Getting true zero volts output from rail-to-rail op-amp

OBW0549

Joined Mar 2, 2015
3,436
Yes, though I don't know whether the LT1677 can get that close to ground.
According to the data sheet the LT1677 is a RRIO op amp whose output can get within a hundred mV of ground (or -Vs).

Emitter followers are a great way to make an oscillator though -- except in simulation.
A lot of people aren't aware of that, or overlook it. Syscompdesign.com has a writeup on curing circuit oscillation, and says this about emitter followers (emphasis mine):

There are some situations that should be an immediate red flag to a circuit designer where oscillation is concerned.

– Emitter followers have a tendency to oscillate. This is counter-intuitive, because the voltage gain of an emitter follower is less than one. But the emitter follower *does* have power gain, and for certain phase conditions it will oscillate. The solution is a resistor in the base lead of the emitter follower: a few hundred ohms for a single emitter follower, about 5000 ohms for a darlington pair.
I've been bitten by the emitter follower oscillation problem myself a few times when I wasn't paying attention...
 

Papabravo

Joined Feb 24, 2006
13,902
You still need to deal with the emitter follower's tendency to oscillate.

See the following based on Jim Williams EDN article "Test Your analog IQ", question #17


The conditions for oscillation are
  1. Loop gain ≥ 1
  2. Phase shift = n*360 where n ∈ {0, 1, ...}
 
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Papabravo

Joined Feb 24, 2006
13,902
The better way to do it would be to have a small negative supply. The input voltages can extend up to 0.3 volts beyond either rail. Put the diode between your virtual ground and the opamp power supply ground. The opamp sees +5VDC and zero. the rest of the circuit sees +5VDC and 0.7 VDC. Another way of labeling things would be 4.3VDC, GND, and -0.7VDC. Now your actual ground is within the common mode range. this will get you to about 100 μV above ground, which is better than a couple hundred millivolts above ground. If you really wan to get to ground the split supply is the way to go, The diode has problems when it transitions from forward to reverse bias.
 

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crutschow

Joined Mar 14, 2008
25,095
The disadvantage with the diode or the transistor at the output is that it prevents the op amp from pulling the load low, so the frequency response/slew-rate in the negative direction is limited by the load RC time-constant.
 

Thread Starter

AlbertHall

Joined Jun 4, 2014
9,986
The disadvantage with the diode or the transistor at the output is that it prevents the op amp from pulling the load low, so the frequency response/slew-rate in the negative direction is limited by the load RC time-constant.
Not a problem. The actual input will be essentailly DC from a current shunt and will feed an ADC. Frequency response is not an issue.
 

Papabravo

Joined Feb 24, 2006
13,902
With the transistor you are still 5-18 mV from ground even if you overdrive the input to -300 mV in the negative direction.

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If the input is limited to ALWAYS be ≥ 0 VAC then using a diode to separate the grounds is the winner in terms of "close to ground". That said I like this solution too.
 
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Papabravo

Joined Feb 24, 2006
13,902
The offset. Of course.
But what I was alluding to, was the ability of a plain vanilla opamp to swing to zero volts.
.. with a unipolar supply. They don't normally have that ability. With a bipolar supply, of course it is no problem. That is why I like splitting the unipolar supply in an asymmetric fashion.

draft (3).png

You can even have the input go a bit below the signal ground, but not too much. You can also do this with any device that allows bipolar supplies and any single supply device where the absolute maximum ratings are not violated. Now getting 0 VDC Input, to produce 0 VDC output is a separate problem from having the output reach 0 VDC. You may be able to introduce a separate nulling current to the summing junction, but as we all know with pots, things can drift. Today's null will be tomorrow's headache.
 
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