# sounding rocket project problem

Discussion in 'The Projects Forum' started by d.sonali20, May 21, 2011.

1. ### d.sonali20 Thread Starter New Member

May 21, 2011
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i am using a meteor smoke particle detector for a sounding rocket project.The output of my detector is current in range pico amps to micro amps.i need to convert it into voltage in the range 0-5v(for the adc).what can i do??

2. ### wmodavis Well-Known Member

Oct 23, 2010
739
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Try googling "low noise current to voltage converter"

3. ### Wendy Moderator

Mar 24, 2008
21,012
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A very simple op amp will do this...

Current must be going to ground though.

4. ### Bernard AAC Fanatic!

Aug 7, 2008
4,551
477
Found Op- amp Electrometer, by Richard Hull, as sub heading @ http://www.amasci.com/emotor/chargedet.html after Googling [ bing really], electrometer. Go to bottom of page to shortcuts, down to Other Links. Use a high value resistor from input to ckt ground to convert I to V.

Last edited: May 21, 2011
5. ### d.sonali20 Thread Starter New Member

May 21, 2011
9
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@bill_marsden
will the circuit also convert current to voltage??and can it amplify current in pico amps??

6. ### Wendy Moderator

Mar 24, 2008
21,012
2,745
Yes, but the current has to go to ground. How much do you know about op amps?

7. ### d.sonali20 Thread Starter New Member

May 21, 2011
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absolutely nothing....i only know that they are used for amplification...i am a first yr engineering student...i will be studying op amps in the next semester...but i've been assigned this work and i need to submit it by today evening

8. ### Wendy Moderator

Mar 24, 2008
21,012
2,745
In this case the formula is really simple.

V = -I * R

So if R is 1000Ω (or 1KΩ) for 1ma (0.001) it will output -1V. You will need a quality op amp that can handle signals near ground if there is only one power supply voltage.

So what power supply voltages do you have? Most traditional op amp circuits need a dual power supply, it would help with this circuit too, and simplify some side issues.

As for op amps, any really quiet high impedance should work OK. Maybe other people can recommend much better, but I'm thinking a TL071 or TL081. The dual versions (TL072 or TL082) could use the second op amp to invert the signal to where the formula is V = I * R. It is up to you.

d.sonali20 likes this.

Feb 5, 2010
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10. ### Bernard AAC Fanatic!

Aug 7, 2008
4,551
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Op amp ckt. with 10 teraohms input Z, Richard Hull, LMC 6081 fromDigiKey.

11. ### Wendy Moderator

Mar 24, 2008
21,012
2,745
Just curious, how accurate and calibrated do you think a design like this would be? It isn't what I would call instrumentation.

12. ### Kermit2 AAC Fanatic!

Feb 5, 2010
4,165
1,121
It would depend on what is used for calibration.

It was offered as a high impedance input, high gain circuit option, that is NOT an opamp. Our poster declared he knew nothing about opamps.

His decision to ask us for help just one day before the 'project' is due, would decidedly make using components he knows nothing about a source of delay and confusion. If you feel I shouldn't offer alternatives to opamps, then you will be less than happy with my input on this topic.

13. ### Wendy Moderator

Mar 24, 2008
21,012
2,745
Actually I did, post #3. One op amp, one resistor, and you have a precision low current to voltage converter. About as easy a project as I've ever seen, the only drawback is the ± power supply (I'm assuming space and weight is tight).

I'm just trying to figure out how that schematic would be applicable or helpful.

14. ### d.sonali20 Thread Starter New Member

May 21, 2011
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can you tell me how to calculate the value of the resistor to be used??i know that it is simple ohms law....but for that i need to know the magnitude of current coming out of the op amp..
but the current magnification factor is not mentioned in the data sheet of TL071

15. ### CDRIVE AAC Fanatic!

Jul 1, 2008
2,223
101
Because of the very Hi Z of the Darlington array circuit it will probably light the LED with just a finger on the input. Problem is, it will require > 1.5V before the LED will begin to glimmer.

16. ### wayneh Expert

Sep 9, 2010
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4,416
Last time I went looking for a high impedance op-amp, I found the LMC6035 or LMC660. I never actually used either of them but I think they'd be fine for this application.

OP, you need to tell us the conversion you want to make, from current to voltage. For instance, you might want zero current to produce 0v, or perhaps some other offset voltage. And you might want 10µA to give a 5v output. A gain of, say, 20x for the op-amp is reasonable. So for 5v/20 = 0.25V to drop across a resistor carrying 10µA, you'd need V = IR 0.25 = 25µA  x. Solving for x gives 10K as the appropriate resistor. Smaller currents will need bigger resistors and/or more gain at the op-amp.

17. ### Wendy Moderator

Mar 24, 2008
21,012
2,745
Post #8 gave the formula. What is the problem?

18. ### wayneh Expert

Sep 9, 2010
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As usual, a total lack of specifications. The range given so far covers about 6 orders of magnitude. It'd be nice to narrow that a bit.

19. ### d.sonali20 Thread Starter New Member

May 21, 2011
9
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as i mentioned in my post, i understood how to apply ohms law to calculate the value of the resistor...
but my problem is.....like in post #16 waynes assumed the gain to be 20 and calculated the rest.
now,when i am looking at the data sheet of an op amp,how do i know what the gain for that op amp is??
i guess its a simple question of how to read the data sheet.

20. ### Wendy Moderator

Mar 24, 2008
21,012
2,745
The formula I gave you works, Ohm's Law doesn't apply directly to it, to understand the circuit you need to understand op amps and virtual grounds. The simplicity of the forumla should have been a clue, as it is not Ohm's Law, and is actually simpler.

You asked, I gave. If you want to understand this circuit some serious reading (more than I am willing to type) is involved. The two key words, op amps and virtual ground, is the key, and would make good words to google. You can also read up on op amps in the AAC text book.

The circuit I gave is very linear, change the one resistor and the gain changes. It is possible it doesn't do what you want, but that is a different kettle of fish. I went with the assumption the current is going to ground. If it isn't a different circuit will be needed.

********************

Thinking about it, I can tell you how Ohm's Law applies to this circuit. Both pins of the op amp are the same voltage, the output of the op amp will assume whatever voltage to balance the current coming into the input. The same current will be going through the feed back resistor. This is where Ohm's Law comes in.

If you want to understand you need to phrase your questions a bit clearer, and expect a lot of reading as homework. Op amps are a full semester course in and of themselves.

Last edited: Jun 19, 2011