Circuit breaker, clearing a fault current

[knowledgeseeki]

Oh really wow!...
The attached circuit is a hybrid circuit breaker.
The voltage controlled switch is used to simulate a mechanical circuit breaker and the Mosfet is to act as my solid state circuit breaker

They are both being controlled by the schmitt trigger

the point of the circuit attached is to interrupt the circuit which is 200Ω supplied by the 20V
to prevent the fault current simulated by the switch in series with the 5Ω closing.

The fault current condition is assumed when the current is at 4.1 and normal condition at 0.1

What i actually expect and this is why i am confused is, if the current in my load(200Ω)
doesnt increase when the switch closes.
how can i say my circuit breaker is interrupting the circuit to prevent a fault condition from remaining in the load.

can you please help me design a Hybrid circuit breaker based on the attachment i sent
i simply need the combination of the M switch +Mosfet to interrupt the circuit which is the 200Ω powered by the 20V supply.
The cause of the simulated fault is the switch in series with the 5Ω closing

I am getting results so to speak but i don't understand them at all.

 

[Carl Pugh]

IRF9530 has a maximum gate voltage of +/-20 volt. Anything higher than this may damage the IRF9530.
See typical curves for IRF9530, +15 volt will probably turn it on.

 

[knowledgeseeki]

Thanks carl
I am not simulating or going to be building with anything higher than +/- 20V

Do you know anything about hybrid circuits
i.e. mechanical switch+mosfet switch
to interrupt a circuit to help protect it

 

[NascentOxygen]

the point of the circuit attached is to interrupt the circuit which is 200Ω supplied by the 20V
to prevent the fault current simulated by the switch in series with the 5Ω closing.

Yes, it will do that (if it works⁂). Your load is simulated by the R25/R26 combination. When the current through them increases beyond 4A the MOSFET switch opens and interrupts the current through them. The fact that R25's current momentarily drops is immaterial, it's the R25/R26 combo that represents your load.

I hope I've managed to explain your design to you.

⁂ I see you have a bipolar Schmitt trigger, even though it can't do any more here than can a unipolar Schmitt trigger, and I can't say whether it is correctly designed to do what you want or not. Have you separately tested its trigger levels?

 

[knowledgeseeki]

Yes, it will do that (if it works⁂). Your load is simulated by the R25/R26 combination. When the current through them increases beyond 4A the MOSFET switch opens and interrupts the current through them. The fact that R25's current momentarily drops is immaterial, it's the R25/R26 combo that represents your load.

I hope I've managed to explain your design to you.

⁂ I see you have a bipolar Schmitt trigger, even though it can't do any more here than can a unipolar Schmitt trigger, and I can't say whether it is correctly designed to do what you want or not. Have you separately tested its trigger levels?

yes i have it triggers correctly
when the input is above the upper trigger point the the trigger outputs a low
when it falls below the lower trigger point a high

so i should change my bipolar trigger to a unipolar trigger
ok
When you say the mosfet switch opens you mean turn off to prevent current from flowing through correct.

the r25/26 combination doesn't represent my load the 5Ω is only put there so that the current in the 200Ω increases when the switch closes. 5Ω+switch= simulated fault
200Ω supplied by 20V is my load

are you saying what i put in bold is incorrect, because i actually just want my load to be the 200Ω

sorry i am just a lot confused.

as now i do not see the point of this project i am doing
also my mosfet refuses to turn off

am i correct in saying that when the mosfet turns off, the current at the drain should be 0?

i get the graph attached below when i use a current probe and measure the current at the drain of my mosfet.
Could you please explain why the graph looks like that. i am quite confused.

 

[NascentOxygen]

yes i have it triggers correctly
when the input is above the upper trigger point the the trigger outputs a low
when it falls below the lower trigger point a high

so i should change my bipolar trigger to a unipolar trigger
ok

I didn't say you must change it. I seem to recall designing a unipolar one for you. Besides triggering at particular levels, Schmitt trigger design must also provide the output voltage levels you need. It looks like the one you are using will swing from +Vcc to -Vcc and this is applied to the gate of your MOSFET. I haven't studied the device closely, but I fear that the FET may not appreciate such a negative voltage V_GS. The unipolar output of the one I came up with delivered +Vcc to 0V and I think that is all that is needed to operate the MOSFET switch. (But I may be wrong, I'm no expert on this.)

When you say the mosfet switch opens you mean turn off to prevent current from flowing through correct.

Yes. (That is a colloquialism newcomers find very confusing. In electronics we say a switch opens and blocks passage through it; on the farm we say a gate opens and permits passage through.)

the r25/26 combination doesn't represent my load the 5Ω is only put there so that the current in the 200Ω increases when the switch closes. 5Ω+switch= simulated fault
200Ω supplied by 20V is my load

are you saying what i put in bold is incorrect, because i actually just want my load to be the 200Ω

I can only repeat what I said before. The 5Ω in your arrangement cannot increase the current in the 200Ω. (Besides, you would need a voltage of 800V if you wanted 4A in the 200Ω!)

as now i do not see the point of this project i am doing
also my mosfet refuses to turn off

am i correct in saying that when the mosfet turns off, the current at the drain should be 0?

Drain current will be 0 if the FET is not conducting and the design is not badly flawed.

i get the graph attached below when i use a current probe and measure the current at the drain of my mosfet.
Could you please explain why the graph looks like that. i am quite confused.

This needs close examination. My first thoughts are that current may be syphoning through protection diodes in the FET, going from +Vcc via R26 to the -Vcc at the output of your Schmitt trigger. I leave it to you to look at the FET's data sheet and determine whether such diodes are present. Can you probe the current in the 47Ω resistor?

Alternatively, to test my theory, change your 47Ω resistor from the Schmitt trigger to something much higher, say 470kΩ and take another look at this rogue current. If I'm right, it's not going into the drain structure, it's skirting around it via protection diodes.

Far from it being a waste of time, I believe you are learning a lot from this exercise.

 

[knowledgeseeki]

I didn't say you must change it. I seem to recall designing a unipolar one for you. Besides triggering at particular levels, Schmitt trigger design must also provide the output voltage levels you need. It looks like the one you are using will swing from +Vcc to -Vcc and this is applied to the gate of your MOSFET. I haven't studied the device closely, but I fear that the FET may not appreciate such a negative voltage V_GS. The unipolar output of the one I came up with delivered +Vcc to 0V and I think that is all that is needed to operate the MOSFET switch. (But I may be wrong, I'm no expert on this.)
Yes. (That is a colloquialism newcomers find very confusing. In electronics we say a switch opens and blocks passage through it; on the farm we say a gate opens and permits passage through.)
I can only repeat what I said before. The 5Ω in your arrangement cannot increase the current in the 200Ω. (Besides, you would need a voltage of 800V if you wanted 4A in the 200Ω!)Drain current will be 0 if the FET is not conducting and the design is not badly flawed.
This needs close examination. My first thoughts are that current may be syphoning through protection diodes in the FET, going from +Vcc via R26 to the -Vcc at the output of your Schmitt trigger. I leave it to you to look at the FET's data sheet and determine whether such diodes are present. Can you probe the current in the 47Ω resistor?

Alternatively, to test my theory, change your 47Ω resistor from the Schmitt trigger to something much higher, say 470kΩ and take another look at this rogue current. If I'm right, it's not going into the drain structure, it's skirting around it via protection diodes.

Far from it being a waste of time, I believe you are learning a lot from this exercise.

increasing the 47ohm resistance to 470kΩ did remedy the situation.

So can you please explain to me the fault condition i tried to simulate, since it is now obvious the fault condition is not the current in the 200Ω increasing.
What is the fault then?

Quick question. Before the fault current at t=0 should my mosfet be normally on or normally off?
i.e. i want to know if my mosfet turns on to interrupt the circuit thus stopping the fault current or turns off to interrupt the circuit

thank you so much for your support

 

[NascentOxygen]

increasing the 47ohm resistance to 470kΩ did remedy the situation.

Well, it may appear to, but that is not a proper fix. No good design places reliance on a device's protection circuit for normal operation.

So can you please explain to me the fault condition i tried to simulate, since it is now obvious the fault condition is not the current in the 200Ω increasing.
What is the fault then?

It is working as you designed. Your load is the combination R25//R26. Normally this load carries a particular current, but if something happens to cause its current to rise to about 4A the MOSFET opens and interrupts the load current. There is no way here that you could cause the current in the actual 200Ω resistor to jump to 4A while still maintaining that resistor as 200Ω. You would need either 800V or else you'd have to somehow make it change its resistance from 200Ω to 5Ω. You can't attack that colourful resistor without spoiling its paintwork, so you have switched a 5Ω in parallel with it to simulate its change of resistance.

Quick question. Before the fault current at t=0 should my mosfet be normally on or normally off?
i.e. i want to know if my mosfet turns on to interrupt the circuit thus stopping the fault current or turns off to interrupt the circuit

The MOSFET has to be ON to be conducting current.

thank you so much for your support

My pleasure.

 

[knowledgeseeki]

Thank you so much once again
I am having an issue though my output keeps oscillating therefore my mosfet doesnt' stay off because of it.

using a bipolar trigger how do i set my hysteresis values to -1V for lower and 3V for upper

bipolar trigger is used because this ensures that the current in my mosfet stays well below zero. no chance of it switching if bipolar is used.

 

[NascentOxygen]

Is that 470 Ohms for R39 correct?

 

[knowledgeseeki]

yes 470
47K or 470k was simply too big and is improper to put such high resistances at the mosfet gate

 

[NascentOxygen]

yes 470
47K or 470k was simply too big and is improper to put such high resistances at the mosfet gate

Why are high values not allowed?

 

[knowledgeseeki]

Because the gate of a mosfet is a capacitor, and the mosfet won't turn on (or off) until it is charged (or discharged). A big resistor will slow it down tremendously.

 

[NascentOxygen]

Because the gate of a mosfet is a capacitor, and the mosfet won't turn on (or off) until it is charged (or discharged). A big resistor will slow it down tremendously.

Okay. Might not be an issue here, where you are allowing the protection zeners to conduct.

I calculate the Schmitt trigger levels here[/color] as already precisely 3V and -1V. Are you saying it is repetitively resetting?

 

[NascentOxygen]

The trigger levels are determined by applying Kirchoff's current law, saying the sum of the currents into pin 3 of the IC = 0.

 

[knowledgeseeki]

yes it repetitively setting
and the reference voltage doesn't even fall to -1
it starts from zero reaches a high point and simply oscillates

 

[NascentOxygen]

I'd like to see more of the operation. Can you provide a graph of the current through the 1Ω, the 470Ω, and the 200Ω, all related one underneath the other?

After you have those plots, change the 470Ω back to 470kΩ and see whether it still oscillates.

 

[knowledgeseeki]

for 200ohm, 1ohm and 470ohm

 

[knowledgeseeki]

470k
200
1 ohm
note: large resistance in front of the gate makes the mosfet switch a lot slower than it should

 

[NascentOxygen]

for 200ohm, 1ohm and 470ohm

Before 1ms, the currents are all zero, so it's not even switched on? It should be showing currents, shouldn't it?

Then we see at 1ms the current in the 200Ω jumps to 0.05A and remains there, and the current in the 1Ω jumps to around 1.75A? So this must be the overload condition, when I was expecting about 3A, so it means the Schmitt trigger (needing 3V) will never be triggered.

Try doubling that 20V supply to increase the current so the Schmitt will switch. Can you choose a colour other than "green" (on my monitor it appears a faint yellow). We need the voltage across the MOSFET during this time, too.

Perhaps we need to take steps to ensure the Schmitt trigger always powers up in a state that keeps the MOSFET on.

Your second graph is inexplicable. That scale is microamperes?

Before you do any of the above, I think you should take that 470Ω off the Schmitt trigger and return it to a 15V source, then to a -15V source, to make sure the MOSFET is being turned on and off by these levels. Monitor the MOSFET current, and the voltage across it.

 

[NascentOxygen]

A capacitor connected from the positive supply to pin 3 of the OP-AMP should ensure the Schmitt trigger powers up with its output high. Try 0.01uF.