# Common Mode Rejection Question

• posted

I'm wanting to get an analog signal output equal to the differential signal IN X gain. In other words, if I have input A = 11V and input B=12V and Gain = 1, then I want an output signal of 1V. Is this what common mode rejection is? ie. it takes the difference in the signals and not the reference voltage of the signals. I want the same output if IN A=1V and IN B=2V or IN A=5V and IN B = 6V. The application involves current sense resistors and mosfets to switch in/out resistances, or perhaps high side resistors. I want to isolate the voltage drop across the resistors and ignore the actual voltage level the resistors are at.

Thanks!

Roger N

• posted

al

and Gain

No, common mode rejection [ratio] (CMRR) is the amount the common-mode signal is attenuated in the output; a measure of how differential a differential amplifier is.

But what you want, yes, is a differential amplifier, and that's exactly what op-amps do (except gain is considerably more than 1).

MOSFETs huh? That sounds like, oh,

Notice that the op-amp's input range MUST include both ends of Rs. Rs mounted on a power supply rail is reason for concern.

Tim

• posted

Roger,

An amplifier that outputs only the difference between two input signals is called a differential amplifier.

Differential amplifiers, while by design only output the difference between the inputs, always output a certain amount of the common-mode voltage of the two inputs. Common-mode rejection refers to the ability of a given differential amplifier to reject the common-mode component of the inputs.

The common-mode rejection of a given differential amplifier depends on the magnitude of the common-mode signal and also the frequency of the common-mode signal. In your first example (A=11V and B=12V) the common-mode component of the inputs is equal to 11.5V.

If you want some help in designing and/or selecting a differential amplifier then you'll have to provide some more details. For example:

- If your input A=11V and input B=12V, but then A changes to 15V and B changes to 16V, how much of a change in the output signal can you tolerate?

Bob

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• posted

Several people make amps specifically to pick off the voltage drop across a high-side shunt. Linear Tech for sure, probably Analog Devices and TI.

"Common mode rejection", as a measurement, defines how well a diffamp actually does ignore the common voltage. If it's perfect, +12/+11 and

+2/+1 produce the same output, 1 volt. If CMRR isn't perfect, ther will be some error between those two cases.

John

• posted

On a sunny day (Sat, 13 Sep 2008 16:34:15 -0500) it happened "RogerN" wrote in :

Yes, correct. You have to make sure your inputs stay within 'common mode range' though. This type of amplifier is called 'differenctial amplifier'. Any normal opamp is one. Also look up 'instrumentation amplifier' perhaps.

• posted

al

and Gain

and IN

sense

no you are not righytt

this ios nit ciommon mode abnnd thiss is differential amplifier.

if you want detail information about common b mode rejection contacyt

• posted

If I remember correctly, college was 25 years ago, if I had 5V and 6V in a

741 OpAmp with a gain of 1, I'd get 6V out, a 1V differential plus a 5V offset of the non-inverting input. I'm wanting rid of the offset to get the differential out with respect to ground.

Is this a 741 problem or would all differential amps have the non-inverting voltage offset?

• posted

a

t the

ng

Take a look at the LTSpice circuit depicted at:

com/SWITCH.bmp

Here you can see a simulation of a high side current sense on a power supply rail of a +12V supply. In this example the design is showing a method to cut off the load using a PFET switch when the load current goes over a pre-defined threshold. The LM324 opamp is used here. It is powered off the source +12V rail and the gain set resistors at the inputs are setup to keep the +/- inputs of U1 within the common mode range of the LM324. Note that the LM324 is not an expensive device with rail to rail inputs and outputs. The second opamp at U2 is operated open loop as a comparator. It compares the current reading from U1 against a reference level produced from a shunt reference device. You adjust the "overload current" level by changing the value of the series R1 resistor or by playing around with the values of R7, R8, R9 and R12 & R13. Note that R9 provides a severe amount of feedback around the U2 comparator to produce a foldback latch type of behavior. As shown here the latched drive of the PFET is removed by cycling off the source +12V rail.

You would use a circuit like this in a design, as opposed to the ready made high side current sense chips that others have mentioned, when you want the lowest cost implementation and when absolute accuracy is not of the highest concern. Another reason to use such circuit would apply when trying to learn about how high side current sensing can be implemented.

- mkaras

• posted

No, What you're looking for is basic differential operations of the + & - inputs.

Common mode rejection ratio is when you apply the same signal to both inputs and vary the level lets say from 0..10 volts, assuming 10 volts will be the max input. An ideal unit would not yield any differences of voltage at the output no matter where the input voltage is. Of course, there is no such animal as an ideal op-amp/comparator etc, so a slight shift in output voltage will take place when the inputs goes from 0 up to 10..

This is both + & - inputs,being driven with the same signal which should cancel each other out in a ideal case.

What your looking for is basically the same thing how ever, since you are using 2 different sources, A unit of 1:1 gain for its +&- inputs will yield what you're looking for.

"

• posted

Use a Dual +/- supply with respect to common. Use an mini DC to +DC/-DC converter to operate the chip.