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# LM2902-Q1: Common mode input voltage problem

Part Number: LM2902-Q1
Other Parts Discussed in Thread: TLV2374-Q1, , LM2902

Dear team,

Thanks & Best Regards,

Sherry

• The absolute maximum ratings table shows when the device could be permanently damaged, but these limits do not guarantee correct operation.

UB_HALL-1V is above VCC-2V, so it might not work. In particular, you might get phase reversal, which would mean that the output is wrong.

There are op amps with a larger input range (e.g., TLV2374-Q1).
• Sherry,

The common mode voltage calculation formula is not correct for LM2902-Q1
The common mode voltage is the lowest voltage input , not the average of the inputs.
If the inverting input is 2.4V lower than the LM2902-Q1 supply, then the output will be correct even if the non-inverting input is higher than VCC - 2V.
Both inputs should be lower than maximum VCC rating for device safety.

• Hi Ronald,

I remember the common mode voltage's formula is  as follows, isn't right?

Thanks & Best Regards,

Sherry

• Sherry,

The Vdiff formula is correct.
The Vcm formula is theoretically correct, but LM2902 and other PNP/PMOS based inputs don't work that way.
For these devices Vcm = lesser [VIN+, VIN-]

The reason is that when Vdiff is large (>0.1V) the lower voltage input get all the internal current flow and the higher input gets no current flow , therefore the voltage of the higher input is not a factor. Non factors should not be used in formulas.
• i Clemens,

Could you please help explain why I will get phase reversal when the input voltage is above VCC-2V? The LM2902's schematic is as below:

Thanks & Best Regards,

Sherry

• Sherry,

There might be phase reversal.  The key point is that the input transistors are cutoff therefore the input voltage doesn't not control output anymore. The output will be based on what the amplifier naturally does. Maybe high maybe low, although the majority will be one or the other.

2V from input to VCC is plenty to support 2 base emitter junctions and a current source. Higher than than that will eventually cause transistors to cutoff, turn off.

• Hi Ronald,

You mean when Vdiff is large (>0.1V) , for example, V+-V->0.1, then I2 will get all the internal current flow, and I1=0, so the higher voltage input can be set any value as long as lower than the maximum VCC rating. Is my understanding right?

Thanks & Best Regards,

Sherry

• Hi Ronald,

Now the condition is V-=VCC-2.4<VCC-2, so the negative input side transistors are on (as below). Due to V+>VCC-2, the IN+ cannot control output, and V+'s value is not sure, maybe high maybe low, so the comparator's output maybe 1 maybe 0, and this phenomenon is phase reversal, right?

• Sherry,

Regarding: You mean when Vdiff is large (>0.1V) , for example, V+-V->0.1, then I2 will get all the internal current flow, and I1=0, so the higher voltage input can be set any value as long as lower than the maximum VCC rating. Is my understanding right?

Yes.

• Sherry,

Regarding: Now the condition is V-=VCC-2.4<VCC-2, so the negative input side transistors are on (as below). Due to V+>VCC-2, the IN+ cannot control output, and V+'s value is not sure, maybe high maybe low, so the comparator's output maybe 1 maybe 0, and this phenomenon is phase reversal, right?

The IN- side can conduct current (and the IN+ side does not) so the output will represent that same result that happens when IN- is 0V and IN+ is 0.1V; That result is output high. No phase reversal risk.

To summarise, only one input needs to be less than VCC-2V for correct operation. The only caveat is that neither input can be less than -0.3V
• Hi Ronald,

Thank you very much!

Thanks &Best Regards,
Sherry