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walid

OVER-/UNDER-VOLTAGE PROTECTION CIRCUIT ANALYSIS

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Some electric motors draw more current when the supply voltage is low. Then if the voltage is too low the motor will burn out.


This is true only under certain load conditions, not without or with light loads!

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Hi
For a comparator : What will be the form of the o/p when exactly equal voltages applied on the inputs.
Also, what the less difference between the two inputs to turn the o/p of the comparator to another, thank you for your attention

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For a comparator : What will be the form of the o/p when exactly equal voltages applied on the inputs?

It depends on the input offset voltage of the comparator.
An LM393 dual comparator has a typical input offset voltage of 1mV (5mV max) and a typical voltage gain of 200,000. Its output would try to be 200V! Its output would be saturated at the positive supply voltage or at ground. If you use a comparator with zero input offset voltage then its output would be full of full-output square-waves noise due to its extremely high voltage gain.

Also, what the less difference between the two inputs to turn the o/p of the comparator to another?

It depends on the voltage gain of the comparator and on its supply voltage.
An LM393 dual comparator has a minimum voltage gain of 50,000. So for its output to change 15V then the voltage between its inputs must be 15V/50,000= 300uV.

Its typical voltage gain is 200,000. Then the input must be only 75uV. But such low levels would be covered up with noise.

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hi guru


(1) Is all of an op-amps suitable for use as a comparator or is that only certain types?
An opamp has frequency compensation so it can work with negative feedback without oscillating. The frequency compensation makes an opamp slow to switch its output.
A comparator doesn't have frequency compensation so its output switches very quickly. Therfore a comparator cannot have negative feedback.
There are many slow applications where an opamp can be used as a comparator. 

It is very good answer.
I read somewhere that op-amps designed to be a comparator has an o/p stage like open collector transistor or a darlington pair. is this true?

AND what is the frequency compensation ? No thanks I found it in: http://en.wikipedia.org/wiki/Frequency_compensation
thank you guru

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Hi guru


Hi Walid,
The transistor is not straining so hard without the LED. With the LED, its load was 60mA for the relay and about 25mA for the LED if it had a current-limiting resistor. The total collector current would have been 85mA so the 8mA base current is fine. Without the LED then you can increase the value of R4 a little for a base current of 6mA. But 2mA doesn't matter.

The transistor was poorly chosen as a BC547 with an absolute max allowed collector current of only 100mA. Its current gain drops above only 30mA. I would have used a relay that uses less current or a higher current BC327 or 2N4401 transistor. Their saturation voltage is much lower so the base current can also be much lower if you want.

BC547 max allowed collector current is 500mA and not 100mA
BC327 is pnp tansistor and not npn
2N4401 is 1000mA but the same power of 625mW
here is a table of some suggested transistors, i see all are suitable.
thank you guru

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Hi Walid,
You are correct, I was wrong. A BC327 is a PNP transistor. I haven't used BCxxx European transistors for about 35 years and I forgot.

Your table does not have correct data. It has the current ratings for the transistors much too high. Check the datasheets for correct max ratings.

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It depends on the input offset voltage of the comparator.
An LM393 dual comparator has a typical input offset voltage of 1mV (5mV max) and a typical voltage gain of 200,000. Its output would try to be 200V! Its output would be saturated at the positive supply voltage or at ground. If you use a comparator with zero input offset voltage then its output would be full of full-output square-waves noise due to its extremely high voltage gain.




It depends on the voltage gain of the comparator and on its supply voltage.
An LM393 dual comparator has a minimum voltage gain of 50,000. So for its output to change 15V then the voltage between its inputs must be 15V/50,000= 300uV.

Its typical voltage gain is 200,000. Then the input must be only 75uV. But such low levels would be covered up with noise.



The LM393 is an open collector device!!!

guru, you may want to clear up a few things in your statements above, in regards to gain and open collector devices, so that walid doesn't get the wrong idea!!

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An LM393 comparator has a typical voltage gain of 200,000 when it has a load resistor (National Semiconductor calls it RL) of 15k or more and a 15V supply.
If the supply voltage for the load resistor is less then the voltage gain will be less. If the supply voltage for the load resistor is more then the voltage gain will be more. The voltage gain changes because of the change in the internal emitter resistance of the output transistor.

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The LM393 is a high gain comparator. It is not a low distortion linear amplifier.

None of its transistors have negative feedback for linearity. The circuit in the IC probably is as non-linear as an ordinary transistor that doesn't have negative feedback like this one:

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Guru

I already know the answers to the questions I asked!!!
I ask them because of statements like...

An LM393 dual comparator has a typical input offset voltage of 1mV (5mV max) and a typical voltage gain of 200,000. Its output would try to be 200V! Its output would be saturated at the positive supply voltage or at ground.


You state that it's output would try to be 200V? I don't think so...even it were to have a Vcc of 1000V!!! You also imply that it's gain is related somehow to what the output voltage will be... nope, not happening!!!

The gain number is useful for one thing however, calculating the "window of uncertanty" as to when the comparator wil "flip".

The point is that this open collector comparators gain has ABSOLUTLY NOTHING to do with how high the output voltage will go!!!

Your the one applying linear properties to non-linear deveices... not me!!

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The voltage gain is too high in a comparator to use its full open-loop gain with slowly changing inputs. It will amplify its input offset voltage and amplify its internal and external noise. It will probably oscillate due to the input picking up the signal from its output.

So hysteresis is used in a comparator circuit to provide a snap-action to its switching. Then noise and oscillation are avoided but the sensitivity is much less.

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By perpetuating your  "smoke and mirror" answers, you demonstrate your "real knowledge" of the subject!! Those that know, know... unfortunate for those trying to learn.

Statements that try to link open loop voltage gain to what level a open collector comparator will go is pure BS!!

Rambling on about

The voltage gain is too high in a comparator to use its full open-loop gain with slowly changing inputs. It will amplify its input offset voltage and amplify its internal and external noise. It will probably oscillate due to the input picking up the signal from its output.

So hysteresis is used in a comparator circuit to provide a snap-action to its switching. Then noise and oscillation are avoided but the sensitivity is much less.


in no way attemps to justify or prove your statement to be true!!

I've said it before... you have a great deal of practical knowledge to share and many appreciate and benifit from it... including myself on some topics, but be careful when you step beyond your own boundaries of knowledge... people that REALY know may be reading!!



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I don't need to prove or justify why the manufacturer of an LM393 comparator specifies that it has a voltage gain of typically 200,000.

I didn't say that its output will go to 200V. I said when it amplifies an input offset voltage of 5mV by 200,000 times then its output will try to go to 200V. My math was wrong, it will try to go to 1000V.

I also said that its output would be saturated at the positive supply voltage or at ground.

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More smoke and mirrors

Where do you see a question that asks that you

...prove or justify why the manufacturer of an LM393 comparator specifies that it has a voltage gain of typically 200,000
?

For some reason you decline to explain why you think the open collector LM393 works the way you think it does...
its output would be saturated at the positive supply voltage or at ground
... not true, with nothing connected to the output, it either FLOATS, or is pulled down to the MINUS RAIL... that's it, no other choices!! Can the output be at Vcc or ground... yes, but that's missleading and doesn't give you the "general case" fpr how it works. You also make no attempt to rationalize why gain has somethig to do with what the output voltage will be... IT DOESN'T... for the reasons above.

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Indulis,
The question about a comparator was, "what is its output (voltage) when both inputs have exactly the same voltage". I answered that it depends on the input offset voltage and will be either at the positive supply (of course it must have a load resistor), saturated at ground or full of noise.
What do you think the output voltage will be?

The open-collector output of an LM393 comparator is supposed to have a collector load resistor to a positive supply. Then the comparator circuit is a linear amplifier with a typical voltage gain of 200,000.

The first part of the circuit is exactly the same an an LM358 or LM324 opamp except the frequency compensation capacitor is removed. The second part of the circuit is just a couple of linear transistors.

Here is the voltage gain test circuit:

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I don't know what is... either I don't know how to ask a question, you can't understand my question, you don't know how to answer my question or I can't understand your reply. Sorry,to me, your answers don't address my questions. Fine, so the intellectual route didn't work, let's try a practical example instead.

Let's take the 0-30V power supply project schematic and for the sake of clarity, remove all the indicator components Q3, R20, R19 etc. and also C8. Now let's replace U3 with a LM393 comparator. The unit will function just fine, and we lte's say we can live with the current limit hic-up

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WHAT WILL THE OUTPUT VOLTAGE OF THE COMPARATOR BE??

It will probably oscillate.
Its output will be floating if the load current is less than the set amount of current regulation.
Its output voltage will drop just low enough to regulate the current if the load current is higher than the set amount of current regulation.

WHAT WILL THE COMPARATOR OUTPUT VOLTAGE BE AS WE ADJUST THE SUPPLY OUTPUT VOLTAGE?

If it is not regulating the current then its output is either oscillating or floating.

IS THE COMPARATORS GAIN GOING TO MAKE A DIFFERENCE AS TO WHAT THE COMPARATORS OUTPUT VOLTAGE WILL BE?

Its very high voltage gain of typically 200,000 makes the current regulation very precise. If its gain is much less (100?) then you would notice that the regulated current amount would change a little if the load's resistance changed.

WHEN THE CURRENT LIMIT SET POINT IS EXCEEDED WHAT WILL THE OUTPUT OF THE COMPARATOR BE?

Just low enough to adjust the output voltage so that the set amount of current occurs.

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Quote from: indulis on February 14, 2007, 02:58:47 PM
WHAT WILL THE OUTPUT VOLTAGE OF THE COMPARATOR BE??
It will probably oscillate.
Its output will be floating if the load current is less than the set amount of current regulation.
Its output voltage will drop just low enough to regulate the current if the load current is higher than the set amount of current regulation.

If it's below the current limit setpoint it WILL NOT oscillate. The only thing a open collector collector can do is "pull-down" if it's active, and it's not active in this state, so the voltage at the output pin will float to around one diode drop, or somewhat less, below the U2 pin 3 voltage. This isn't because the comparator has an output, it's the value of the U2 pin 3 nodal voltage.  

Quote
WHAT WILL THE COMPARATOR OUTPUT VOLTAGE BE AS WE ADJUST THE SUPPLY OUTPUT VOLTAGE?
If it is not regulating the current then its output is either oscillating or floating.

The comparator has NO ability to regulate the current, it can only limit it. If it's not in current limit, it will float to/track the U2 pin 3 voltage, again around one diode drop, or somewhat less. The "somewhat less stems from the amount of leakage current in the D9 diode.


Quote
IS THE COMPARATORS GAIN GOING TO MAKE A DIFFERENCE AS TO WHAT THE COMPARATORS OUTPUT VOLTAGE WILL BE?
Its very high voltage gain of typically 200,000 makes the current regulation very precise. If its gain is much less (100?) then you would notice that the regulated current amount would change a little if the load's resistance changed.

The gain will not make ANY difference whatsoever... the comparator only has two states. It CAN NOT be made to behave like a op-amp! Having a high gain isn't going to change the threshold level, however, the "window of uncertainty" will get smaller... i.e. for a 1 volt threshold and a lower gain unit might have a window of 1V

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You cannot use a comparator to replace an opamp in most circuits. We were discussing what might happen if a comparator replaced the U3 opamp in the regulated voltage and regulated current power supply.

The output of the comparator would be linear during current regulation, its collector resistor is the resistors from the +11.2V reference in series with forward-biased D9. The only time it won't be linear is when it is not regulating the current or if the output voltage is set very low.

With a load resistor, the DC voltage gain of an LM393 comparator is exactly the same as a similar LM358 opamp, from 50,000 to over 200,000.

Its output would be full of high frequency noise or it will oscillate, with its noise triggering the oscillation An opamp operating open-loop without negative feedback would also have its output full of noise.

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