(Moderator's note: This was split from another thread at atferrari's request.)

I do not intend to derail this thread but every time I run across a circuit as in the OP, I am disconcerted by the resistor R5.

I happily learnt that the R3/R4 divider (with no other components) sets the voltage of the summing point. Since depending of the actual value of R5 that voltage could be certainly altered, my "model" goes haywire. How to deal with it?

If the OP does not accept this digression I will delete my post.

EDIT/
Now I recall seeing that resistor used to set the hysteresis in op amps used as comparators.
Still disoriented anyway.
/EDIT

MOD: Added original image.

 

Hi there,

I do not intend to derail this thread but every time I run across a circuit as in the OP, I am disconcerted by the resistor R5.

I happily learnt that the R3/R4 divider (with no other components) sets the voltage of the summing point. Since depending of the actual value of R5 that voltage could be certainly altered, my "model" goes haywire. How to deal with it?

If the OP does not accept this digression I will delete my post.

EDIT/
Now I recall seeing that resistor used to set the hysteresis in op amps used as comparators.
Still disoriented anyway.
/EDIT

R5 can also be used to raise input impedance in some circuits.

To deal with R5, simply model the op amp as a voltage controlled voltage source with high gain like 10000 or more. That will show how R5 influences the gain and input impedances presented to V1 and V2 input sources.

If R5 is made too low, and there is a test for that, then the op amp will latch up. The output saturates either high or low and stays that way.

We can look at this in more detail.

 

Hello again,

It looks like the thread was split so you can talk about whatever you wanted to talk about here. I guess it pertains to that same circuit, but if not, simply post the new circuit and that should be fine.

 

(Moderator's note: This was split from another thread at atferrari's request.)

I do not intend to derail this thread but every time I run across a circuit as in the OP, I am disconcerted by the resistor R5.

I happily learnt that the R3/R4 divider (with no other components) sets the voltage of the summing point. Since depending of the actual value of R5 that voltage could be certainly altered, my "model" goes haywire. How to deal with it?

If the OP does not accept this digression I will delete my post.

EDIT/
Now I recall seeing that resistor used to set the hysteresis in op amps used as comparators.
Still disoriented anyway.
/EDIT
View attachment 143805
MOD: Added original image.

I think of it as, R2 (negative feedback) limits the gain so the output can track the input instead of the output slapping rail to rail with the hypothetical infinite gain of an ideal op amp.

Conversely, R5 adds to the gain (positive feedback).

The calculations can be simplified if (R3 & R4) << R5.

 

I do not intend to derail this thread but every time I run across a circuit as in the OP, I am disconcerted by the resistor R5.

Positive feedback does have its uses, such as in a Howland current source circuit (to create a circuit node having infinite impedance) or in circuits to linearize resistive sensors, such as in TI's XTR105 RTD-to-current loop interface chip (see Fig. 1 and note the positive-feedback effect of Vlin and Rlin1).

I happily learnt that the R3/R4 divider (with no other components) sets the voltage of the summing point. Since depending of the actual value of R5 that voltage could be certainly altered, my "model" goes haywire. How to deal with it?

Paraphrasing H. L. Mencken, “Every normal man must be tempted, at times, to spit on his hands, hoist the black flag, and begin writing nodal equations.” Sadly, sometimes it's the only way.

 

I think of it as, R2 (negative feedback) limits the gain so the output can track the input instead of the output slapping rail to rail with the hypothetical infinite gain of an ideal op amp.

Conversely, R5 adds to the gain (positive feedback).

The calculations can be simplified if Rp << (R3 & R4) << R5.

Hi,

I have to ask, why did you want to restrict Rp like that?
That's just the output resistor to ground right?
Doesnt change the gain or anything in the usual circuit.

 

Positive feedback does have its uses, such as in a Howland current source circuit (to create a circuit node having infinite impedance) or in circuits to linearize resistive sensors, such as in TI's XTR105 RTD-to-current loop interface chip (see Fig. 1 and note the positive-feedback effect of Vlin and Rlin1).


Paraphrasing H. L. Mencken, “Every normal man must be tempted, at times, to spit on his hands, hoist the black flag, and begin writing nodal equations.” Sadly, sometimes it's the only way.

Hi,

That's funny
That's quite different from what he said though apparently
His being a much darker idea.

 

Hi,

I have to ask, why did you want to restrict Rp like that?
That's just the output resistor to ground right?
Doesnt change the gain or anything in the usual circuit.

True. I'll correct it.

 

True. I'll correct it.

Hi,

Looks better now

What i found was that the solution can be tricky because a given steady state solution might look ok when really it's not. This could mean that there could be more than one solution depending on how the voltages come up on power up.

To check this out a little more, what i did was i placed a small capacitor across R2 like 200pf, then analyzed the circuit and looked for the real part of the pole crossing into the right half plane of the complex plane. Since this happens when the root becomes positive, i got:
(R1*R4*R5)/(R4+R3)+(R1*R3*R5)/(R4+R3)-(R2*R3*R4)/(R4+R3)<=0

as the criterion for the output going toward infinity.

The interesting part of this though comes in when we look at the practical circuit which can not have an output that goes to infinity. The output can only go up to maybe +10v (or down to -10v) and maybe a special high voltage circuit output can go as high as +100v, but interestingly once we hit that limit, it may not be considered a latch up anymore because if the input voltage V2 can be brought down low enough, then it could pull the circuit out of that pseudo latch up state. Of course that might mean that it latches up into a state of opposite polarity (from +10 to -10) but that's not exactly the same as a continuous latch up.

So maybe this circuit is not as simple as it seems at first. We could look into it more i guess. I can see though that limits have to be set that are not apparent from the schematic as shown in the first post. We'd have to establish the power supply rail voltages for one thing, something that we dont always hve to do directly with the theoretical circuits. This one though depends more highly on these other parameters too.

As a side note, with R4=R3=R2=R1 then the criterion reduces to:
2*R5-R1<=0

or simply:
R5<=(R1)/2

 

I think of it as, R2 (negative feedback) limits the gain so the output can track the input instead of the output slapping rail to rail with the hypothetical infinite gain of an ideal op amp.

Conversely, R5 adds to the gain (positive feedback).

The calculations can be simplified if (R3 & R4) << R5.

That is more or less what I had in mind. Busy here so just now starting to model what I hope could help me to revisit the concepts.

@MrAl
Your suggested modeling with a gain of 10000. Not what you use in op amps actually, isn't it?

 

That is more or less what I had in mind. Busy here so just now starting to model what I hope could help me to revisit the concepts.

@MrAl
Your suggested modeling with a gain of 10000. Not what you use in op amps actually, isn't it?

Hi,

Well you may want to clarify your question, do you mean what i myself use in op amps or what we should use in op amps?
I use 10000 to 100000 sometimes higher for regular simulations where i want to see the results where the gain may have an effect. When i dont want to see how the gain changes things i go to a gain of infinity which also can greatly simplify the transfer function and allow us to see the basic theory of how the circuit works.

LATER:

A couple examples...

The expression for the output of a simple non inverting op amp circuit is:
Vout=(Vin*A*(R2+R1))/(R2+A*R1+R1)

and with A=100000 we get:
Vout=100000*Vin*(R2+R1)/(R2+100001*R1)

and with A allowed to approach infinity we get:
Vout=Vin*(R2+R1)/R1

The first shows how the gain effects the output while the second shows the basic theory of how the op amp circuit works. We often use the second because the difference between the practical and the theoretical isnt that much in many cases. When it matters, then we must use the first.

 

Ok MrAl. For modeling I used a gain higher than 10.000.
The asc file is attached.

Higher the positive feedback resistor, input current seems to become smaller but the hysteresis span becomes smaller as well.

Are my conclusions correct?

 

Are my conclusions correct?

Yes. The higher the feedback resistance the less effect the feedback has.

 

Ok MrAl. For modeling I used a gain higher than 10.000.
The asc file is attached.

Higher the positive feedback resistor, input current seems to become smaller but the hysteresis span becomes smaller as well.

Are my conclusions correct?

View attachment 144347


View attachment 144349

Hi again,

If all you are asking is if raising the feedback resistor decreases the hysteresis, then yes that's correct.
The feedback resistor takes part of the output and feeds it back to the non inverting input so that makes it harder for the inverting input to again reach the same level as the non inverting terminal, and when it finally reaches the non inverting terminal level the output state changes again and then it makes it harder for the inverting terminal to reach the next level. So the feedback is changing the level that the invertinbg input must reach in order to change the output state again. The smaller the feedback resistor, the more it affects the level, and the higher the resistor value the less it effects this change of level behavior. Higher resistor value, less current, less current, less effect on the required input voltage level to get another output change.

 

Just for practice, I tried to calculate voltage gain, input impedance of the circuit.
From the gain expression, we can see that it agrees with Mr.Al on the criterion for the output going toward infinity post #9.
I used superposition method to calculate these expressions.
A: gain of OpAmp

1. V1 is active, V2 is set to zero ( inverting)



2. V2 is active, V1 is set to zero ( non-inverting)

Higher the positive feedback resistor, input current seems to become smaller but the hysteresis span becomes smaller as well.

Is there a way to infer the hysteresis property by using these expressions above?

 

Just for practice, I tried to calculate voltage gain, input impedance of the circuit.
From the gain expression, we can see that it agrees with Mr.Al on the criterion for the output going toward infinity post #9.
I used superposition method to calculate these expressions.
A: gain of OpAmp

1. V1 is active, V2 is set to zero ( inverting)

View attachment 145847

2. V2 is active, V1 is set to zero ( non-inverting)

View attachment 145848


Is there a way to infer the hysteresis property by using these expressions above?

Hello again,

Is there any reason why you have to use them?

If you have a comparator with hysteresis then the output would either be high or low.
For a first approximation, assume the output is rail to rail. Set the output to either high or low, then solve for the two inputs to the comparator. Find the point where the output switches state which will be when the two become equal. As soon as the output changes state, the non inverting input will change and that means you'll solve for one more set of voltages to find out when they are equal again.

You can use Vout as a variable and solve for the two inputs, and that would give you a starting point. Make Vout equal to either Vcc or ground and solve for when the two inputs become equal for both conditions. This is easier than calculating gain.

 

Is there any reason why you have to usethem?

Because I didn't understand your explanation about hysterisis above so I would like to take another approach and then come back.

However, I've got it now after watching the video.
R5 is same as R2 in the video.

 

Because I didn't understand your explanation about hysterisis above so I would like to take another approach and then come back.

However, I've got it now after watching the video.
R5 is same as R2 in the video.

Hi,

Oh great. We could run though the calculations here too if you like.

 

First, I'd like to start with this simple example. I am somewhat confused with the calculation of VH and VL here.
At the crossing points, for that ideal opamp Vout is undetermined. So Vout at these points can be anything from -VoL to VoH.

From the picture, I can infer the author calculation as below:

VL = (-R1/R2)* VoH
VH = (-R1/R2)* VoL

So for calculating VL, why VoH instead of anything between -VoL to VoH?

 

First, I'd like to start with this simple example. I am somewhat confused with the calculation of VH and VL here.
At the crossing points, for that ideal opamp Vout is undetermined. So Vout at these points can be anything from -VoL to VoH.

From the picture, I can infer the author calculation as below:

VL = (-R1/R2)* VoH
VH = (-R1/R2)* VoL

So for calculating VL, why VoH instead of anything between -VoL to VoH?



View attachment 145907

Hi again,

That's nice i was going to suggest starting with simpler circuit first.

The reason is because we assume that the output is bi stable. It is either at +Vcc or -Vcc.
We assume that because we have a comparator with hysteresis, which forces that in many cases.

Unfortunately not all cases, so we have to look at it from a linear standpoint too and see if we can find a linear response. The linear response will be one that does NOT force the output to go to +Vcc or -Vcc. The bi stable response will force one or the other.

This means in the more complicated circuit you probably have to solve the linear equation too and see where it starts to top out. Because there is a gain that is smaller than the assumed infinity, there may be some inputs that cause a linear response while other inputs causes a bi stable response, with in between responses happening when one or both input voltages are varying slowly (goes from bi stable to linear, to bi stable, etc.).

With some of these circuits you really have to think about them, when there are a number of variables that are not yet specified. With circuits where all the components are specified, it's usually much easier because you dont have to solve for every possibility. This means it may be good to start a more complicated circuit out with some constant values for the resistors and go from there. You can always go back and try new values later.
All this takes more time than usual for a circuit, but the information gleaned will be more applicable too then.

One of the conditions i found in the work place a long while back was that sometimes an input can cause the hysteresis to change. That is, we might have a 2v hysteresis with one level input, then 1v with another level, where it varies between 2 and 1v. This caused a problem so the circuit had to be redesigned to get a more constant hysteresis. With the more complicated circuit you were looking at this might happen also.