op-amp comparator, Floyd13-3

[PG1995]

Hi

Please have a look on the attachment. Please help me with it. Thanks a lot.

Regards
PG

 

[duffy]

The differential input voltage is negligible because an opamp always drives it's output in such a way as to try and achieve that. That's how it works, that's the main thing to remember with opamps. The input current is negligible because an opamp has a very high input impedance, approaching infinity.

The 5.4V comes from the 4.7V zener voltage (reverse voltage) plus the .7V forward voltage of the other diode. The voltage is referenced from A to B, not from either of those to something else.

 

[MrAl]

Hi

Please have a look on the attachment. Please help me with it. Thanks a lot.

Regards
PG

Hi,

Q1:
The voltage across D1 and D2 here is the absolute value of the voltage measured in either direction.
That would be the voltage measured if you put a volt meter across A and B.
So it is not a voltage measured to ground, but across both diodes, and the way they state it here
it is the absolute value of that voltage. They are just saying that the voltage drop is always
going to be 5.4v and that's about it without specifying a polarity yet. It's like saying, "I have
a 9v battery" where you didnt specify how you would hook it up yet.

Q2i:
The voltage at the minus input is always going to be Vout plus or minus 5.4v if the zeners
are conducting because the zener gets biased between the input and the output voltages,
with the series resistor Ri. This is just like if you took a zener and a resistor and hooked them
up to a 9v battery with the proper polarity, you'd see a voltage across the zener equal to the
zeners voltage. With one backward zener, you see the zener voltage plus the other zener forward
voltage so together they equal 5.4 volts. Since Vout is measured to ground, at the minus input
we'll see Vout-Vz volts because the zener is always biased properly because when the input is
positive the output is negative, and when the input is negative the output is positive.
So if Vout where say 10v, then the minus input would be at 10-5.4=4.6 volts, and if
Vout was at -10v then the minus input would be at -10+5.4=-4.6 volts.
However, the output will follow the input up to the point where the zeners conduct,
according to the gain set by R1 and R2. Since the ratio is about 3:1, the output
will be 3 times the input until th zeners conduct. After they conduct then the
output will be clamped because of the feedback through D1 and D2. This happens
because then the zeners conduct they prevent the input to the minus terminal from
rising any farther so the output can not rise anymore either.

Q2ii:
The voltage between the non inverting and inverting terminals is considered to be 0v
for these kinds of problems. This happens only when the op amp is operated in the
linear mode, which is the case for many of these problems. You do need to check for
that though to make sure that is always the case when there is feedback because
sometimes the op amp is driven as a comparator and that takes it out of the
linear mode.
In the linear mode, the circuit is always arranged so that the output can drive the
input difference to zero volts, or nearly so. It isnt really zero but is really
an error voltage of some small value. The amplitude depends on the internal
gain of the op amp. The higher the gain, the lower this differential voltage will
be, but in a simpler analysis it is considered to be zero volts. In some analysis
you can not assume this because it would defeat the whole purpose of the analysis.

Q2iii:
The author here is using the fact that the non inverting terminal is at Vout+/- 5.4v,
and so is subtracting the non inverting voltage from the output to get the
difference voltage across R1. They can thus calculate the current through R1 that way.

Q2iv:
The input current is considered to be low enough to not matter because the input
impedance to the op amp is considered much higher than the external impedances.
This is usually the case but there are exceptions.

NOTES:
1. If you have never analyzed a zener diode in series with a resistor in series with a
voltage supply then i suggest that you study that first and understand how that works
in its entirety. That understanding is paramount to understanding this op amp circuit.

2. If the supply voltage can not support the 'proper' operation of the linear mode of the
op amp, the op amp goes out of the linear mode and the operation becomes non linear.
This means the input differential is no longer negligible although the circuit may still
operate just fine for this particular application because it squares up the input which
is a nonlinear function anyway. So an assumption for this circuit is probably that the
supply voltage is always high (and low) enough to support linear operation.

 

[duffy]

However, the output will follow the input up to the point where the zeners conduct,
according to the gain set by R1 and R2. Since the ratio is about 3:1, the output
will be 3 times the input until th zeners conduct.

R1 and R2 do not set the gain. They are connected to the positive input. They are used to set Schmitt trigger levels.

 

[PG1995]

Thank you, duffy, MrAl. My special thanks to you, MrAl, for all the explanation. I think I understand it now.

The voltage between the non inverting and inverting terminals is considered to be 0v for these kinds of problems. This happens only when the op amp is operated in the linear mode, ....In the linear mode, the circuit is always arranged so that the output can drive the input difference to zero volts, or nearly so.

So far I have studied the use negative feedback in inverting, non-inverting, and voltage follower configurations of op-amp. Two days ago I studies simple comparator with positive feedback where Vout of op-amp is connected to (+) input. Does the differential voltage generally become almost 0v in both types of feedbacks? Please let me know. Thanks.

Best regards
PG

 

[MrAl]

Hi again,


duffy:
Yes i must have misread something the first time around because i assumed that if the paper said "the differential voltage is negligible" then it must have had negative feedback. I'll have to read this again and zoom in on the schematic to see it better, then correct my previous post for those issues.

PG:
The strange thing about this circuit is that it is not operated as a linear circuit after all, so why they would say the differential voltage is negligible i cant say. This circuit is non linear and thus does not have a near zero differential input.
The differential voltage generally becomes 0v when there is negative feedback so that the op amp can operate as a linear device. The output tries to reduce the differential input to near zero and that's what makes it work. In the non linear mode anything is possible.

 

[PG1995]

The strange thing about this circuit is that it is not operated as a linear circuit after all, so why they would say the differential voltage is negligible i cant say. This circuit is non linear and thus does not have a near zero differential input.
The differential voltage generally becomes 0v when there is negative feedback so that the op amp can operate as a linear device. The output tries to reduce the differential input to near zero and that's what makes it work. In the non linear mode anything is possible.

So, an op-amp is generally said to be in linear mode when there is negative feedback and it drives the differential voltage to almost 0v. But an op-amp isn't said to be in linear mode when it has positive feedback instead.

Doesn't the circuit under discussion use negative feedback? The output of the op-amp connects with inverting input and zeners function as a 'resistor' with voltage drop of 5.4V. Please let me know. Thanks a lot.

Best wishes
PG

 

[duffy]

The differential voltage WILL be negligible in this circuit. Vin will largely be dropped across Ri. The output will be adjusted so there is no appreciable difference between the inverting and non-inverting input. The circuit has both positive and negative feedback. The gain is infinite before the diodes conduct, and unity afterwards.

 

[PG1995]

The differential voltage WILL be negligible in this circuit. Vin will largely be dropped across Ri.

1: Why is so? Why would Vin largely be dropped across Ri? Could you please tell me?

The output will be adjusted so there is no appreciable difference between the inverting and non-inverting input. The circuit has both positive and negative feedback.

2: So, even when both types of feedbacks present, the output would still adjust itself to make the differential voltage 0v (I was under the impression that it only happens when there is only negative feedback present).

The gain is infinite before the diodes conduct, and unity afterwards.

3: I think the diodes won't take long to conduct but why is the gain unity?

Please help me with the queries above. Thank you.

Best regards
PG

 

[duffy]

So, even when both types of feedbacks present, the output would still adjust itself to make the differential voltage 0v (I was under the impression that it only happens when there is only negative feedback present).

Cover all the feedback with your monkey-paw. The op-amp is STILL trying to make its inputs the same. That's what it does, that's how it works, like I said in my first reply. If the negative input is bigger than the positive, the output swings negative to try and balance them. That's all it knows.

Why is so? Why would Vin largely be dropped across Ri? Could you please tell me?

I think the diodes won't take long to conduct but why is the gain unity?

This concept's a little tricky. Semiconductors don't have appreciable resistance. So there's nothing to give a proportional DIVISION in conjunction with Ri - there is only a voltage SUBTRACTION from the drops across the two diodes. Whenever you have zero ohms from the output to the negative input, you have what is called a "voltage follower", or "unity gain amplifier". Usually in this type of circuit there is no Ri to confuse the matter, the voltage present on the (+) input is reflected on the output as the opamp drives itself to balance the (+) and (-) inputs.

In this case we have Ri. Consider the battle between Vi and the output of the amp for control of the (-) input. Vi is presented with Ri, some unspecified resistance. The output effectively has no resistance, and that is seen through the diodes at the (-) input. So the output of the amp wins, and voltage at the (-) input is wrested to the output of the amp (minus the diode drops) and the difference between that and Vi is developed across Ri.

 

[MrAl]

Hi again,

Taking a better look at the circuit, it appears that the op amp works in both the linear and non linear modes.
The following discussion assumes a total zener voltage of 6v (zenering plus forward zener voltage) and resistors
R1=100k and R2=50k to form a perfect gain of 1/3. This makes the numbers simpler.

During startup with the output at zero and the input going positive, when the input gets
even up to only some small positive voltage the output starts to swing negative. The
non inverting terminal sees about 1/3 of this output voltage so it too goes negative.
Since the input is positive and the zeners are not conducting yet, this pushes the
output even more negative, which in turn makes the non inverting terminal even more
negative also.
With the input just over 0v, the zeners start to conduct when the output reaches
-6v. With the input still near 0v the non inverting terminal is at -2v. Thus
the op amp is in the non linear mode.
With 0v in the inverting terminal and -2v on the non inverting terminal, the output
continues to fall negative. This pulls the inverting terminal down lower and lower,
more negative. The inverting terminal thus follows the output plus 6v so as the
output reaches say -7v the inverting terminal is at -1v, and the non inverting
terminal is at -2.333v. So we see as the output falls the two inputs change, but
for that one extra -1v the inverting terminal fell by -1v and the non inverting
terminal fell by only -0.333v. Thus the inverting terminal will eventually equal
the non inverting terminal. When the output falls to -9v the inverting terminal
is at -3v and the non inverting terminal is at -3v, thus they becomes equal.
The op amp now enters the linear mode, where if the output tries to fall lower than
-9v the inverting terminal will become less than the non inverting terminal and the
output would ramp up slightly, but then that would make the inverting terminal
higher again so the output would fall again, and this is where the op amp negative
feedback keeps the output of the op amp at a constant -9v.

As the input goes higher it doesnt change anything too much (although with non ideal
zeners it would slightly). But as the input goes negative, it eventually reaches the
point where it becomes less than -3v and this causes the zeners to stop conducting,
as well as brings the inverting terminal down lower than the non inverting terminal.
This causes the output to rise and the whole process starts again only with a
positive output instead of negative and the alternate zener being the one that
zeners and the other forward biased.

So it is clear to see that the op amp is being operated in both the linear and non linear
modes. The non linear mode is used to obtain the switching action and the linear
mode being used to regulate the output amplitude. Both the switchpoint voltage and the
output amplitude are controlled by the total zener voltage and the resistors R1 and R2:
The switchpoint plus and minus voltage is:
Vsw=(R1+R2)/R2-1
and the output amplitude is:
Vpk=Vtz+Vsw
where
Vtz is the total zener voltage which is the zenering voltage plus the forward zener voltage.

It's obvious now that the goal of the design is to not only control the switching point voltage
but also to set the plus and minus output voltage for the next stage. This is a little more
unusual than most squaring circuits that use the power supply rails as the output limits.

 

[PG1995]

Thanks a lot, duffy, MrAl.

@MrAl: I don't think I understood everything which has been said here but still I have made some progress.

Best regards
PG

 

[MrAl]

Hi again,

Well actually you should be looking at a simpler circuit first. A circuit with ONLY positive feedback, and getting a firm grip of that first. Then you can move to the more complicated circuit with zeners and you will understand it much easier.
We can do a simple circuit with just positive feedback if you want so you can see how that works first. That's the way you should really learn this. It's not right to just jump in with both kinds of feedback in the same circuit if you've never done the simpler circuits first because you build on the understanding of those simpler circuits.

 

[PG1995]

Thank you for the advice, MrAl.

Soon, I will start studying comparator circuits which have positive feedback, then I will resume this thread.

Best wishes
PG

 

[PG1995]

op-amp integrator, FloydExamp13-10

Hi again

Please help me with these queries, **broken link removed** and **broken link removed**, enclosed in the attachments. Thank you.

Regards
PG

 

[duffy]

Q1. Why are you ignoring the second half-cycle? Use the whole wave. The area under the positive going square wave is clearly equal to the area under the triangle wave. You can solve it numerically, you can solve it geometrically using a construction, or you could just take a good long look at it.

Q2. Hysteresis means they change the trip points. Imagine you scored a goal and then they moved the goal posts behind you. Now you have to go back the other way. As soon as you do, do they move the goal posts ahead to where they were the first time. That's hysteresis.

 

[throbscottle]

Question about notation

I'm not very mathematical, but I've been looking through The World of Voltage Comparators which gives some formulae containing the || symbol, such as (R1 || R2) where are R1 and R2 are part of the hysteresis setting network. Does this mean I have to calculate the value of R1 and R2 in parallel?

TIA

(edit: Sorry this was meant to be a new thread - don't know how it ended up on the end of here)

 

[duffy]

Yes, the || is used to indicate the resistance should be calculated in parallel. I like to use the reciprocal of the sum of the reciprocals, it works with any number of resistors, and there's usually a button for reciprocal.

 

[Winterstone]

Hi again

Please help me with these queries, **broken link removed** and **broken link removed**, enclosed in the attachments. Thank you.

Regards
PG

Hi PG1995,

Only today I have discovered your inquiry dated April 23th.
My question: Is your problem solved? (Because I am not too happy with the last answer you got from duffy).
Regards.
W.

 

[WTP Pepper]

I think the circuit has been drawn incorrect. It can latch in a state determined by the power supply voltage which is a crap design. The Op-amp will only try and balance it's inputs in a negative feedback config which this isn't. You can remove the feedback diodes for a low supply voltage and then try and analyse. You will get nowhere. It's just a comparator and with real world components and the signal provided will just add to the confusion.

PG. Never be afraid to contradict academics. Most have never built anything.