OPA1652 INPUT COMMON MODE VOLTAGE RANGE

From the attached White Paper:

Normally there is a voltage that is common to the inputs of the op amp. If this common mode voltage gets too high or too low, the inputs will shut down and proper operation ceases. The common mode input voltage range, specifies the range over which normal operation is guaranteed.

See pages 25 - 26 and let me know if you have more questions.

Hi John,

  " VCM Common-mode voltage range (V–) + 0.5, (V+) – 2"

There are two separate statements there...(you are missing a seporating comma).

 "(V–) + 0.5" and "(V+) – 2"

"(V–) + 0.5" means "Vminus plus +0.5V", or +0.5V above V-

"(V+) – 2" means "Vplus minus 2V", or 2V below V+.

They are stating the maximum voltage in relation to the nearest supply rail (V+ or V-).

The parentheses are confusing because they are separating the "signs" in the supply name - not an actual mathematical function.

 "Vee + 0.5V" and "Vcc - 2V" is a little less confusing...and some datasheets have it this way.

So for the OPA1642:

If you have a ±5V split supply, the "legal" common mode input range would be -4.5 to +3V, or a half volt above V- and 2V below V+.

If you have a single supply of +10V and GND(V-), the "legal" common input range is +0.5V to +8V.

The OPA1652 is not a "Rail to Rail" input device. Rail to Rail input devices may have opposite signs, which meas they can accept inputs slightly outside the supply voltage range.

For a R-R input device, you may see "(V–) -0.3" and  "(V+) +0.3" , which means you can exceed each rail by 0.3V.

If you have a ±5V split supply, the "legal" input common mode range range would be -5.3 to +5.3V.

If you have a single supply of 10V, the "legal" input common mode range is -0.3V to +10.3V.

These are the rated limits. In reality, there is a narrow grey zone between the limit and complete failure.

This "grey zone" area varies over temperature and supply voltage - so it is not constant and is not guaranteed. When operating in the "grey zone", the device seems to function, but some specifications will start to degrade quickly (CMRR, Offset, PM, Gain, BW, Slew, etc) until they completely fail at the extremes (the output rails, oscillates, or "sticks" in a random position).

It is always best to stay within the limits...of course....

The output swing can also be presented in the same format - just that in 99.999% of the cases, the output will not exceed the supplies...

Hope this helps...

Regards,

Oops...I corrected a few typos in the single supply ranges...Ignore the email version...

Jeffry and Paul,

That is excellent information from both of you.  All of this makes sense. I now understand that if both inputs of an op-amp simultaneously reach or exceed either the min or the max common mode voltage of the op-amp that current will stop flowing and the input stage will shut down resulting in distortion in the output such as clipping or phase reversals. 

Now I need to understand how this condition could happen.  I expect that if both inputs could approach or exceed the min or max common mode voltage range then it would be possible they could do this simultaneously and cause the input stage to shut down resulting in distortion in the output. 

As an example take an inverting amplifier with + input tied to ground and with a single supply of +10V as you described above.  The "legal" common input range is +0.5V to +8V.  With the + input tied to ground it has already exceeded the minimum common mode voltage specification.  Any time the – input is at +0.5V or lower the input stage would shut down causing distortion on the output.  Is this correct?

So, in summary any time the voltage at both the + input and – input of an op amp simultaneously reach or exceed the common mode voltage specifications of the op amp the input stage will shut down causing distortion on the output.  Does it seem like I understand the concepts correctly?

Regards,

John

Actually - an inverting amplifier is a good BAD example... :^)

The VCM on an inverting amplifier generally does not change because of the signal going into the false summing node through the resistor, which is "fixed" by the non-inverting pin reference level through feedback. You generally do not see large common mode excursions in inverting mode (which is why the audio guys like to use inverting gains to reduce distortion).

A non-inverting, and in particularly in the buffer configuration will see large excursions on the input since the input signal is applied directly to the non-inverting input, and the inverting input follows the non-inverting input through feedback.

Since lower gains mean a larger input signal for a given output swing, lower gains will have a larger swing on the inputs than higher gains. The follower ("buffer") is 1:1.

A buffer is obvious - the output is capable of swinging rail-to-rail, but the input CMR will limit the actual output swing, since the inputs are 1:1 with the output.

At high supply voltages you do not run into the common mode limit problem too much. But you can and will run into issues at low supply voltages.

Say you have a low voltage, non-R-R input op-amp on a 1.8V supply, in a non-inverting gain of +2, and the upper VCM limit of the amp is "V+ - 1V" (common mode range shrinks as the supplies get lower...beware!).

That means the highest voltage on the input can only be +0.8V...(can you see the problem coming?)....

In order to drive the output to a full 1.8V - you must apply +0.9V....Uh-oh...

And...if the output does swing to the rail (say 1.75V), there will be 0.875V on the inverting input - Uh, oh again...

So in order to get full swing on the output - you may violate input VCM at low gains.

This is a very common mistake and has caused problems for many customers. Because you are partially operating in the "grey zone", the circuit seems to work fine - until you push one of the parameters to extremes, like temperature, speed or a lower supply voltage. Then errorz start to appear..

Also - a clarification. Many texts define the Common Mode Voltage as the "average" voltage of the two inputs. While academically true - it is NOT how we spec the VCM limits.

The VCM specs are PER INPUT, not the average of the two.

Why the big deal? Comparators are designed to accept and operate with large voltage differences across their inputs. If one input was well within the VCM spec, and one input was slightly outside VCM spec, then the average is within VCM spec, right? Nope! One of the inputs outside VCM spec and is therefore violating the VCM spec! We have had customers make this assumption...and they assumed wrongly...

So that is probably more than you needed to know about Common Mode Voltage.

Hope this helps...

Regards,

Thanks for your patients Paul.

It seems like I have much more to learn.  Something that is not making sense is that VCM seems to imply both the + & - inputs are involved, but you stated "The VCM specs are PER INPUT, not the average of the two."  We already looked at the OPA1652 and found that its specification for VCM is (V–) + 0.5 min,  (V+) – 2 V max.  So what you are saying if either pin reaches of exceeds this the input will shut down causing distortion in the output.  Is this correct.

In the case of a non-inverting amplifier using the OAP1652 and +/-2.5 power rails if either input gets to [(+2.5V) - 2] = 0.5V (or higher) or [(-2.5V) + 0.5V] = -2.5V or lower the input will shut down causing distortion on the output.  Is this correct? My original understanding was both pins had to reach or exceed the limit together for the input to shut down.

Thanks,

John

Hi John,

The spec is the same for for both inputs - per input.

Yes. If either input goes outside the specified VCM range, then you will see errors.

Under normal (linear) op-amp operation, the voltage difference between the two inputs (called the "differential voltage") will be almost zero (microvolts to millivolts) due to the feedback action.

Only during extreme non-linear operation events (fast slewing, overload, etc) will the inputs be more than a few dozen millivolts apart. During that time, the output is in error anyways (slewing or railed), so the point is moot...So both inputs can be thought of being together when looking at the VCM spec.

Again, if EITHER pin exceeds the limit, it is in violation of the VCM spec and errors can occur. But as I said, for an Op-Amp, the inputs would rarely be more than a millivolt or more apart from each other.

Regards,

Hi Paul,

I understand that because the gain of an op-amp is large that the voltage difference between the + and - inputs will be small in normal operation.  I thought that VCM was beginning to make sense because I was thinking of is as a voltage common to both pins and because it was common would not be amplified, but could prevent any signal riding on top of the common mode voltage to be distorted.  This appears not to be the case though. 

Regards,

John

HI John,

There are two voltages associated with the input:

 "Common Mode Voltage" and "Differential Voltage".

The "Common Mode Voltage" is the average of the two input voltages referred to the system reference point (usually ground). The op-amp is *supposed* to *reject* any changes in Common Mode Voltage (hence the CMRR specification). But remember what I said about "average" and per-input when it comes to the specs!

The "Differential" voltage is the difference *between* the input voltages. This is the voltage that the op-amp is *supposed* to amplify.

You can think of the differential voltage as "riding on top of" the common mode voltage, but really the differential signal is the *difference* in the common mode voltages.

Nothing is perfect, so the Common Mode Rejection Ratio (CMRR) spec describes how much the output offset will change as you change the common mode voltage changes.

I think you have it now - the amplifer sees two voltages on it's inputs - the differential (which it amplifies), and the common mode voltage (which is is supposed to reject).

Regards,

Hi Paul,

One last question and I'll mark this post answered. I have attached a schematic of an op-amp circuit based on the OPA1652 op-amp.  The input signal is +/-0.5V RMS.  With this input signal is there a possibility of violating the Common Mode Input Voltage range of the OPA1652.

Thanks for your help!

Regards,

John

John,

I'm a bit puzzled by the input waveform you have drawn, labeled as it is with "RMS." Assuming this waveform is as would be seen on an oscilloscope, the magnitude is 0.5V peak, or 353mV RMS. This signal would exactly equal the specified common-mode range of the OPA1652 on the positive peak. It would have room to spare on the negative-going peak.

It appears, however, that this signal would overload the output with the indicated circuit gain (assuming the frequency is within the intended amplifier bandwidth).

Regards, Bruce.

 Hi Bruce,

Sorry, I labeled the signal plot wrong.  I should have labeled it 2.828V p-p so that the RMS value was 1V.  I have attached an updated drawing. The OPA1652 data sheet states:

     Common-mode voltage range (V–) + 0.5 to (V+) – 2 V 

In the case of the attached circuit the Common Mode Voltage range would be  (-2,5V + 0.5) to (2.5V -2V) or -2.0V to 0.5V.

Considering just the input signal at Vin which is 2.828V p-p it definately exceeds the  Common Mode Voltage range of the OPA1652; however, the full amplitude of Vin is not going to be seen at the input pins of the OPA1652.  Also, we have discussed that because the open loop gain of an op-amp is large (106 min for the OPA1652) the difference in voltage at the input pins themselves is very small; otherwise, the output of the op-amp would go to one rail or the other.  How can I estimate what  Common Mode Voltage would be at the input pins so I can determine if there is a possible problem that needs furhter investigation?

Thanks for your help!

Regards,

John

0876.OPA1652 Schematic.pdf

John,

Sorry but this makes no sense to me. The new drawing attached to your latest post shows a 1V p-p input waveform, correctly labeled as 0.353V RMS. Your message states a very different input voltage--2.828V p-p, 1V RMS.  Which is correct, your words or the attached drawing?

You say "the full amplitude of the input is not seen at the input of the OPA1652?" For frequencies above 10Hz or so, the full input amplitude (or 99%, anyway) is applied to the non-inverting input terminal. Thus the input common-mode voltage is equal to whatever you decide your input waveform is going to be.

And again, your circuit gain appears to be too high. For any frequency between 7kHz and 100kHz you have a gain of about 8.5. The input signals that you are suggesting (even the smaller one in the drawing) will drive the op amp output into the rails. This will render any concerns about common-mode range mute.

Regards, Bruce.