Clamping an op-amp output

[ThomasHenry]

Hello all,<p>Here's a problem I've been struggling over. I'm hoping someone might have an idea that will put me back on the straight and narrow.<p>I have a single op-amp stage, configured as a summer. It's duty is to mix a variety of control voltages, any of which might range from -15V to +15V, and present the sum to a voltage controlled filter.<p>The gain elements in the filter must never see a voltage below ground or the entire unit becomes unstable. So, I'm trying to come up with an elegant way to ensure the op-amp output (representing a sum of various control voltage inputs) never falls below ground.<p>In a nutshell: as long as the various inputs sum to 0V or higher (after the inversion of the op-amp mixer) the output is passed along. But if the mix creates something that would be negative, then I want the output clamped to 0V and have it stay there until the mix goes positive again.<p>If diodes and op-amps were perfect, then a diode in the negative feedback loop of the op-amp summer would do the trick (anode to the inverting input). However, with a real-world diode like a 1N4148, the output actually goes to around a negative half-volt before the clamping action kicks in. Even -0.5V is enough to cause the following VCF to become upset. (I'm using gain stages in the VCF that attenuate for positive control voltages and boost---up to +20dB) for negative control voltages. I'm trying to stay away from the boost, for stability.<p>Obviously, I'm hoping for an elegant solution (translate: no more op-amps).<p>Any ideas?

[Dean Huster]

It may work to simply make the -V supply of the op amp ground rather than -15v or whatever you have, assuming you're running a bipolar supply. All the op amp can do then is output positive voltages. In this regard, the op amp output will be self-clipping, going from zero to whatever +V is. I think the summing operation will still work OK, even with negative inputs.<p>Dean

[Chris Smith]

Did you try pre charging the diode, so that it sits at a slightly higher than zero voltage all the time?

[ThomasHenry]

Many thanks to Dean and Chris for their rapid replies.<p>Dean: I'd like to use multiple packages if possible (like the four-banger TL074), of which the control voltage summer would be one element. So, using a single supply instead of a bipolar one doesn't work well in this application (since the filter itself requires bipolar action). However, I'm going to think this over a bit---your idea would certainly be elegant, if I could get the rest of the circuit to cooperate.<p>And to Chris: The diode is in the feedback loop of the op-amp, so it's not clear to me how to pre-charge it. I have used that idea before, where the output of an op-amp was driving a diode into one supply or the other and it works great. (I used it in a ramp-to-triangle converter a couple weeks ago to minimize diode switching time). But like I say, with the diode in the negative feedback loop, I'm not sure how to bring it closer to conduction what with a variety of mixed voltages possible at the input.<p>To be more precise: imagine an op-amp with a 100K to inverting input, non-inverting input grounded, 100K in parallel with a 1N4148 in the feedback loop. How can we keep the diode closer to conduction under these circumstances?<p>Again, thanks for the suggestions. I was afraid I hadn't made the description of the problem clear enough, but both of your ideas indicate you understand exactly what I'm up against.

[cato]

I think you're looking for a precision rectifier. It'll cost you an extra diode. So, take off one elegance point if you will. Here's a web page:<p>http://www.ecircuitcenter.com/Circuits/ ... tifier.htm

[MrAl]

Hello there,<p>Great ideas.<p>Another idea, if the circuit following the
op amp can handle a voltage offset is
to use a resistive divider from the output
to a regulated source supply. If the two
R's are the same, when the op amp swings
negative 14 volts or so the junction of
the two resistors swings to zero volts.
The ac gain of the two resistors would be one
half.<p>There's hardly anything simpler then two
resistors, but the resulting output will
have a large offset dc value that the circuit
following will have to be able to handle.
If it's ac coupled, there's no problem <p>If you use the precision half wave circuit or
any other circuit that causes the op amp to
go out of the linear region then you must also
check for response time when the device tries
to come back from 'saturation'. This response
time can be MUCH longer then the speed of the
op amp because the speed specs are produced while
the op amp is in the linear mode. This extra
time can produce very bad distortion or prevent
the circuit from working at all depending on
operating frequency and op amp bandwidth.<p>I guess in addition to the two resistors idea
(which keeps the op amp in the linear range
for signals -14 to +14 volts)
you could add a level shifter using one of the
remaining op amps which would then give
you roughly 0 to 14 volts output for -14 to
+14 volts input.<p>Take care,
Al

[ThomasHenry]

Hi gang,<p>(Slap)...if you can hear that, that was my hand slapping my forehead. Sherlock Holmes (in one of those cheesy Rathbone movies) said "There's nothing so deceptive as the obvious fact."<p>Well, Cato went right to the nub, and as soon as I saw his comment, the slapping began...<p>When I started thinking about this project, the notion of precision rectifier came to my mind early on, but I dismissed it at once---for I was imagining a full-wave rectifier which clearly wasn't what I needed. I then moved on to more complicated ideas, completely missing the notion of half-wave rectifier.<p>As Cato notes, that's exactly what I was looking for even though I stated the problem in a far more wordy way that concealed the true nature. In short the solution to the problem is easy: just build a precision half-wave rectifier. All it takes is two diodes, plus the usual input and feedback resistors. Now that's elegant to my mind!<p>I have always tried to follow the dictum: never open a can of beans with a stick of dynamite. I am grateful to Cato for stomping out the fuse and handing me a can opener instead!<p>Many thanks to all who contributed, and especially to Cato for drawing my attention to the precision half-wave rectifier.

[cato]

de nada