Circuit breaker, clearing a fault current

[knowledgeseeki]

hi could someone please explain the attached circuit to me in detail.

From what i understand:

the open circuit caused by the opened switch in series with the resistance causes a fault current in the system.
This higher than normal current across the 200 ohm resistor causes a high voltage to flow through the mosfet and into the inverting input.

the input is then compared to a reference( using a schmitt trigger config Threshold values set to 1 an 5V. )

when input increases past the high threshold value set the output is low, when input is below high threshold output is high.

The low output will turn the mosfet off and the high output will turn it on.

When the mosfet turns on, a voltage is sent to the switch closing the switch.

This switch should not close until the fault current is cleared and the switch doesn't re-open unless reset.

based on what i explain is my design correct.

Think of what i am doing as a way to clear a fault current using a schmitt trigger configuration+ a mosfet+ a mechnical switch

please help.

 

[NascentOxygen]

From what i understand:

the open circuit caused by the opened switch in series with the resistance causes a fault current in the system.

An open switch doesn't cause a fault current. The closed switch causes higher than normal current through the MOSFET.

This higher than normal current across the 200 ohm resistor causes a high voltage to flow through the mosfet and into the inverting input.

the input is then compared to a reference( using a schmitt trigger config Threshold values set to 1 an 5V. )

when input increases past the high threshold value set the output is low, when input is below high threshold output is high.

The low output will turn the mosfet off and the high output will turn it on.

When the mosfet turns on, a voltage is sent to the switch closing the switch.

It is not apparent how this last action happens. In any case, after clearing the fault you want the switch to OPEN and the MOSFET to turn ON and that puts you back to normal operation.

 

[knowledgeseeki]

Thank you, so when the switch is closed is what causes the more than normal current.
When the mosfet turns on how does it make the switch to turn on?

thank you though you explain things really well.

 

[NascentOxygen]

Thank you, so when the switch is closed is what causes the more than normal current.

When the switch is closed, it places a 5Ω resistor in parallel with the normal 200Ω, so current can be approximately x40 larger.

When the mosfet turns on how does it make the switch to turn on?

We aren't told. Presumably, someone or something will independently turn the switch back on or off.

The MOSFET itself will be automatically turned back ON immediately after being turned OFF, the speed of the Schmitt trigger causing this. The MOSFET switching will repeat over and over until the switch is opened to remove the fault current. The circuit doesn't seem to accomplish much.

 

[knowledgeseeki]

hey how do i get a 0.2V as my reference if i know that R1 is 10K

i would like to get a 0.2V voltage set but i cnt figure out how to do it please help.

based on the posted schematic.
how do i get my mosfet to switch on and off?

it doesn't seem to switch, because the voltage at the source of my mosfet is just a straight line.

Would my voltage varry when the switch goes from an open position to a closed position? if so why isn't my mosfet reflecting this.

 

[NascentOxygen]

hey how do i get a 0.2V as my reference if i know that R1 is 10K

i would like to get a 0.2V voltage set but i cnt figure out how to do it please help.

The Schmitt Trigger comparator has two switching levels. Do you want to make the lower one 0.2V, or the higher one?

based on the posted schematic.
how do i get my mosfet to switch on and off?

it doesn't seem to switch, because the voltage at the source of my mosfet is just a straight line.

With R₅ being 0.1Ω, and the comparator requiring 5V to switch, you need 50A through R₅ to trigger the MOSFET OFF.

Would my voltage varry when the switch goes from an open position to a closed position? if so why isn't my mosfet reflecting this.

What value is your voltage at R₅ when the switch is (a) OPEN, and (b) CLOSED?

Are you constructing this using hardware, or are you using a computer simulation?

 

[knowledgeseeki]

hi, in the the schematic below i would like my upper trigger point to be at 3 and my lower trigger point at 1. How do i choose my resistors appropriately?
i can't seem to get the hysteresis right, so my output keeps switching rapidly.

i am going to construct this design on a veroboard after i run a few simulations and i am satisfied with it.

My R5 (currently R6) has been changed to a 1W power resistor now. Wouldn't want to burn anything.

i am currently supplying my comparator with +15V

 

[NascentOxygen]

If you construct this on breadboard, you may find the tracks or contacts are damaged by 5A.

If you supply the comparator with a single-ended 15V supply, not ±15V, with [strike]R9=120k, R8=22k, and R10=68k, the upper switching level will be 5V.[/strike]

EDIT:
With R9=560k, R8=22k, and R10=100k, the upper switching level will be 3.1V

 

[knowledgeseeki]

If you construct this on breadboard, you may find the tracks or contacts are damaged by 5A.

If you supply the comparator with a single-ended 15V supply, not ±15V, with [strike]R9=120k, R8=22k, and R10=68k, the upper switching level will be 5V.[/strike]

EDIT:
With R9=560k, R8=22k, and R10=100k, the upper switching level will be 3.1V

Thanks, built it today. Burnt a few things, even though i was sure only 4A was supposed to run through my system.

thank you let me try that now.

actually those values don't work even if i only use a single ended 15V supply
My trigger output switches too quickly from its low state to high state and then takes progressively longer to switch from high back to low state.

what formula did you use?

 

[NascentOxygen]

Thanks, built it today. Burnt a few things, even though i was sure only 4A was supposed to run through my system.

With R13 being 5Ω, if you were to have 4A through it that amounts to 16*5=80watts
so R13 would need to be a big power resistor, or else something like 15 resistors, each 10W and in parallel.

Actually those values don't work even if i only use a single ended 15V supply
My trigger output switches too quickly from its low state to high state and then takes progressively longer to switch from high back to low state.

Maybe it is working, but just not how you'd imagined it would?
I expect the MOSFET will switch on and off a few hundred times per second. When the comparator switches, it turns off the MOSFET causing current to fall towards zero. As the level of current falls below what will trigger the comparator, the comparator output returns high and turns on the MOSFET. Current returns, and triggers the comparator, etc..

▻ When the comparator output is +15V, the R8-R9-R10 resistor network forms a simple potential divider where two of the resistors return to +15V.

▻ When the comparator output is 0V, the R8-R9-R10 resistor network forms a simple potential divider where two of the resistors return to ground.

The voltage trigger levels are 3V and 0.5V with R8-R9-R10 as 22k, 560k, and 100k, resp.

 

[knowledgeseeki]

With R13 being 5Ω, if you were to have 4A through it that amounts to 16*5=80watts
so R13 would need to be a big power resistor, or else something like 15 resistors, each 10W and in parallel.

Maybe it is working, but just not how you'd imagined it would?
I expect the MOSFET will switch on and off a few hundred times per second. When the comparator switches, it turns off the MOSFET causing current to fall towards zero. As the level of current falls below what will trigger the comparator, the comparator output returns high and turns on the MOSFET. Current returns, and triggers the comparator, etc..

▻ When the comparator output is +15V, the R8-R9-R10 resistor network forms a simple potential divider where two of the resistors return to +15V.

▻ When the comparator output is 0V, the R8-R9-R10 resistor network forms a simple potential divider where two of the resistors return to ground.

The voltage trigger levels are 3V and 0.5V with R8-R9-R10 as 22k, 560k, and 100k, resp.

my comparator has 2 output states though
a high of +15V and a low of -15V. i am using dual supply
i can't find any examples of a dual supplied schmitt trigger.

i need to use one so that my on and off periods are equal, with a single supply i noticed that the off period is always quite abit longer than the off period.

also slowing the period and pulse width improved my output response.

 

[NascentOxygen]

my comparator has 2 output states though
a high of +15V and a low of -15V. i am using dual supply
i can't find any examples of a dual supplied schmitt trigger.

i need to use one so that my on and off periods are equal, with a single supply i noticed that the off period is always quite abit longer than the off period.

I doubt that you do need to change the Schmitt trigger. Are you still using the basic circuit in your post #7? Because, in that circuit there is no designed "on period" or "off period". There is no monostable multivibrator. There is no periodic signal whose duration you have considered or calculated.

Maybe you are referring to the switching rate you see the MOSFET is being subjected to? But that is not something you have designed for; it's due to the slew rate of the op-amp and the battery/load characteristics, and is something I expect you were not anticipating. If you want to control the rate of automatic switching, you will have to incorporate a monostable multivibrator, as was used in one of your original schemes. (I'm curious, how many times per second does your circuit, as it stands, switch the load?) What delay would you like before the load is switched by the MOSFET?

also slowing the period and pulse width improved my output response.

What do you mean by "output response"?

 

[knowledgeseeki]

I doubt that you do need to change the Schmitt trigger. Are you still using the basic circuit in your post #7? Because, in that circuit there is no designed "on period" or "off period". There is no monostable multivibrator. There is no periodic signal whose duration you have considered or calculated.

Maybe you are referring to the switching rate you see the MOSFET is being subjected to? But that is not something you have designed for; it's due to the slew rate of the op-amp and the battery/load characteristics, and is something I expect you were not anticipating. If you want to control the rate of automatic switching, you will have to incorporate a monostable multivibrator, as was used in one of your original schemes. (I'm curious, how many times per second does your circuit, as it stands, switch the load?) What delay would you like before the load is switched by the MOSFET?

What do you mean by "output response"?

hey sorry, to be honest i was confused for a while and i was doing the wrong thing

here is what my complete circuit design looks like.

The mosfet starts on at t=0
At t=1ms the switch closes causing fault current which gets cleared.
At t=3ms, the breaker is reset.

How do i design a drive circuit supplying ±20V to my mosfet in order to turn it on or off.

The input coming into the push pull circuit attached below is ±15,V i need to boost the 15V input to give a 20V output.

 

[NascentOxygen]

The mosfet starts on at t=0
At t=1ms the switch closes causing fault current which gets cleared.
At t=3ms, the breaker is reset.

Who or what performs the 3ms timing for the breaker reset? (No circuit is shown.) That is a very short time within which an overload can be expected to be removed.

How do i design a drive circuit supplying ±20V to my mosfet in order to turn it on or off.

Isn't the op-amp's output driving your MOSFET switch? I'd have thought 15V and 0V should suffice.

The input coming into the push pull circuit attached below is ±15,V i need to boost the 15V input to give a 20V output.

Are you sure you need this?
Anyway, if you correctly copy a working amplifier circuit from somewhere, it should work.

 

[knowledgeseeki]

I need something to boost the ± 15V coming into the Mosfet
because when i am building it the Mosfet i am using requires ±20V
thats why.

the reset button is the switch timed at t=3ms connected to 0V.

 

[NascentOxygen]

the Mosfet i am using requires ±20V
thats why.

I doubt it. I think you should begin building parts of this circuit, starting with just the MOSFET switch and a drain load of the LED and associated resistor. Don't bother about the Schmitt trigger, leave that out at this stage, and instead manually switch the gate of the FET to either 0V or 15V and see whether that switches it. (Keep a voltmeter constantly connected between drain and source to monitor how well the FET is switching.) If this works well, then close U2 and keep your finger on the FET to make sure it doesn't get hot. It shouldn't get hot whether it's ON or OFF.

Don't leave the gate connected nowhere; it should go to either 0V or 15V. If you leave it floating it may half turn on and cause the FET to get hot and burn out. The safest approach would be to put a resistor of a few 100kΩ from the gate to ground, by default that connects the gate to 0V. Leave the resistor there, and whenever you need 15V just connect 15V to the gate.

Report back what you find.

BTW, what is your reason for adding R6 in series with R13?

 

[knowledgeseeki]

I doubt it. I think you should begin building parts of this circuit, starting with just the MOSFET switch and a drain load of the LED and associated resistor. Don't bother about the Schmitt trigger, leave that out at this stage, and instead manually switch the gate of the FET to either 0V or 15V and see whether that switches it. (Keep a voltmeter constantly connected between drain and source to monitor how well the FET is switching.) If this works well, then close U2 and keep your finger on the FET to make sure it doesn't get hot. It shouldn't get hot whether it's ON or OFF.

Don't leave the gate connected nowhere; it should go to either 0V or 15V. If you leave it floating it may half turn on and cause the FET to get hot and burn out. The safest approach would be to put a resistor of a few 100kΩ from the gate to ground, by default that connects the gate to 0V. Leave the resistor there, and whenever you need 15V just connect 15V to the gate.

Report back what you find.

BTW, what is your reason for adding R6 in series with R13?

put R13 in series with R6 because i wanted the R13+Diode connection to be my fault indication
From what i now, know R13 and the diode should be in parallel with R6 is this correct.

I will report back, however your suggestion will only work with a Mosfet that requires less than 20V to turn on or -20V to turn off. the IRF9530N requires 20V.

 

[NascentOxygen]

That's an impressive power MOSFET you've got hold of. Go cautiously, don't want to damage it. For the IRF9530N, its drain should be negative, so reverse your 20V supply.

V_gate of ±20V is the maximum rating. From Fig. 1 (see datasheet), a gate voltage of –15V should be ample to turn it fully on. It will be fully off with about –4V, but try it with 0V since that level may be more convenient. (Looks like the Schmitt trigger will need to be modified.)

data sheet: http://alltransistors.com/pdfview.php?doc=irf9530n.pdf&dire=_international_rectifier

 

[knowledgeseeki]

That's an impressive power MOSFET you've got hold of. Go cautiously, don't want to damage it. For the IRF9530N, its drain should be negative, so reverse your 20V supply.

V_gate of ±20V is the maximum rating. From Fig. 1 (see datasheet), a gate voltage of –15V should be ample to turn it fully on. It will be fully off with about –4V, but try it with 0V since that level may be more convenient. (Looks like the Schmitt trigger will need to be modified.)

data sheet: http://alltransistors.com/pdfview.php?doc=irf9530n.pdf&dire=_international_rectifier

so it would seem. what is weird is that i cannot find any good examples of dual supplied schmitt triggers. To achieve the 0V to turn the Mosfet off, i would have to change my current schmitt trigger to a single supply.

One thing i noticed with my design is that after my fault current has been cleared it, the current doesn't return to its original value

on another note: what do you know about zero current/voltage crossing??
I am trying to design my circuit breaker to have this feature.

To me zero crossing is when the voltage or current crosses below zero a switching device will switch. Zero switching won't happen before that. is this correct??

 

[NascentOxygen]

With your load powered by DC there will be no zero crossings.

Only when the load is purely resistive does the zero crossing of the AC voltage co-incide with the zero crossing of the current. With a partly inductive load if you can switch at a zero crossing of the load voltage, this is not a zero crossing of the current. I think you should defer that enhancement until you have perfected the basic switch.

While testing the current design, keep the load resistive. If you add inductance you may damage the device. For inductive loads steps need to be taken to deal with voltage spikes; I think you had this in the very first (of your many) schematics.

 

[knowledgeseeki]

With your load powered by DC there will be no zero crossings.

Only when the load is purely resistive does the zero crossing of the AC voltage co-incide with the zero crossing of the current. With a partly inductive load if you can switch at a zero crossing of the load voltage, this is not a zero crossing of the current. I think you should defer that enhancement until you have perfected the basic switch.

While testing the current design, keep the load resistive. If you add inductance you may damage the device. For inductive loads steps need to be taken to deal with voltage spikes; I think you had this in the very first (of your many) schematics.

sorry I've sent soo many schematics thanks for your patience though
found something out today. My built circuit wasnt working because i somehow thought i was using an n-channel mosfet meanwhile i was using a p-channel.

Would test that on monday and let you know if that makes a difference.

ON pspice though, the current through my load seems to fall as opposed to increase when i close my switch.

 

[NascentOxygen]

found something out today. My built circuit wasnt working because i somehow thought i was using an n-channel mosfet meanwhile i was using a p-channel.

Your first schematics showed a n-channel, but then without warning you changed to a p-channel.

You need to reverse the direction of the LED indicator in the load when you reverse the supply voltage.

 

[knowledgeseeki]

Your first schematics showed a n-channel, but then without warning you changed to a p-channel.

You need to reverse the direction of the LED indicator in the load when you reverse the supply voltage.

sorry.

 

[knowledgeseeki]

hi, do you know anything on CMOS Quad bilateral switch.
I am using this switch instead of a timed switch since the time switch doesn't respond to the Mosfet turning off.

About the Mosfet i am using an n-channel i was connecting the mosfet wrong that's why it wasn't working the way i wanted it to.

its all fixed now.

Just the switch, to cause the fault. Then am all done. thank you very much for your help.

 

[NascentOxygen]

hi, do you know anything on CMOS Quad bilateral switch.
I am using this switch instead of a timed switch since the time switch doesn't respond to the Mosfet turning off.

The 4066 has an ON resistance of typically 125Ω for 15V operation. This isn't going to be any use for switching your 20 ampere overload, if that's what you plan to use it for. CMOS is a low power family of gates and switches.

Is this final circuit going to do anything useful, or is it just something you worked on as a learning exercise?

 

[knowledgeseeki]

no just an exercise
still deciding if i want to be an engineer
the quad switch was only to be used in simulation but seeing its limitations it still won't have worked.
i am using a push switch as my mech switch causing the overcurrent
i am using the CD4013B as my flip flop
and LM393 as my comparator

 

[knowledgeseeki]

hi, sorry i have been out of touch for a while could you please tell me why
in this circuit the 200ohm resistor current drops from 100mA to 80mA when the switch closes at 1ms
meanwhile it should increase
My entire circuit attached to it would make no sense unless the current in the 200ohm resistor increases

 

[NascentOxygen]

hi, sorry i have been out of touch for a while

We've missed you.

could you please tell me why
in this circuit the 200ohm resistor current drops from 100mA to 80mA when the switch closes at 1ms
meanwhile it should increase
My entire circuit attached to it would make no sense unless the current in the 200ohm resistor increases

The switch will increase the current through the 1Ω resistor, and isn't that what you are wanting? But nothing you do can increase the current through the 200Ω. The switch and R26 divert some of R25's current around R26, so the current in R25 inevitably drops.

You could do the resistor current calculations and see how paper figures compare with what you measured. When the switch is open the circuit is 20V applied across 200Ω and 1Ω (plus unknown MOSFET resistance). With the switch closed, a 5Ω resistor is placed in parallel with the 200Ω. Current through the 1Ω increases, but how is the current shared between the parallel pair?

 

[knowledgeseeki]

We've missed you.


The switch will increase the current through the 1Ω resistor, and isn't that what you are wanting? But nothing you do can increase the current through the 200Ω. The switch and R26 divert some of R25's current around R26, so the current in R25 inevitably drops.

You could do the resistor current calculations and see how paper figures compare with what you measured. When the switch is open the circuit is 20V applied across 200Ω and 1Ω (plus unknown MOSFET resistance). With the switch closed, a 5Ω resistor is placed in parallel with the 200Ω. Current through the 1Ω increases, but how is the current shared between the parallel pair?

Thanks
yes i want the switch to increase the current through the 1ohm resistor when the switch closes.
so that a voltage in the opposite direction is induced and goes into the negative input of my opamp

What i actually expected when i caculated it on paper was that when the switch is open the current coming into the Mosfet from the drain should be at 0.1
When the switch is closed the current coming into the mosfet should be at 4.1

Based on the parallel combination of the 200Ω and 5Ω resistor being 4.8Ω
20/4.8 = 4.1 appx
20/200= 0.1 appx

Are you saying my assumption that the current in the 200Ω is supposed to increase is completely false.
If so i am really really confused.

the Mosfet i am using is the IRF150 with an rds on of 0.055Ω

in a simple circuit such as the one attached could you please tell me why the current in my 200Ω
resistor doesn't change when i close the switch at 1ms
the current in the 200Ω just stays at 100mA through out

i don't think that is what should happen

is my assumption that the current in my 200Ω should increase wrong?

 

[NascentOxygen]

What i actually expected when i caculated it on paper was that when the switch is open the current coming into the Mosfet from the drain should be at 0.1
When the switch is closed the current coming into the mosfet should be at 4.1

Based on the parallel combination of the 200Ω and 5Ω resistor being 4.8Ω
20/4.8 = 4.1 appx
20/200= 0.1 appx

Those seem about right.

Are you saying my assumption that the current in the 200Ω is supposed to increase is completely false.

Completely false.

in a simple circuit such as the one attached could you please tell me why the current in my 200Ω
resistor doesn't change when i close the switch at 1ms
the current in the 200Ω just stays at 100mA through out

The 200Ω resistor has 20V across it all the time, regardless of what the switch is doing in that parallel circuit, so nothing about the 200Ω changes to cause its current to change.

 

[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.