Driving mcp601 opamp to near ground is possible?

[Flyback]

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:
https://ww1.microchip.com/downloads/en/devicedoc/21314g.pdf

 

[crutschow]

Actually, a lower value of resistance to ground should help pull the output closer to ground. You might try adding a 5k ohm resistor to ground.

 

[ericgibbs]

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:
https://ww1.microchip.com/downloads/en/devicedoc/21314g.pdf

hi,
I would say you could have a problem getting down to 10mV, look at this clip from the d/sheet.

You could try as crutschow suggests to get the 10mV, but it may reduce the close to Vdd rail output.
Try it and it let us know.

E

 

[Flyback]

I will have a look, but I worry that even if I can get down to 10mV on the mcp601 I have here.....what will the next batch be like?
I am not sinking any current into the mcp601 output, so from the graph Eric kindly provided, it does look like I can get down to 10mV...of course, I need it to be in the linear region at 10mV too.

 

[ericgibbs]

I will have a look, but I worry that even if I can get down to 10mV on the mcp601 I have here.....what will the next batch be like?
I am not sinking any current into the mcp601 output, so from the graph Eric kindly provided, it does look like I can get down to 10mV...of course, I need it to be in the linear region at 10mV too.

hi F,
Your cautionary note: I have here.....what will the next batch be like?, is a sensible approach.

If you are building a number of units, its a pain carrying AOT or SOT measurements.
I would be interested in what you do find regarding what the spread/typical output range really is during your testing.
This type of question often comes on ETO.
Eric

 

[audioguru]

It is a Cmos "rail-to-rail" opamp so of course the output goes to ground when it has no load or when its load connects to ground.
But the spec's and graph in the datasheet show the load connected to half the supply voltage (VL= VDD/2) which causes the output voltage swing to be reduced.
You have opamp U17 driving the inverting input of opamp U7 which is NOT a load to ground so the output of U17 cannot go to ground.

 

[Flyback]

thanks AudioGuru, I see your point, but I thought the bottom FET in the output totem pole of the mcpp601 had its source connected to ground?...and if that fet switches on fully, then the output of the opamp should be ground, in spite of the fact that its connected to the inv input of the other opamp?

 

[ericgibbs]

hi F,
I must admit I had not see your actual circuit diagram.!
I downloaded your PDF named 21314 ie: the actual MCP601 datasheet, but I thought the file you named MCP601 was just another copy of the datasheet.

I can see clearly the output will not pull to ground as 'agu' also states.

E

 

[Flyback]

So is the source of the bottom fet in the mcp601's totem pole not actually connected to ground?

 

[MrAl]

Hello there Fly,

There is a trick you can use with any op amp even one that can not get down to 0.1 volts, to get down very close to 0.000 volts, that works when the output load does not source any current or very low current.

The trick is to connect a diode to the output of the op amp and then use a pull down resistor to ground to provide the low level output voltage rather than having the op amp itself do the work.

With a diode on the output when the op amp output goes positive, the circuit output goes positive, as usual. When the op amp output tries to go to ground, it reaches a level of about 0.5v in the linear mode and with the diode drop that means the circuit output is at 0.000 volts. The op amp feedback comes from the circuit output not the op amp output, so the op amp keeps the op amp output at 0.5v not at 0.000 volts, but again the output is 0.000 volts so it provides a near zero output.
The output will still not be 0.000000 volts though because the feedback resistor will keep the output biased a little higher, so the tradeoff is to use as high impedance feedback as possible and as low a pull down resistor as possible. With a typical 100k feedback and 1k pulldown and 2.5v op amp input, we might see around 0.025 volts on the output, so you'd have to do better than that. With 500 ohms pulldown, that comes down to about 12.5mv, and with 200k feedback that comes down to 0.006mv, so you get the picture.

Oh yeah, the other tradeoff is we loose output at the top end. If we originally had a full 5.000 volts output max when the op amp output goes to a high level, now we only get about 4.5 volts output when using a Schottky diode for the output diode.

 

[Flyback]

Hello Mr Al,

So you mean like in the following "SEPIC" schematic, (where the concerned opamp is now U5) ....I could change it to the other schematic "SEPIC with diode"? (shown below)

 

[audioguru]

Opamp U7 is inverting so when its output goes to ground then its inverting input is positive and sources a current through R16 and R19. Then the output voltage of U17 is above ground.

Wait a minute. Opamp U7 does not have any DC negative feedback so there is no DC current in R16 and R19.
Since U7 has no DC negative feedback then its input offset voltage is amplified a lot and forces its output to be saturated high or low.

 

[Flyback]

U7 is an error amp, its required to have extremely high gain at DC.

Since U7 has no DC negative feedback then its input offset voltage is amplified a lot and forces its output to be saturated high or low.

...but the negative feedback stops that......the circuit does work, its been running for months now in trials.

 

[audioguru]

U7 is an error amp, its required to have extremely high gain at DC.

...but the negative feedback stops that.......

But U7 has no DC negative feedback. The maximum input offset voltage of 2mV is amplified 100dB which is 100,000 times. The output will be deeply saturated high or low.

 

[Flyback]

I don't think its that high due to GBW limit etc...that cct works fine, the micro can use the error amp to give led output of 1w, 2w, 3w, 4w, 5w, and that's all we need.........seriously it works fine....has been for months and months...the error amp is just like any other in any smps...no dc fdbk path.

 

[audioguru]

Isn't this a linear circuit? It is different when it is a switching circuit.

 

[Flyback]

yes its a linear opamp control circuit....the smps is a sepic

 

[MrAl]

Hello Mr Al,

So you mean like in the following "SEPIC" schematic, (where the concerned opamp is now U5) ....I could change it to the other schematic "SEPIC with diode"? (shown below)

Hi Fly,

Yes, that is the basic idea. However, as i mentioned in my previous post there are some important issues that have to be dealt with.

First, the output resistor to ground has to be low enough to swamp any effects from the output load because the op amp section no longer has the ability to pull the output down. The equivalent resistance could be as high as 500 megohms depending on the diode used (like 1N4148).
So that means we have to pay strict attention to the detail of what exactly is going to pull the output down now that the op amp output no longer has any pull down control over the output node we are now using as the op amp 'output'.
This could be as simple as calculating the current through a resistor and making sure the pull down resistor can sink that current and still provide 0.005 volts at the output. But with capacitance as part of the load, we'd have to make sure it can discharge the cap in a fast enough time to not bother the dynamic aspects of the design, which could be dramatic depending on the required feedback response.

For your circuit with U5 and R14 and C6, you should calculate the max voltage that can appear across C6 and then calculate the max current through R14, then make sure that R4 can sink that current and still provide say 0.005v output level. SO if the max voltage for C6 was 3.3 volts with the 3.3k resistor that means a max current of 1ma. And 1ma through 51k is 51 volts, so that's not going to cut it
To get 0.005v with 0.001 amps, a 5 ohm resistor would be required. Obviously that's too low for the U5 op amp to handle, so we'd have to impedance scale that part of the circuit. For example, with R14=33k, we'd need a 50 ohm resistor. With R14=100k, we'd need a 150 ohm resistor. So you see how that works. But then the time constant is altered, so we'd have to change C6 to a smaller value. But then the following time constant is altered, so we'd have to increase R15 and recalculate R16 and whatever other parts are associated with that circuit section.
So it gets a little more complicated with this circuit, but you get the idea