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Current to Voltage Converter

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Hello there,

I am interested on a Current to Voltage Converter, that basic says, "A current to voltage converter will produce a voltage proportional to the given current".
This configuration can be used equally well for devices that source their current via some positive excitation voltage.

But I am disappointing what this circuit works ?

Could you kindly explain it ?

current_to_voltage_converter.png

Take a look

1. Diodes are used for some purpose, alternatively.
2. In ideal case inverting input is a virtual ground, in real case any photo diode is used. but here how the things working!
3. Negative feedback contains RC combination.
4. Not sure why R2 and C2 has add here, suppress noise?
 

dknguyen

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I was always told a current-to-voltage converter was a resistor. Then add op-amps to buffer and amplify the resulting voltage as required
 

ronsimpson

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I am not happy with the circuit on post #1.
We all know a "volt meter" can measure current.
current-to-voltage converter was a resistor.
In my circuit a 1V meter can measure current with the help of a 1 ohm resistor.
upload_2017-12-14_15-49-46.png
I do not like the diodes in the circuit from post #1.
These little diodes will explode if high current passes through them. It is not input protection.
I make meters like this:
If the current is high for a long time the fuse will blow. (long time in electron time)
Rm and the diodes save the meter. Rm will limit the current so the diodes will survive.
upload_2017-12-14_15-53-59.png
About the op-amp:
I am not certain it works in circuit #1, but.....
Here is the circuit with a simple 10X amplifier.
So the voltage across the current sense resistor needs to be only 0.1V to drive the meter to full scale. (1.0V)
upload_2017-12-14_15-55-45.png
I did not include protection, power supplies, etc.
 

schmitt trigger

Well-Known Member
The circuit looks like the type which appeared on National Semiconductor application books.

The description would thoroughly explain the circuit.
I have access to hardcopy data books, but require to know where you found the circuit for me to help you.
 
I am not happy with the circuit on post #1.
We all know a "volt meter" can measure current.
Very common and constrain rule of electronics, as long as you increase input impedance you will get more current.

I do not like the diodes in the circuit from post #1.
These little diodes will explode if high current passes through them. It is not input protection.
Yes, but keep in mind its for LF356 application, those diodes are clamping or useful for their leakage current


If the current is high for a long time the fuse will blow. (long time in electron time)
Rm and the diodes save the meter. Rm will limit the current so the diodes will survive.
Accept them for protection.

About the op-amp:
I am not certain it works in circuit #1, but.....
Here is the circuit with a simple 10X amplifier.
So the voltage across the current sense resistor needs to be only 0.1V to drive the meter to full scale. (1.0V)
Does it make sense for ideal voltage to current converter ?


An expert says,
"That is the kind of current to voltage converter that would be found in a floating instrumentation application like a bench multimeter. Take special notice of the table shows which parts are shorted or removed for different input current ranges."
Current to voltage converters which use feedback are also known as transimpedance amplifiers so a search for that will turn up lots of practical theory.
 
Last edited:

MikeMl

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To me, this is the prototypical "current-to-voltage" converter:

CV.png
 

KeepItSimpleStupid

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That circuit is really, really cool! I have it nailed except for the purpose of R2 and C2.

Normally R1*C1 =k; and helps stability.

The Diodes are for input protection. They can also shorten recovery times from overloads.
 

schmitt trigger

Well-Known Member
Unfortunately, the link does not provide with an app note number.
But let me see what I can find.

As others have mentioned, I'm also very curious about the purpose or R2, other than it is used for ultra-low current ranges.
 

ronsimpson

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To me, this is the prototypical "current-to-voltage" converter:
upload_2017-12-15_6-35-41.png
Thanks Mike, I could see that the circuit from #1 has two different modes of operation. For high current it watches the voltage across a resistor, amplifies and outputs. I just could not see that in low current mode it works like what you posted. (#8)
 

KeepItSimpleStupid

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A real easy way of visualizing the above post (standard transimpeadance amplifier) is:

The (-) terminal is held at ground. I and -I1*1M sums to keep the (-) terminal at ground (Vos really). The current into the op amp is CLOSE to zero. The OP amp must be capable of sourcing/sinking the current being measured. Vos and Ib are important parameters. Ib is a function of temperature.
 

KeepItSimpleStupid

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In the other mode: (take the 2 Ma range)

2 mA is impressed across R4 (10 ohms); generating (2 mA) or 20 mV accross R4. R2 is zero and C2 is zero.

20 mV is impressed on R3 (5K). Since the (-) terminal is at virtual ground, you have I=20 mV/5K, you have 4 uA
4 uA is converted like in the transimpeadance amplifier (4uA * [email protected]) yielding 0.2 V Full scale for 2 mA in.

So, cool !
 
Last edited:

tomizett

Active Member
It had me foxed for a while there, too. As Ron has pointed out above, it's really two different circuits, because not all the resistors are actually present at the same time (some will be replaced by shorts or opens).

It would probably help if you where to re-draw the circuit without the shorted or opened resistors.
 
In the other mode: (take the 2 Ma range)

2 mA is impressed across R4 (10 ohms); generating (2 mA) or 20 mV accross R4. R2 is zero and C2 is zero.

20 mV is impressed on R3 (5K). Since the (-) terminal is at virtual ground, you have I=20 mV/5K, you have 4 uA
4 uA is converted like in the transimpeadance amplifier (4uA * [email protected]) yielding 0.2 V Full scale for 2 mA in.

So, cool !

Yes, its interesting indeed.
But did you notice the capacitance of the photodiode is a major problem , http://www.electronicdesign.com/analog/whats-all-transimpedance-amplifier-stuff-anyhow-part-1
 

KeepItSimpleStupid

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But did you notice the capacitance of the photodiode is a major problem ,
Been there. Done that.

I did a design of a 4-terminal I-V converter as a front-end of a lock-in amplifier for measuring the Quantum Efficiency of solar cells.

For +-10 V full scale out, with ranges of 100, 10, 1 and 0.1 mA full scale with up to +-10 V bias unless +-50 mA of suppression was used; then +-5V bias.

My design went south when our calibration cells were used. Silicon, 1 cm, about 25 mA at AM 1.5 G. Major tweaking to get it to work.

AC performance was excellent and could be calibrated to get better than 1%. DC had about 40 pA of bias current and less than a few mV of offset with no adjustments. I planned on an electronic adjustment, but it was unnecessary for the objective.

So, modes of Voc, 2 Terminal//4 Terminal, Zero Check, Zero correct

A red/green clipping indicator. Two HP system DMM's for voltage and current and a 4-channel D/A converter and some digital I/O. So it was IEEE-488 (HP-IB) controlled.

What I forgot to account for was I could not get exactly 0 V out from the D/A's.
 
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