the IVC102, but just saturate it. With 
an extra capacitor it may be possible to use larger currents - still a 
large background current makes interpreting the output difficult. Such a 
current is also accompanied with a significant amount of noise. So the 
best would be to get to reduce background light as much as possible.

To get a low noise current to compensate for the background, one can use 
a rather large resistor (metal film, not carbon or semiconductor e.g. 
near 100 K) and a well stabilized/filtered voltage. To adjust the 
voltage and thus the current a digital pot is possible before filtering, 
after a good ref. source.

It is reasonably easy to get the noise of the current lower than the 
inherent noise due to the background photo-current. So reducing this 
current may be essential.
If this is really background light, there may be extra variations in the 
light, and second photo-diode to measure the background may be a good 

von Enrique Perez (Guest)

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Yes Ulrich, you are in the right direction.

In a first instance we are able to keep the dark current to 100uA always 
with a temp of -60ºC. Now I was thinking to use a high quality digi pot 
like x9118tv14z-2.7 to compensate the dark current. But I was wondering 
what do you mean when you say "rather large resistor (metal film, not 
carbon or semiconductor e.g. near 100 K) and a well stabilized/filtered 
If you use this pot: x9118tv14z-2.7, Is it also neccesary to use the 
large resistor that you say? For sure that I am able to provide a well 
stabilized/filtered ref. voltage for this pot....


von ulrich (Guest)

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It depends on the type of digital pot, but usually they have more noise 
than a resistor made from metal. So it is better to have the metal (at 
the high values metal film) resistor and a low noise voltage. To get a 
low noise variable voltage, a good (especially low frequency noise) ref 
source, a D/A convert or digital poti and a filter is the usual way. So 
any noise from the D/A or digital pot an  reference is filtered first.

It may work with just a voltage and the digital pot, as the requirements 
are not so very high (the noise of the diode bias is already quite high) 
- but a metal film resistor in the 100 KOhms range is a rather cheap 
thing, so why take chances for just a few cents.

What type of detektor is this, that it needs cooling to -60 C to get 100 
µA dark current ? Must be rather large or infrared i guess. Dark current 
also depends on the bias voltage - often a compromise between dark 
current and speed / capacitance related noise. So you may have to 
recalculate that choice.

von Enrique Perez (Guest)

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Hello again!

Thank you for you advise I will use a metal resistor compensate the dark 

Yes. It a detector manufactured by Hamamatsu and it is an experimental 

On the other hand I would like to have another option in my prototype, 
because could be that the IVC102 won't work in the way I would like. 
Then I was thinking to set up my own reset circuit + transimpedance Amp. 
I mean i would like to design something like the picture attached

But the main problem of this circuit is to find a good switch with low 
noise and good characterisitics...

Any advise?


von ulrich (Guest)

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The switch really is a difficult part, because it needs to have low 
charge injection and low leakage. However a very low resistance in not 
needed - this could be in the 100 Ohms range. The obvious choice is a 
CMOS switch. Against leakage it may be possible to use 3 switches in T 
form: 2 in series and one to ground, so that the last switch will be 
between the input and quasi- GND if off. So charge injection is the 
really critical parameter.

With the rather large current from the detector the circuit is not that 
critical as in original IVC102. Likely that capacitor will be larger and 
the amplifier may be lower noise and faster, even if this would mean 
more leakage current. So the requirements for the switch may be less 

I don't have an advice on a special part - just look at the 
manufacutures (e.g. LTC, AD, Ti, Maxim) websites.

An alternative may be using a photomos switsch, though I don't know 
exactly how good they really are. Also 2 JFETs may be possible, one as a 
switch and one to compensate.

von Enrique P. (flote21)

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Thank you so much ulrich for your advise. I was taking a look and this 
kind of switches seems to be good: ADG821.

They have low noise and combained with a good transimpediance OA could 
give good results.

On the other hand there is another switched integretor + reset IC very 
similar to IVC102 that but with a lower input offset voltage. This part 
is ACF2101.

I will try several options in my prototype and then I will see which is 
the best...


von ulrich (Guest)

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The ADG821 is a good switch for some applications that need a low 
resistance, bud a really bad choice for such an integrator for small 
currents. Low noise may be important with the amplifier, but low bias is 
also important.

Something like a ADG1201 or ADG202 would be more suitable, though charge 
injection is still much worse (at least guaranteed over full range) than 
in the IVC102.

von Enrique P. (flote21)

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Right the switch ADG1201 is really a better option. I will try with this 
and the I will tell you something.


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