Non-Inverting Open Collector Comparator

Thread Starter

pinkyponky

Joined Nov 28, 2019
351
Hi all,

Can you anyone help me to understand the below formulas. Please can you explain only formulas, how they are derived?.

1640707194411.png
 
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Audioguru again

Joined Oct 21, 2019
6,690
We call it a comparator because its inputs compare voltages.
You call it a compactor maybe because your circuit compacts voltages because the resistor values are too low.

Ohm's Law is used in the formulas.
 

MrChips

Joined Oct 2, 2009
30,802
An analog comparator takes the difference of two inputs.

You can change from inverting to non-inverting output by swapping the two inputs (i.e. the Vref can go on either inverting or non-inverting input).

The purpose of R2 is to introduce hysteresis.
 

Thread Starter

pinkyponky

Joined Nov 28, 2019
351
Then, which value (of V+) to be used in the 2nd equation (which is in 1st post), either the V+ value is Vc or Vp (from the below image).
 
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Thread Starter

pinkyponky

Joined Nov 28, 2019
351
Hello,

1) In the derivation below, I haven't considered the Rp (Pull-up Resistor) in this context, because I understand that the Rp is only used to not keep the output signal is floating. So, I'm not considered in this context. What I understand is Right?.

2) Please can anyone correct me If I'm wrong either in above point 1 or below derivation?.

Thank you very much for the verification.
 
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ericgibbs

Joined Jan 29, 2010
18,848
Hi,
If the Rp is omitted for a Open Collector output, where is positive hysteresis feedback current coming from to drive R2.???
E
 

ericgibbs

Joined Jan 29, 2010
18,848
hi pp,
This shows the plots for a Comparator with Rp and one without Rp.

Note the change in switching points, on the falling edges of p1 and p2.
E

Update:
If it helps, post the threshold and hysteresis component values, using your calculated values, and I will run a simulation.

EG 1175.png
 

ericgibbs

Joined Jan 29, 2010
18,848
hi pp
I think from Rp only. Correct

I want formula. Please correct me If I'm wrong.


As I suggested, calculate and post actual component values for a Comp using the equations you have derived in post #9, and I will use those values and run a LTS simulation.
This will prove/disprove your Equ's.

E
 

Thread Starter

pinkyponky

Joined Nov 28, 2019
351
As I suggested, calculate and post actual component values for a Comp using the equations you have derived in post #9, and I will use those values and run a LTS simulation.
This will prove/disprove your Equ's.
E
Can you give me the clarity that do I need to use the Rp for the derivation in this context?.
 

ericgibbs

Joined Jan 29, 2010
18,848
hi,
It is all in Post #3. PDF.

As you may know, most Comparators have some inbuilt small hysteresis level, so that PDF covers the question you are asking. ie: adding hysteresis to a comp.

E
 

crutschow

Joined Mar 14, 2008
34,418
Can you give me the clarity that do I need to use the Rp for the derivation in this context?.
You need to consider Rp, since current flows through it to R2 when the output is high, which causes a voltage drop through Rp.
Typically Rp is selected to be a much lower value than R2, so the voltage drop is low.
 

Thread Starter

pinkyponky

Joined Nov 28, 2019
351
You need to consider Rp, since current flows through it to R2 when the output is high, which causes a voltage drop through Rp.
Typically Rp is selected to be a much lower value than R2, so the voltage drop is low.
Can you verify that, Is the derivation is right?. If not, please let me know where I did the mistake.
 
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ericgibbs

Joined Jan 29, 2010
18,848
hi pp,
I have changed the Node labels to match your equations.
All you need to do is plug the actual component values into those equations and see if the results match close to the simulation plots.
E
EG 1179.pngEG 1180.png
 
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Thread Starter

pinkyponky

Joined Nov 28, 2019
351
hi pp,
I have changed the Node labels to match your equations.
All you need to do is plug the actual component values into those equations and see if the results match close to the simulation plots.
E
View attachment 256404View attachment 256406
As you requested, I have placed all component values into the equations and the results are as follows:

Vth_high = 3 + ((10k/4.7K) * (10-9.8) = 3.425 V
Vth_low = 3 + ((10k/4.7K) * (10-0) =24.276 V

Vhys = (Vth_high - Vth_low) = - 20.851 V

when compare simulation and theoretical results, both are not matching at all. I'm not sure where is mistake, either in Derivation or in substitution ?. Could you tell me where the mistake is...
 
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ericgibbs

Joined Jan 29, 2010
18,848
hi pp,
Where has the 4k7 come from, it should be 200k.
Also the Vin [ Vref =1]

Vth_high = 3 + ((10k/4.7K) * (10-9.8) = 3.425 V

E
 
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