Practical Op Amp Fault Finding

Could you give some numbers--part numbers, component values, and some numerical indication of what the problem is?

If you're worried that you only get 0.25 V to 5.75 V on a 6V supply, that's probably an output loading issue--either the load resistance or possibly the feedback resistor is too small (not enough ohms). Check the data sheet for a plot of output voltage vs. current.

If it stops responding when its inputs are within a couple of volts of one or other supply rail (i.e. +6 or ground), your op amp is running out of input common-mode range, i.e. it's R-R output but not R-R input. (There are lots of those.) Check the data sheet for a plot of input offset voltage vs. common-mode voltage, or "Input common-mode voltage range" in the 'DC specifications" section.

One tip on understanding feedback amplifiers--don't fall into the common trap of thinking "the op amp tries to make its input voltages equal" or that sort of thing. It's the feedback connection that determines that--the op amp just takes its differential input and amplifies it by a large factor. If the inputs are wired up backwards (feedback to + instead of -) the op amp will 'try' to make its inputs as unequal as it can!

Cheers

Phil Hobbs

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"Lincoln Davies"

** Try posting this nut case shit on " sci.electronics basics " .

Least you will provide more entertainment for the trolls over there.

... Phil

It would be nice to know the Op-Amp number you are trying to use? And don't make that mistake of using a comparator as a op-amp. It'll work, as long as you pull up the output and account for some oscillation problems..

If the load on the Op-Amp is exceeding it's abilities, the op-amp can thus clip. This can be seen for example, if you were to drive a low R load or a load that has a large cap on the input that will require an initial surge of current before it settles down..

The device you are driving could also have a open collector device attached to it else where that is clamping low while your op-amp is running high.. The driven device could also have a limited clamp on the input..

Have you tested this output with it not connected to the device or next stage ?

Other issues are, oscillation in the op-amp that is causing gain issues at your DC or near DC inputs. That could be solved with a small loop back cap in the gain circuit of the op-amp. You really should scope this area looking for high frequency artifacts that shouldn't be there.

Also, there is the possibility that if you're using a bipolar type input op-amp, you are getting R.F. introduced in the input and it's getting rectified, thus offsetting the values. A good shielded lead that is only grounded on one end, can help with this, especially if this lead in a race way with other conductors that are inducing currents, which really isn't a good idea which is why they keep low voltage low currents in a separate race way. The currents will induce into the input of your op-amp circuit.. Just like having induction coupling (transformers) etc..

If you are seeing this at the bench in a clean environment, I would say you may want to have a bypass cap at the Vcc rail of the op-amp and a small R from the supply feeding that circuit to start with.. And maybe a small loop back cap in the (-) gain circuit of the op-amp.

This should get you started..

Thank you for your helpful reply. Parts like the TLV246X , OPA340 series, etc.

With these, I see no reason for a half-supply bias. So it seems more straightforward to my limited experience.

Generally, I am just outputting, DC coupled, to a Darlington type current driver, with feedback taken from the load connection.

Best regards,

Lincoln Davies

Then you're already loosing a fighting battle..

I can assume that you are using the darlington transistor with the Emitter as the output and feeding the collector from the (+) rail with the base from the op-amp output ?

Assuming this is what you are doing, you take a big hit with losses via the emitter junctions, especially the darington types..

Each Emitter path is doing a drop output voltage of ~0.7 volts. Using the darington, this most likely will add up to 1.4v loss since the double emitter paths, found in most darlingtons.

If you request demand on the op-amp for full output at the power junction of the darlington emitters, you won't get it.. The op-amp can Rail but then you need to subtract 1.4 volts on top of that ..

So, 12 volt supply may reach ~ 10.3v output at the darlington emitter how ever, if 10 volts is your goal for example, you'll to have enough current there to maintain it. You need at least 1.4 Volts or so driving the base above what you expect to get out of the emitter of the darlington.

You need a supply of like 15 volts at least if you want like a 0..10 V output etc..

What you need to do is use a PNP high side for a power output and drive that with a NPN, pulling down the base of the PNP with a pull up R for the base of the PNP. The base of the NPN is driven from the output of the op-amp and your loop back comes from the Collector of the PNP as this will also be the main output. Emitter of the PNP connects to your raw supply (+) voltage..

This is a current mode type of regulator and requires much less to drive it.. With this, you can make the PNP reach saturation if needed.

I suppose a schematic would be better how ever, LTSpice is all I have that would be the most possible thing I can offer because I don't know how those other guys get the ASCII text to work like that and I aint doing that by hand. :)

jamie

This introduces an extra delay around your feedback loop, and makes it much more likely to oscillate.

Darlington pairs often benefit from a resistor between the base of the second - driven - transistor and its base. Without it, when you are trying to turn the Darlington off, the Miller (collector-base) capacitance of the driven transistor can only discharge through the base-emitter junction, and this can take quite a while.

6k8 is a popular value - most people can afford to throw away 100uA to get a bit more speed out of their Darlington driver.

-- Bill Sloman, Nijmegen

Yep, that explains it ... exactly was I was getting. Thanks for explaining. Now I can stop trying to do the impossible. Electronics is hard enough for noobs as it is ;-)

I have access to LTSpice if you could possibly send the circuit. This is something I will use all the time. I can provide my email address off-group for the attachment..

Many thanks,

Lincoln

OK I've done some research. Appears to be a "high side" Sziklai pair. Wonder why you never see these useful things in the standard texts.

I am only expecting a 0.7 volt drop though, not lower. Will see when I breadboard it.

Thanks again,

Lincoln

VBE of the PNP plus VSAT of the NPN. ...Jim Thompson

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