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Voltage to current using IC/circuit

Hi,

I'm having a problem not much related to LabVIEW but circuit design. Part of what I want in this project is to send a certain current, which could be in the range of 0.1uA to 10uA and -0.1uA to -10uA, to a biochemical sample. One week before, this problem didn't exist when we were connecting to a current source which could send out any current we wanted. Now the problem comes up because we want to get rid of this current source. By using DAQ card we can output a positive or negative voltage certainly. So the problem is, how to convert this voltage to current?

The easiest way you may think of is to put a fixed resistor across the output voltage. However, when we try to achieve such tiny current, using a resistor could have a couple issues, such as temperature-dependence and lack-of-precision. I thought of another approach that takes voltage as Vgs of a NMOS and then generates a Id which is very stable. In this way, how to generate a negative current is a unsolved question for me so far. How about those commercial voltage-to-current ICs? Do those provide only positive currents?

How about this one:
http://www.edn.com/article/CA321804.html
This circuit can generate a positive or negative current depending on input voltage.
Do you have an easier way making the circuit?

The answer for this might be a simple one for you guys who are familiar with circuit design. Please give me a hint. Thx a lot!!


Regards,
Scottie
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Message 1 of 5
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I have found this one:
http://www.allaboutcircuits.com/vol_3/chpt_8/7.html

If I replace that 250 ohm resistor with a 1M ohm resistor, then we can convert 0.1~10 Volt input to 0.1u~10u Amp current; -0.1 - -10 Volt to -0.1u ~-10u Amp current.

Can any expert provide your opinion? Thanks^^


Scottie
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SCOTTIE wrote:
> I have found this one: <br>http://www.allaboutcircuits.com/vol_3/chpt_8/7.html
> If I replace that 250 ohm resistor with a 1M ohm resistor, then we can convert
> 0.1~10 Volt input to 0.1u~10u Amp current; -0.1 - -10 Volt to -0.1u ~-10u Amp
> current. <br><br>Can any expert provide your opinion?

In theory this will work. In practice however you will get all kinds of
effects but rarely the actual signal. Noise will be a significant
problem too at those levels. But the first and foremost problem is the
fact that the sample you want to apply some current to will have a
resistance too. This resistance will be in series to your current
limiting resistor and therefore the actual current flowing will be some
more or less significant amount smaller than what you think it should be.

With careful selection of the OpAmps to reduce their input offset
voltages the schematic on the TI application note seems the most simple
solution to me. In addition you may have to consider a fully galvanic
isolated amplifier to reduce noise problems. This would involve an
isolation amplifier before the VtoI converter and also a DC-DC converter
for the power supply of the output stage of that isolation amplifier as
well as the op-amps of the VtoI converter.

Rolf Kalbermatter

Rolf Kalbermatter
My Blog
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Message 3 of 5
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Scottie,

that may work, however some points to regard:
- choose the appropriate OP-amp with very low bias current (some pA) and noise
- use a good PCB-material
- your R_i_sense (250ohm in your mentioned example circuit) determ your accuracy, look for some with low ppm/K, in order to use smaller values, you can divide down your control voltage.

Your probe has to be isolated (!) with this circuit and not grounded. Other circuits can be found in application notes of TI or Analog devices (AN-573 for example) and National Semiconductor App.note 242!

Also have a look to 'linear-technology' AN 67 page 82 🙂 should have noted this first, all you will need is in there 🙂
Greetings from Germany
Henrik

LV since v3.1

“ground” is a convenient fantasy

'˙˙˙˙uıɐƃɐ lɐıp puɐ °06 ǝuoɥd ɹnoʎ uɹnʇ ǝsɐǝld 'ʎɹɐuıƃɐɯı sı pǝlɐıp ǝʌɐɥ noʎ ɹǝqɯnu ǝɥʇ'


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Take a look at this web site

http://members.tripod.com/michaelgellis/howland.html
Randall Pursley
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