# Driving opamp output to near ground?

#### eem2am

Aug 3, 2009
429
Hello,

I need our (single supply) MCP601 opamp follower (U17 in below schematic) to be able to drive its output down to as low as about 10mV above ground….and still be operating in the linear region.

However, page 2 (near bottom) of the MCP601 datasheet states that MCP601 can only drive to within 100mV of ground..However, this is when there is a resistance to ground of 25K.
In our schematic, ,we have at least 100 times more than 25K resistance to ground, so I presume I will be able to drive the MCP601 output down to 10mV above ground?
(I presume the “VL” in the datasheet means “ground” for single supply connection?)

MCP601 datasheet:

Schematic:
http://i40.tinypic.com/2nusiys.jpg

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• mcp601.jpg
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#### duke37

Jan 9, 2011
5,364
To get the output closer to ground, you will need to lower the resistance to ground, not increase it.

I could not find 2.5M resistance on the output of the op-amps in your circuit.

U17 has the two inputs connected with a 1k resistance, how is this supposed to work?

#### eem2am

Aug 3, 2009
429
U17 has the two inputs connected with a 1k resistance, how is this supposed to work?

...sorry,i don't see that?

"2.5M was just a guess at the resistance it will probably see....ie a big number"

#### Laplace

Apr 4, 2010
1,252
With a 25 KΩ load resistance the output current will be 100 μA and the specified headroom is 100 mV. Yet Figure 2-20 in the datasheet is a plot of output headroom versus current showing an output voltage of less than 10 mV with output current at 0.1 mA. Why the difference? The 100 mV is specified and tested whereas the 10 mV value is 'typical'. So I would venture to say that if the output current is kept below 10 μA then this op-amp's output can be driven to within 10 mV of either supply rail.

#### KrisBlueNZ

##### Sadly passed away in 2015
Nov 28, 2011
8,393
I think it's a miscommunication in the data sheet.

In the first two lines of the "Output" section, the author of the data sheet has combined VOL and VOH onto the same line, and the second line, with the lower RL resistance, has the greater dropout voltage for both VOL and VOH. That is, 45 and 60 mV from the appropriate supply rail, compared to 15 and 20 mV for the higher RL in the first line.

But the "Conditions" field says that RL is connected to VL (i.e. 0V for a single-supply circuit). That doesn't fit with the lower RL value causing a higher dropout voltage for both VOL AND VOH. Increasing the load current drawn towards VL would increase the dropout voltage for VOH, but would DECREASE the dropout voltage for VOL.

I think the "Conditions" on those lines should read "RL = xx kilohms to THE OPPOSITE RAIL", not "... to VL" because that is the important factor for dropout voltage - the current that the load draws AWAY from the rail that the output is trying to drive itself to.

Current that the load draws TOWARDS the rail that the output is trying to drive itself to will HELP the output get closer to that rail. So if the specification table was correct, and the load resistance is really connected to VL in all cases, using a lower load resistance would REDUCE the VOL figure; it would not increase it, as the table shows.

This error could have easily come about if a draft or earlier version of the data sheet had the VOL and VOH numbers on separate lines, and they were combined onto single lines to save space without thought for the different direction of output current in each case.

If I'm right about this, if you load the output towards 0V with a 5k load resistor, the guaranteed maximum VOL will be even lower than the 15 mV specified on the first line in the "Output" section, where the output is pulled towards V+ with a 25k load. In fact, even using a 25k load resistor to VL would probably be enough to guarantee VOL less than 10 mV. (Assuming there is no other circuitry connected to the output that feeds current into the output; all loading must be towards VL.)

If you want, you could email this explanation to Microchip and ask them to confirm it and correct the data sheet. I think they have an email address for reporting data sheet errors.

#### Laplace

Apr 4, 2010
1,252
Check more closely. The Electrical Specifications heading clearly states that VSS=GND and VL=VDD/2.

#### eem2am

Aug 3, 2009
429
"VL=VDD/2" confuses me, because most users are using the mcp601 as a single supply opamp, and won't have a vdd/2 rail.

incidentally, I had a telephone interview a few months back, and he asked me if I could drive an opamp's output all the way to either rail with a cmos opamp...I said no, and he said I was wrong.....that you could drive all the way to the rails at output. I don't believe that one can ever drive all the way to the rail, even say if you are driving to gnd on a single supply opamp, and sinking zero current into the opamp output, I still don't believe you can drive it all the way to ground. Also, its important to know where linear operation stops, because I don't want to drive it out of the linear region.

#### KrisBlueNZ

##### Sadly passed away in 2015
Nov 28, 2011
8,393
Ah, thank you Laplace. That explains it. So there's no error in the data sheet.

I think it's fair to assume that if your load resistance is returned to the op-amp's negative supply rail, the output will be able to get closer to that rail than the specifications show.

eem2am, many single-supply designs DO have a VDD/2 rail, but yours doesn't. Still, they have specified the output characteristics using a load that's returned to VDD/2.

I guess it should be possible for a device with a MOS output stage to pull fully to either supply rail with absolutely no load current. It would depend on the design of the output stage. Also, the op-amp might not be operating linearly at that point. This might be useful in practice if the op-amp is being used as a comparator, perhaps?

That's a good point about the linear region limitation. The spec shows that the linear region only extends to 100 mV from each supply rail. But again, if you load the output with a resistor to the negative supply rail, I would expect the linear region to extend closer to the negative rail as well, especially if the load resistor is a relatively low value.

Of course, loading the output heavily to the negative rail will increase the dropout voltage from the positive rail.

#### eem2am

Aug 3, 2009
429
Thanks KrisBlueNZ , Duke and Laplace..i see now.

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