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Hysteresis in comparator (IGBT protection for voltage inverters)

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QubeZX

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Hi,
currently im making an IGBT protection for voltage inverters - by turning off PWM signal in right moment. Now im making PCB and because its my first big project I have some questions:
1. I want to use TLC3702 comparator, they say that it has build in push-pull.. so i dont need any other resistor for it to work (besides current regulations ofc)?
2. My input signal is sinusoidal 0 to 3.3V range. My power supply is 3.3V.
I've made hysteresis based on this link: https://www.maximintegrated.com/en/app-notes/index.mvp/id/3616 because TLC3702 has no built-in hysteresis.
Vref in my case is 2.8V ,Vthr is ~2.97 and Vthf is ~2.95. And this 2.965 is what i wanted to switch around (strange that Vref is not the voltage that comparator switches at - is that ok?)
3. Now i want to get on second comparator output logic true when my input is below value 0.55V. I know i should switch + and - of op-amp but what with hysteresis ? I should leave positive feedback i guess. Meaby i shouldnt worry at all and just do exacly the same only switch inputs ?
 

Hi,

Your description confuses me somehow.

Please post your circuit. Show where you measure those given voltages.
How did you measure the threshold voltages, why they differ from Vref?
Why 2.965V and not 2.960V?

Klaus
 

Thanks for reply,
I rechecked my calculations and ofc I found mistake. Proper value at which I want "1" form comparator is 2.97V and turn it off at lets say 2.88V.
So now my Vref would be 2.87V. I calculated those threshold voltages from the link in 1st post. My Vref voltage is from voltage divider.
Ok so to make things easier: pcb1.JPG
L1 L2 L3 those are signals from current converter (+- 70mA), those are converted to voltage signals by 22Ohm resistor, so now signal is +-1.54V, next Im adding offset (1.64V) on op-amp to get around 0-3.3V range ( 0.1-3.18V).
On U2 comparator 1st In I want 2.97-2.88V hysteresis.
Now on 2nd in I want to turn output on when signal is below 0.33V and back to off when signal is greater than 0.42V.
 

Most resistor values are off feasible resistance range, the circuit can't work with the given dimensioning. You need particularly to consider available output current of LM124 and LMC3702. Also the input offset network is apparently not correctly calculated (besides too low resistance values).

- - - Updated - - -

Also consider that LM124 is no rail-to-rail OP. Its output voltage range ends far below Vcc, e.g. 1.5 to 2 V, even with moderate load current. Read a datasheet!
 
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    QubeZX

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Hi,

I completely agree with the above.

Some recommendations:
* read EAGLE tutorials
* use standard Opamp / comparator symbols
* the use of "junction" dots. Usually EAGLE sets them on it's own. It is very, very seldom that one has to set or delete them manually.

Klaus
 
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    QubeZX

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Ok, so I should:
Change op-amp to rail-to-rail, calculate correctly resisotrs, use bigger values. But is it a good idea to use Vref on comparator from voltage divider but with. 0.5% resistors for example? I want output to be 3.3V any other voltage can be scaled. Any advices?
 

Hi,

Regarding using a resistor divider from a supply line for a reference voltage, I would only do so if I had no other options, and even then be grimacing at the thought.

For 2.87V from a 3.3V supply, an option is to find a 3V reference IC, if that value exists (??) as a 3.3V ref won't have the headroom for the series resistor to set IK, and use a trimpot or a fixed resistor divider inside the Vref.

The image is from the Intersil 7135 ADC datasheet, but shows the general idea.
 

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Hi,

Regarding voltage divider from Vcc.
I thought you want to build a protection circuit. It seems you want to protect against overcurrent.
Does an overcurrent protection really need 0.5% accuracy? I don't think so. Here VCC should be OK.

Often I see schematics using VCC as reference of an ADC. Then I recommend to use a voltage reference IC.

****
To your first post:
You say protection "IGBT protection on voltage inverters".
--> Please describe more exact what and how your "protection" should work.

Then you say your signal is sinusoidal.
--> what signal is this? What frequency?
If you want to protect an IGBT, then you should measure it's current, but I don't think it's sinusoidal.

Please post the whole schematic. Your snippet doesn't show the whole signal flow.

Klaus
 
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    QubeZX

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Ok so after simulating my project i understood how terrible I am at electronics:
I rethought it so forget about previous values.
I'll be using Current Transducer placed on output phase of inverter.
KlausST you are right IGBT current wont be exacly sinusoidal but at least its peak value will be coresponding to phase peak value.
On PCB i will only: measure and condition signal from transducer and then use window comparator to know when signal is above or below proper value.
Protection will be realised at FPGA - input of fpga will be pwm signals for transistors and this error signal from PCB (thats why i want 3.3V), output will be gate signals for transistors.
I will use closed loop Transducer. It output will be +-125mA (at 125A).
So what I think I need to do (but obviously i dont know exacly how):
Change current signal from transducer to voltage signal (via a resistor I guess?) (so my signal would be -Vt to +Vt)
Add an offset to signal so my signal will be 0-3.3V. (Can I just simply do:sources1.JPG - connect two sources via the same resistance? So do i even need op-amp?)
Then on this signal use window comparator. Here's general idea LINK, but im confuse how should i connect it.
And from comparator get 3.3V (logic 1) or 0V (logic 0) to connect to FPGA. In link logic 1 will be when my signal is between the values (better option would be logic 1 for outside - should I only swap + and - of comparators ?)
Thank you very much for previous posts showing me im a noob :<
But last thing I have to ask, could you show me a simplified scheme how should it be done ?
 

Hi,

For the window comparator, you can just swap the In+ and In- connections to get logic 1 for outside the values, reversing the connections in the window comparator schematic in the link you posted.

I may not have understood what you're asking for. A 26.4 ohm resistor will convert the transducer 125mA to 3.3V ACROSS the resistor terminals (dissipating about 0.4W) - I think I would do that with a difference amp or an in amp, but there are no doubt other/better ways.
 
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Ok little by little im making progress. I've done diffrence amp, thanks d123
scope1.JPG
So with this scheme im getting what I want 0-3.3V output. My gain is equal to 1. But my DC voltage is 1.65V in this exapmple, and I only have 3.3V from supply. So my first idea is ofc voltage divider:
scope3.JPG
Obviously gain is chainging for DC voltage somehow. I managed to guess 1.8k values but they are not exacly 0-3.3V but really close. I can live with that as long as it will work but is there any good equation ?
 

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Look up thevenin equivolence.

The answer is that if your divider was two 5.4k resistors from 3.3V it would be the same as the 1.65V with the 2.7k. So you'd have the two 5.4k resistors and no additional series resistor between them and the opamp.

In general you take the two resistances in the divider and calculate their impedance as if they were in parallel. Then you replace the source they were dividing with a source equal to the voltage they divided too. If that makes sense. Or vice versa (as was the case for what you were trying to do).


Or another general approach would be to have a strong divider, say 100 ohm resistors which would be equivalent to 50 ohm impedance and then you could put your 2.7k in series with that divider (the 50 additional ohms would have negligible impact here).

Or a third approach is to have your 50/50 divider using any resistors you want and then buffer it with an opamp in a simple voltage follower topology. This gives you a solid 1.65V source which you could use just like the 1.65V source in your first screenshot.
 
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