Hi,

I am in the process of buliding a dc motor performance testing station.In which I measure the motor current through a 10 amps 100 millivolt shunt.To convert the milli volt output from the shunt to my PLC anlog input( 0~10 Volts) I need an amplifier.Can someone help me in this regard please.

 

Hello,

For converting 100 mVolt to 10 Volt, you will need an amplifier that amplifies the signal by 10000.

Take a look at the divided feedback section of the Vol III.
http://www.allaboutcircuits.com/vol_3/chpt_8/5.html

Here is explained how to amplify a signal.

Take an opamp with very low offset, because this may influence the accuracy.

Greetings,
Bertus

 

Bertus, you've misplaced your decimal point.
If 100mV (0.1v) represents 10A, you would want to multiply by 100, not 10,000.

Low offset voltage and low drift is important for accuracy. If you have dual supplies available (+/-12V to +/-22 or perhaps +12/-5) have a look at LT1007 or LT1037 from Linear Technology.

If you only have a single-ended supply, you'll need to look for a single supply opamp that's rail-to-rail for both input and outputs, low offset, and low drift.

 

Thanks a lot for guiding me.I will try to implement.

 

I am confused a little bit. My amplification ratio is 1:100.By various combination I can get the R2 and R1. I am an Electrical Engineer with little knowledge in Electronics.Please help.Guide me how to choose right combination with respect to input voltage and output voltage.I have a dual power supply +/- 15 ,+/- 12 volts.Can I use OP07?I am using Keyence 12 bit analog input unit PLC.

 

[eta]
Removed; posted in the wrong thread! I don't know how this happened.

 

Hello,

Sorry for the calculation mistake.
For 100 times R1 can be 1K and R2 can be 99 K.
The 99 K does not exist so take a 100 K.
The input can be corrected with a pot of 10 K and a resistor of 100K.
This gives you a gain range from 101 (pot at the top) and 92 (pot at the bottom.

Greetings,
Bertus

 

Yes, you can use your OP07 with the +/-12v supply.

See the attached; it's using a different opamp, but it will work similarly.

 

I am extremely grateful to you all.

 

SgtWookie,

Your circuit seems to be good for me.Will it work for pure dc signal?You have mentioned 6Hz signal.Correct me If I am wrong.I need to learn a lot in electronics.

 

Hello,

The 6 Hz is used to show its behavour.
It will work for DC too.

Greetings,
Bertus

 

I feel great for your support,Bertus.

 

How can I introduce zero and span adjustment in this circuit, Bertus.

 

1) Remove the signal at point A, and ground it. You will then read just the amplified offset voltage at the output of the opamp. You can use a 20k potentiometer with one end connected to pin 1, the other end connected to pin 8, and the wiper connected to V+ to adjust the offset to zero. See the datasheet for the offset nulling circuit (page 3).

2) Connect a 100mV signal to point A. Adjust R3 until you read exactly 10V out.

3) Remove the 100mV signal and ground the input. You should again read 0v at the output.

4) If you wish, connect a 50mV signal to point A. You should read 5v on the output. This will give you an indication of the linearity of the circuit.

The offset will drift over time and temperature. You will need to periodically check/readjust for this drift. As the opamp ages, it is likely that the drift will tend to stabilize somewhat; however the temperature drift will remain relatively consistent.

 

Wookie's circuit will work, but there is a better circuit for this application.

You really should be using a differential amplifier. This is because there is a voltage generated from the bottom of your sense resistor and ground. Secondly, as the copper heats, it changes its resistance significantly (compared to a good sense resistor).

http://www.allaboutcircuits.com/vol_3/chpt_8/9.html

Steve

 

Not a bad idea, Steve.

However, attempting to build a differential amplifier out of discrete components is problematic, at the very least. Resistors have to be matched with a great deal of accuracy, or CMRR goes right out the window.

Intersil published a good intro to Instrumentation Amplifiers (which is a differential amplifier) AN1298, available on their website.
Direct link: http://www.intersil.com/data/an/an1298.pdf

Have a look at INA118 and INA128 on Texas Instruments' website. These are reasonably inexpensive amps with nice specs that are internally trimmed; you just add a single external resistor to set the gain. A minimum of fuss to get an amp with a very high CMRR.

 

Thanks Wookies and Steve. I am in the process of buiding the circuit.Let you know the outcome.

 

Wookie and Steve:

I am escalated. Circuit works fine and linearity is excellent. My senior asked me to introduce a buffer after the Vout.Why?

 

A buffer will be necessary if the load is of low impedance or reactive. I had simply made the assumption that your load on Vout would be high impedance (ie, DVM). With a high impedance load (Rload > 10k Ohms) the output can typically reach the supply less 2v. Since your supply is 12v, you are right at the limit.

If your load is reactive (inductive, capacitive) then you should consider using something like a Linear Technology LT1010, which is a fast 150mA power buffer. This buffer goes in the feedback loop, so there isn't an additional offset error introduced.

If you used a traditional buffer (ie: opamp with output tied to inverting input) you would need to compensate for it's offset as well.

 

Thanks Wookie. I will introduce LT1010 after after Vout.I have connected the Vout to the analog input of the PLC which has high input impedence.