[SOLVED] amplifier for current-to-voltage converter

Good day everyone!

I am in need of an amplifier that can provide 0-5V output
of the voltage equivalent of the current I need to measure..

initially I had


in which I had a 46-49mV output...
I am planning to input this to Arduino
so I thought I still need to amplify this

so I added an amplifier


The results was unexpected. the 10k ohms Rf for LM358 had something around 400mV,
the 10k ohms Rin for TL072 had something around 350mV
and the output was 3.65 V

what op-amp parameter could have caused this?
Thank you for your help!

Let us analyze LM324 first.
Its output cannot be lower than ground. At best it would be close to 0V.
So in your circuit, the two resistors 1M and 10K are actually like a voltage divider between 12V and opamp output that should be close to ground in this case.
This explains why you got about 50mV
5 * 10 / ( 1000 +10 )

Just for clarification, we're you referring to figure 1 (1.PNG) ?
The op-amp is LM358 by the way.

Thanks for the explanation about the voltage divider principle between the 1M and 10k resistors.
It gave me a different perspective about the circuit.

Actually, what I intended to implement in 1.PNG is for the op-amp part is to act as a current-to-voltage converter.
From current-to-voltage principle
Vout= -IRref

P.S. (I forgot to indicate that I neglected the voltage sign of the output, it was negative (-)46 to (-)49 mV), Sorry.

Taking 10k as Ref and -49 mV as Vout
I= 4.9 uV

which is near what I expected as the current through the 1M resistor having a 5V across it

You are right, I was referring to 1.png
I usually hear from "current-to-voltage converter" that, at the input of the converter, there is an external current to be monitored by getting at the converter output a voltage proportional to it.
It seems to me you have another idea about this. Would you please elaborate it more?

I am trying to implement something similar with item 3 from this link
**broken link removed**

The 1M resistor is a substitute for a sensor that will measure skin conductance (1/R).
For testing purposes, I placed the 1M resistor there whose value is within range of typical skin resistance.
So in actual setting...
I have to know the resistance across the sensor
which I plan to get using Ohm's Law V=IR
Voltage supplied to the sensor is kept constant at 5V
So now, I need a way to get the current through it.

I think I got your point now, but I am not sure why I couldn't open your link.

Let us update your 1.png
(1) Note: For LM324 and Vcc=5, the maximum linear output voltage is 3.6 V ( Vcc - 2*Vd = 5 - 1.4 )
(2) Let us replace the 10K resistor (R2) with a 1M one (between output and the negative input).
(3) By a voltage divider, the positive input could be set, for example, at Vref = 500 mV (but not less than 200 mV).
(4) The skin resistance (R1) is between the negative input and ground.
(5) If the skin resistance (R1) is infinity, the voltage gain of the opamp becomes 1 and the output would be 500 mV.
(6) If the skin resistance (R1) is 1M, the voltage gain of the opamp becomes 2 and the output would be 1 V

Vout = Vref * ( 1 + R2 / R1 )

Vout_max = Vref ( 1 + R2 / R1_min )
R1_min = R2 / ( Vout_max / Vref - 1 )
R1_min = 1000 / ( 3.5 / 0.5 - 1 )

The minimum skin resistance in this case is 166 Kohm.

The voltage supplied on the skin is Vref = 500mV

Vout = Vref + I_skin * R2
where
I_skin = Vref / R_skin
R_skin = R1

Note: We didn't consider here the possible noise interference, mainly due to the mains powerline (50 or 60 Hz).

here is a snapshot from the link


I followed your instructions in creating a new schematic
Did I follow it right?

I see now why you were confused.
From the circuit of the snapshot, it wasn't clear that the opamp is supplied by dual sources as +Vcc and -Vcc. In this case, Vout can be negative. But I think you like using +Vcc only (5V for example).

You did well in drawing the schematic.
There is a minor error.
There is no connection between the positive and negative inputs of the opamp.
I think you connected them based on

The voltage supplied on the skin is Vref = 500mV

Click to expand...

When an opamp runs linearly (in case of LM324, its output voltage is between the upper limit, Vcc-2*Vd, and lower limit, close to Vss), its positive and negative inputs have almost the same voltage since its voltage gain, Vout/(Vin_pos-Vin_neg), is rather very high, more than 500,000. So if we set Vin+ = 500 mV, Vout increases till Vin- becomes, via the feedback resistor R2, very close to 500 mV. I hope you get the idea why the voltage on R1 (sensor) is also equal to Vref (connected to Vin+).

About the values R_vDiv1 and R_vDiv2, you may increase them (to reduce their current) since the input impedance (or resistance) of Vin+ is very high and doesn't load the voltage divider.

Added:
For instance, if you have a regulated 12V supply, we can use the previous formula of Vout and choose its parameters so that we can get a wider dymamic range.
I wonder how you like to display your measurement in this project.

Thanks for guiding me through...
I am afraid I didn't understand why the voltage supplied on the skin is Vref = 500mV as I got confused with the statement you quoted previously

I temporarily took R1 from the circuit


I did my homework and I think I have an idea why V_R1 = Vref...
*negative feedback will do whatever it can to make the input voltages equal
*almost no current will be drawn in by both inputs

I hope I get R1's proper connection now


BTW, according to the datasheet of LM358, only 1 positive supply is needed for the op-amp

Sorry again for not being clear when I said

But I think you like using +Vcc only (5V for example)

Click to expand...

So, I went on using a single supply, +5V as we did since the start.

As you said, only 1 positive supply could be used for LM324.
Do you know why?

The range of its linear ouput voltage is between Vcc-2*Vd, and close to Vss (pin 4).
If Vcc = 12V and Vss is ground, Vout could be between 10.6V and close to ground (about 0.2V).
Not all opamp outputs could be varied linearly close to Vss (pin 4) as it is the case of LM324.
Note: The pin 4 of LM324 (labelled as GND on your pic though on some other ones Vss) could be supplied by a negative voltage, say Vss = -12V. And if Vcc=5V, the linear range of the output volatge would be from -11.8V (-12V+0.2V) up to 3.6V (5V-1.4V). I hope you get the idea.

Your last drawing is right. You need removing the -5V supply only and use ground instead of it for LM324.

Please note that you didn't tell me, if you can use a regulated voltage higher than 5V or not. Increasing Vcc helps us increase Vref to get a better accuracy.

On the other hand, I can't know what to do next since I have no idea of what the end idea of this project is. I mean how the measurement of the skin resistance should be displayed. For example, will it be by an analog meter or digital display? But perhaps, the circuit ouput will be used to control something else.

I think I will proceed with using just a Vcc...
Right now, I am thinking of using 9V supply for the op-amp

The output (hopefully filtered) will be fed into Arduino where it will be analyzed

- - - Updated - - -

I think I will proceed with using just a Vcc...
Right now, I am thinking of using 9V supply for the op-amp

The output (hopefully filtered) will be fed into Arduino where it will be analyzed

I used designing complete boards. I am not familiar with Arduino, I just hear of it. It seems you have no more questions. Good luck

follow-up question...
about the filter for the output signal
can I use typical parameters or
a new custom design is needed since the signal that will be filtered is already amplified?

by typical parameters I mean the standardized ones
commonly used in the application of my circuit

time constant: 6 s
bandwidth: 0.03 Hz - 1 Hz

Which Vcc did you decide on (5V or 9V)?
If you chose 5V, is the range of 500 mV to 3.5 V suitable for the following stage?
I mean what is the optimum range of the next input (to be connected to the opamp output)?

Sorry for the late reply..
I thought it through.
I'll have 9V as Vcc
but I have to make sure that the output will not exceed 5V (Arduino input pin can only accept up to 5V)
the resistance range will be (according to my readings) 100k ohms - 10 M ohms

Please note, the ratio 10M/100K = 100 is relatively high for good accuracy but you can keep the circuit simple now and if it will be necessary it could be updated later.
I suggest, for the time being in the least, that Vcc is 5V.

From the formula:
R1_min = R2 / ( Vout_max / Vref - 1 )
Vref for R1_min = 100K is 318 mV. Therefore we can assume Vref (from the voltage divider) could be about 300 mV.

Vout = Vref * ( 1 + R2 / R1 )
Vout = 0.3 (1 + 1000 / R_skin )

For R_skin = 100K
V_max = 0.3 (1 + 1000 / 100 ) = 3.3 V

For R_skin = 10M
V_min = 0.3 (1 + 1000 / 10,000 ) = 0.33 V

When you will get a reasonable result from this simple interface, you will know which parameters need to be updated though the circuit will likely have to be updated as well (like adding another opamp and resistors, if not CMOS switches if you have one or more free digital outputs to split the range of the R_skin measurement).

Good day!
After rummaging into some literature...
I reconsidered my potential range...
it's now 800k ohms - 10 M ohms.

I came up with this



which was in alignment with these

green = V+_input; red = V-_input


green = V+_input; blue = Vout