comparator circuit

[ramuna]

Hi Joe,
As far as the Ramura's problem of the circuit not working ... I'd like him to post his circuit.
See attached schematic.

 

[MrAl]

Hi again,

Yes you are right Joe, i read the whole thing wrong
I also did not see the first statement up top.
I also just noted that they seem to be indicating two separate IC packages even though each package contains 4 comparators each.
The fact that it turns OFF with too low a voltage (too negative) means it can not be that the opto output is shorted.

Anyway, it looks like the bottom comparator is bad, being stuck 'on' and thus always applying a negative cathode drive for the input of the opto.

 

[JoeJester]

Ramuna,

You are missing the input to the comparators. You have the reference established, but no input signal.

Like most school assignments, they are not using standard resistor values, like the E24 resistor series.

 

[JoeJester]

Al,

That was my conclusion, the bottom comparator is stuck in the "saturation" mode, the negative 15 Volts.

 

[ramuna]

Hi Joe,
Correct me if I'm wrong, but I interpreted post #10 to mean that the faulty component is R4. That when R4 is increased to 90 Mohms (ie an open circuit in place of R4), CR1 will light. Thats why I said what I did in post #13, did the physical test whose result was reported in post #15, using the circuit posted in post #21.

You are missing the input to the comparators. You have the reference established, but no input signal. Indeed so. How else can we test the assertion made in post #10, of an open circuit in place of R4 (please correct me if my interpretation of that assertion is wrong) ?

 

[MrAl]

Hi,

Well if R4 is open how do you explain that we get the LED to turn on with some inputs and off with other inputs? I dont think the circuit would respond in any way at all with very high R4. That was what Joe was pointing out.

 

[ramuna]

Well if R4 is open how do you explain that we get the LED to turn on with some inputs and off with other inputs? I dont think the circuit would respond in any way at all with very high R4. That was what Joe was pointing out.

I only wish Joe had pointed this out in his post #11 ... that would have meant I had no need to do the time-consuming physical test. Anyways, I'm glad that we are all in agreement, even if it took a short detour and a slightly round-about way to get there in the end .

 

[MrAl]

Hi,

Well that's the most likely, or else we have to believe that the resistor R4 can fail in a way that is perfectly within a range of high resistances that cause only one comparator to switch from saturated output to open output (the upper one). It would also have to allow the correct turn off. That seems less likely. With completely open R4 though the two outputs should be saturated (about -14v out of both comparators).

Were you saying that you actually hooked up this circuit with real parts to try it? That would be great.

 

[ramuna]

MrAl said: Were you saying that you actually hooked up this circuit with real parts to try it? That would be great.

Yes that is exactly what I did, as mentioned in post #15, which is why I got a bit confused with all the JoeJester & MrAl posts since then. I mean simulations have their place, but the ultimate proof is the physical test. And per that test, the R4 being busted = CR1 lights theory was killed straightaway.

 

[Zabb Csaba]

If the LM339 model is the same, the different simulation software give the same results, so there is no difference here.
The main point that the input bias currents of the comparators generate voltage on the examined resistor. These currents are quite low so it needs a high value (but not infinite) resistor to have this voltage between 6V and 7V.
In practice some of them differ from the simulation because the specified current range is wide and it is something else by every piece (?-25nA-250nA). So the value of the resistor will be different but there is always a range where the voltage is between 6V and 7V. That time the LED is active.
This can be some hundred Mohms. So everything works fine, the comparators are not faulty.
The bias current in the simulation model is ca. 50nA. The indicated resistor range is related to this.

It is quite difficult to try it out with high value resistor. But you can use a good quality (film) capacitor instead of it. The bias current charges the capacitor. You can see the result here.



**broken link removed**

Regards,
Csaba

 

[JoeJester]

I will agree that nano amps through a hundreds of megaohm resistors will product the necessary voltage to energize the lamp. R4 would be the last of the suspects on the list.

The circuit as designed, you can sweep the DC from Vee to Vcc. The window is specified by the divider string. With that in mind, sweep the circuit with R4 at 90 Megaohms. You will find the LED energized from Vee to Vcc. The first part of the problem statement stated the LED was not energized when the input was more negative than the window, but was energized at zero volts.

If you want to build it, just remove the lower comparator and place the diode at the negative voltage. You should be able to replicate the problem statement.

If you change Vee or Vcc, you might want to recalculate the window resistor.

If I were troubleshooting the problem, I would sweep the circuit to confirm the specifications. R4 resistance being changed is low on my list of suspects. Confirmation of the others would take precedence.

 

[Zabb Csaba]

The LED is not energized when the input is more negative than the window (6-7V), but is energized at zero volts.(?)

I'm sorry, I don't understand exactly...

Regards,
Csaba

 

[MrAl]

The LED is not energized when the input is more negative than the window (6-7V), but is energized at zero volts.(?)
View attachment 87052
I'm sorry, I don't understand exactly...

Regards,
Csaba

Hi,

Yeah, but doesnt that switch too soon? I would think that "too negative" means -7v or more, but with the 90M resistor it switches at around -5v, which in itself would not be working then. -5v is not "too negative", only -7v is too negative. That's the way i see it.

But if you want to define "too negative" as -5v, then you have to tell us why they designed the circuit to switch at -7v instead of -5v.

If you insist, then you should also realize that it would be very hard to get the resistor to fail within a certain range of resistance even if we accept the -5v switch point idea. I guess it could happen though. Resistors i have seen from the past usually go out of tolerance, but not by more than 100 percent. If the resistor breaks for some reason then we have to believe that it still makes some contact. Could happen i guess, but this is still if we accept the -5v idea as being ok.

 

[Zabb Csaba]

Hi Al,

Thank you very much. It's clear to me now (my English...).
I thought that the -5V is "too negative" compared to zero for the 1V window comparator.
Sorry.

All the best,
Csaba

(I have seen broken SMD resistors in this range /100Meg/.)

 

[JoeJester]

Csaba,

By broken do you mean cracked?

 

[Zabb Csaba]

Hi Joe,

Yes, but "invisible" cracked.

Regards,
Csaba

 

[ramuna]

Hi,
This discussion is beginning to resemble a farce ! Please refer to the first line of post #9, which, for your convenience, is reproduced below:
According to R5, R6 & R7, U2A output is high & U3b output low only when the voltage at U2a pin 4, V_in satisfies 6v < V_in < 7v (both bounding voltages w.r.t. ground).
Now, with reference to the original Problem 13 schematic posted by the OP, with TP1 = 0V, U1a's output will be 0V. You can vary R4 from 0 to infinity, but this would make not the slightest bit of difference to U1a's output of 0V. And since this output of 0V fails to satisfy the boundary condition of 6v < V_in < 7v, neither the LED in the opto, nor CR1, will turn on.

Just to ensure that this is indeed the case in reality, I physically connected the joined inverting inputs of my real live & practical (as opposed to simulated) test circuit (schematic loaded in post #21), to ground via a 20 Mohm resistance. As expected, the LED did not turn on.

Hopefully this result will end this futile discussion once and for all (amen) !

 

[audioguru]

The inputs of an LM339 comparator uses PNP transistors. With the inverting inputs floating then the inputs float high causing the output to go low. Both outputs will be low so the LED does not light.

 

[MrAl]

Hi Al,

Thank you very much. It's clear to me now (my English...).
I thought that the -5V is "too negative" compared to zero for the 1V window comparator.
Sorry.

All the best,
Csaba

(I have seen broken SMD resistors in this range /100Meg/.)

Hi again,

Well i think your view was still interesting as to what you found about the resistor value. It's usually good to know more about something than less

 

[MrAl]

Hi,
This discussion is beginning to resemble a farce ! Please refer to the first line of post #9, which, for your convenience, is reproduced below:
According to R5, R6 & R7, U2A output is high & U3b output low only when the voltage at U2a pin 4, V_in satisfies 6v < V_in < 7v (both bounding voltages w.r.t. ground).
Now, with reference to the original Problem 13 schematic posted by the OP, with TP1 = 0V, U1a's output will be 0V. You can vary R4 from 0 to infinity, but this would make not the slightest bit of difference to U1a's output of 0V. And since this output of 0V fails to satisfy the boundary condition of 6v < V_in < 7v, neither the LED in the opto, nor CR1, will turn on.

Just to ensure that this is indeed the case in reality, I physically connected the joined inverting inputs of my real live & practical (as opposed to simulated) test circuit (schematic loaded in post #21), to ground via a 20 Mohm resistance. As expected, the LED did not turn on.

Hopefully this result will end this futile discussion once and for all (amen) !

Hi,

I myself did not see it as futile but actually interesting. Csaba brought up the point that maybe with the right value resistor this problem would look almost the same. The resistor must be higher than 20M however to see anything near this, so it may be hard to test for this condition. Maybe 100M or 150M might start to show this 'other' problem, but the question is then we need a very large resistor value which im not sure we can get in real life. Maybe some reversed biased si diodes in parallel ?