# What does it mean to AC couple a lm358?

Discussion in 'General Electronics Chat' started by doug08, Oct 31, 2011.

1. ### doug08 Thread Starter Member

Jan 30, 2011
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I have a circuit that has the tendency to become unstable at times/temperature. I was told that if I modified the circuit to AC couple the comparator stage of the lm358, instead of DC as it currently is.....that would be the solution. Please explain. Thanks.

2. ### crutschow Expert

Mar 14, 2008
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What do you mean "unstable"?

AC coupling may or may not help it depends upon the circuit application.

The LM358 does not have a "comparator stage", it is an op amp. Are you referring to the differential inputs?

Post a diagram of the circuit you are using,

Dec 26, 2010
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AC coupling most often refers to capacitive coupling, where the DC component of a signal is blocked by a series capacitor, leaving only the AC content.

Whether or not this would be at all useful in your application would depend on what it is. Please post a schematic so that we can see what you are doing.

4. ### doug08 Thread Starter Member

Jan 30, 2011
153
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Here ya go. Works very good, but can become unstable. The designer said to AC couple the comparator.

Dec 26, 2010
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Well, you could try a blocking capacitor of say 10nF between the output of the first amplifier and the input of the second. You would then need to supply a DC bias path for the input at pin6, perhaps a potential divider made of two 47kΩ resistors to put it at Vs/2.

I don't really know if that would work, and the values are more or less guesses, but I would think this is the sort of thing he had in mind.

Whether that circuit will really work as a detector is another matter. Is the BPW34 supposed to detect radiation by itself, or is it "seeing" scintillations from something?

Jan 30, 2011
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7. ### ErnieM AAC Fanatic!

Apr 24, 2011
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The designer states the BPE34 is doing the detection. As drawn, JFET PN4117 is normally full on, so a gamma ray must cause a burst of current to turn the FET off, giving a positive pulse to pin 3 of the op amp. That amp stage has a very high gain of 1M/1K = 1,000.

Such an amp can be upset by many things. The LM358 has an input offset voltage of 3 to 7 mV at room, send that thru a gain of 1,000 and you get 3 to 7 Volts into the comparator. And while the input bias current is pretty low the inputs are seeing different impedances (56K vs 1K).

For a cap to help I would place it between the 56K and pin 3 of the op amp. At the same time I would add a 1K from pin 3 to ground to give the input current a path.

8. ### doug08 Thread Starter Member

Jan 30, 2011
153
2
I have all the parts. I will put it all together later. I'll let you know! I did look at the data sheet for the lm358.....it shows the ac coupling, but I still did not have a clue before I posted this. What you said to do, almost matches the data sheet. Thanks for the help.

Dec 26, 2010
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I suspect that as low a resistance as 1k to ground to from pin3 would load down the pulses from the FET far too much, and might stop the circuit working.

The DC resistance seen by the bias current from the inverting input is actually 1.5Meg, so there is no reason to set the non-inverting ground return as low as 1kΩ in a quest for DC balance.

Jan 30, 2011
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from pin 3.

11. ### ErnieM AAC Fanatic!

Apr 24, 2011
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Yeah, 1K is a big load from a 56K source. It's never good idea to drive a cap without a place for the DC bias current to go. As complete fix would mean a lot of redesign (two op amp gain stages and a transistor for the comparison) just try another 56K here. The DC bias is still imbalanced but just as bad as where it started (which works most of the time anyway), though it does half the gain to that point.

(doug may wish to ignore the following
The (-) input impeadance is the 1K in parallel with the 1.5M (not 1M as I said previously), which is still about 1K. It's a Thevenin equivalent where the output voltage is treated as a voltage source and thus shorted to ground, just leaves the 2 resistors.

Dec 26, 2010
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@ErnieM: As far as AC impedance goes that's right, but the 1kΩ is DC blocked, so the bias current for the LM358 all has to come via the 1.5MΩ. This thing started out so far from being balanced that it should be no surprise if it drifts.

I would be more interested to know how anyone could guess the actual sensitivity of the BPW34 photo-diode as a detector of anything other than the visible light or near infra-red for which it was designed.

http://www.vishay.com/docs/81521/bpw34.pdf

13. ### MrChips Moderator

Oct 2, 2009
12,622
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I believed the BPW34 is not being used as a photo-detector but I'm only guessing. The diode is reverse biased so the FET is turned off. I can only guess that the diode is supposed to momentarily conduct sending a low going pulse into pin 3 of the LM358.

If I have time I could breadboard this circuit and report back.

Ding-dong! There goes the door bell again. Darn goblins. I hate halloween!

Last edited: Oct 31, 2011
14. ### Ron H AAC Fanatic!

Apr 14, 2005
7,050
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PN4117A Idss is 90uA max, so Vd(min) is about 3.25V, if the diode dark current is zero (which it won't be). BPW34 dark current (Iro) can be as high as 30nA. If Idss and Iro were both max, the JFET COULD be turned on hard, but with typical values of Idss and Iro,
I would think that ionizing radiation would cause a burst of leakage in the diode that would turn the JFET on even more, resulting in a negative pulse at the drain.
The DC gain of the op amp is one, due to the 15nF capacitor. Input offset is therefore basically irrelevant. The big problem for the direct-coupled comparator is temperature-induced leakage in the diode, and drift in the JFET.
I would put the cap between the op amp and the comparator. See attachment.

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15. ### doug08 Thread Starter Member

Jan 30, 2011
153
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Question. Why is there an up facing arrow on the right side of the 100k pot(now where your 10k pot is)? Just ignore it...correct?

Thanks.

16. ### SgtWookie Expert

Jul 17, 2007
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That up-facing arrow means it should be connected to +9v

17. ### doug08 Thread Starter Member

Jan 30, 2011
153
2
but usually a schematic would show a +9v next to the arrow, or a triangle with a +9v next to it instead of just an arrow.

18. ### Ron H AAC Fanatic!

Apr 14, 2005
7,050
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You originally posted the schematic. I just edited it. It didn't occur to me that you didn't know what the arrow meant.
At the top of the schematic, there is an arrow labeled +9V.
When I generate a schematic, I label all power supply nodes, and I use a more common symbol for ground.

19. ### doug08 Thread Starter Member

Jan 30, 2011
153
2
Does not work right....very irritating. I have pics. The tube assembly is perfect(according to the creator), could not be made better. I have the rest of the circuit assembled on a breadboard. Tried 2 different times making it....same result. poor detection if any.

20. ### doug08 Thread Starter Member

Jan 30, 2011
153
2
It now works fine. I kept playing with the 100k pot until I got it right. Next step is the modifications. I can post a video link.