Solid state DC circuit breaker, used to protect a load

[knowledgeseeki]

My choice of solid state Dc circuit breaker is the Mosfet, easily available( i got loads!)
i want the mosfet to interrupt the circuit to protect the load from overvoltages
This is what I've done so far, please help its not working the way i want it to.

the resistor connected to the inductor is the load i want to protect

 

[NascentOxygen]

You have breadboarded this circuit, or are you saying the simulation doesn't work?

You have managed to get the DC supply for the op-amp mixed in with the 24V for the load. Separate these. You intend operating the 324 on a single supply?

Is that really a 10 Ohm resistor in series with the FET's gate?!

I suggest that you remove the inductor and its associated zeners, etc, and the 24V supply, and concentrate on getting just the op-amp and the MOSFET working. Perhaps use a LED and a 2.2k resistor to represent the load, powering it off the 10V, so you have a visual indication of the MOSFET switching correctly. Only when that is working and the voltages all tally with your designed values, should you bring in the load and its associated power supply.

 

[knowledgeseeki]

You have breadboarded this circuit, or are you saying the simulation doesn't work?

You have managed to get the DC supply for the op-amp mixed in with the 24V for the load. Separate these. You intend operating the 324 on a single supply?

Is that really a 10 Ohm resistor in series with the FET's gate?!

I suggest that you remove the inductor and its associated zeners, etc, and the 24V supply, and concentrate on getting just the op-amp and the MOSFET working. Perhaps use a LED and a 2.2k resistor to represent the load, powering it off the 10V, so you have a visual indication of the MOSFET switching correctly. Only when that is working and the voltages all tally with your designed values, should you bring in the load and its associated power supply.

I obviously suck at this! thank you.
i was trying to design a mosfet switch to protect a load from an overvoltage by interrupting the circuit
I will do as you have recommend
but if u were to design such a switch how would you go about doing it

 

[NascentOxygen]

but if u were to design such a switch how would you go about doing it

I wouldn't. I would do a google search for a similar arrangement, and base my design on that. Better brains than mine have succeeded at this task.

I noticed that your circuit does nothing to sense the overvoltage, but I presumed you were concentrating on getting the basic switch working first.

 

[knowledgeseeki]

I wouldn't. I would do a google search for a similar arrangement, and base my design on that. Better brains than mine have succeeded at this task.

I noticed that your circuit does nothing to sense the overvoltage, but I presumed you were concentrating on getting the basic switch working first.

yes i was
however i haven't found anything on google that might help
what do you think about this
http://www.linear.com/product/LT4356-3
this is the best design i have found

however my pspice package doesn't have LT43256 or any surge protector so i guess ill just start building

 

[knowledgeseeki]

You have breadboarded this circuit, or are you saying the simulation doesn't work?

You have managed to get the DC supply for the op-amp mixed in with the 24V for the load. Separate these. You intend operating the 324 on a single supply?

Is that really a 10 Ohm resistor in series with the FET's gate?!

I suggest that you remove the inductor and its associated zeners, etc, and the 24V supply, and concentrate on getting just the op-amp and the MOSFET working. Perhaps use a LED and a 2.2k resistor to represent the load, powering it off the 10V, so you have a visual indication of the MOSFET switching correctly. Only when that is working and the voltages all tally with your designed values, should you bring in the load and its associated power supply.

hi please could you give me a quick sketch of the type of circuit you just described in this post
10v to to power the mosfet and the opamp and the LED being there to give me an indication that my mosfet is switching.

thank you.

this is what I've got so far, are my connections correct.
i want the gain of my differential amplifier to be 2.
so i can decide R2 ,R1, R3 and R4 pretty easily, my question is though, is it necessary to have those resistors in place?

 

[NascentOxygen]

I think you will need those 4 resistors around the op-amp. Don't forget that the op-amp needs to be supplied with power, too.

You may need to experiment when you build it, to get the switching working how you want it to.

 

[knowledgeseeki]

I think you will need those 4 resistors around the op-amp. Don't forget that the op-amp needs to be supplied with power, too.

You may need to experiment when you build it, to get the switching working how you want it to.

I want V1 to be where i introduce my voltage pulse
so does that make V2 where i introduce for 12V power supply?

 

[NascentOxygen]

The power supply pins are often not shown on basic schematics. V2 is not for the power supply (although you may need to connect V2 to the power supply as well, to allow the op-amp to operate off a single supply.)

Your original attachment showed power supply connections. Download that op-amp's data sheet and check that you have it correct.

 

[knowledgeseeki]

Following your advice i made my load the LED and a 2.2kohm resistor and connected it, the LED turned on and off indicating i connected them properly to my 10V supply.

Having moved on from that could you please tell me if the design attached is correct
i.e does it function as a circuit breaker protecting the LED and the 2.2kohm resistor

I am trying to design it in such a way that when the LED remains off there is no fault condition, there is no difference between the overvoltage and the reference voltage (feedback from the mosfet source resistance back to the negative input of the comparator) no signal is sent to the 555 timer, which leaves the mosfet off.

When there is a voltage difference at the comparator a signal is sent to the 555 timer, a square wave is generated and the 555 timer determines when the mosfet turns on or off.
the LED, would flicker indicating that there is a fault condition being resolved

When the LED remains ON till it burns out this means, that the fault condition could not be resolved. (Unlikely scenario)

please send me your email NascentOxygen if interested the file size is too big to attach
ive tried everything.

edit: ps: why is the LED connected a 2.2kohm resistor? how would this affect my mosfet is the resistor was 2Kohm?

 

[knowledgeseeki]

in this circuit how do i prevent the voltage at the source of my Mosfet from being distorted

i am trying to feed it back to the inverting input of my opamp but i can't do that for now because it is awfully distorted

 

[NascentOxygen]

I would have said the source is grounded. Maybe you mean the MOSFET's "top" terminal? But that's not what you want to feed back either. When the FET is ON its drain voltage is low and it likely stays low even when the 15V shows some voltage spikes. The drain voltage will never trigger overvoltage protection.

Are you really powering pin 8 of the 555 from your OP-AMP output? Is that intentional?

What is the purpose of the pair of 2N2222's?

Don't you want the MOSFET to be ON and conducting during normal operation? Then if the voltage supply (shown as 15V here) goes too high the comparator switches the MOSFET OFF and no current flows from the 15V through the LED to ground. That's how I picture it to operate.

 

[knowledgeseeki]

I would have said the source is grounded. Maybe you mean the MOSFET's "top" terminal? But that's not what you want to feed back either. When the FET is ON its drain voltage is low and it likely stays low even when the 15V shows some voltage spikes. The drain voltage will never trigger overvoltage protection.

Are you really powering pin 8 of the 555 from your OP-AMP output? Is that intentional?

What is the purpose of the pair of 2N2222's?

Don't you want the MOSFET to be ON and conducting during normal operation? Then if the voltage supply (shown as 15V here) goes too high the comparator switches the MOSFET OFF and no current flows from the 15V through the LED to ground. That's how I picture it to operate.

sorry should have been more careful before i sent a circuit design
the pair of 2N2222's is actually supposed to be 1 NPN and 1 PNP transistor to give a push-pull output, this serves to source some current to the mosfet to kick start it.

the LM324 is supposed to be a LM393 but i don't have that comparator in my package.(should have stated that! sorry)

yes the output of the comparator is to be connected to pin 8 and 4, if its a comparator the output signal should be strong enough to connect it directly

this is how i see it working.

The surge at the inverting input( is going to be replaced with the feedback from the source of the mosfet) when Vin(overvoltage) > Vref the output at the comparator is low
vice-versa the output is high
when the output is high a signal is sent to the 555 timer which sends it to the mosfet

in short, this means that when the input voltage is above the Vref no signal is sent from the comparator( this config limits the current in the load=2.2K+LED )

seeing that an overvoltage would cause a fault current in the load, i think the design might be able to respond to an overvoltage and protect the load from a fault current

 

[NascentOxygen]

sthe pair of 2N2222's is actually supposed to be 1 NPN and 1 PNP transistor to give a push-pull output, this serves to source some current to the mosfet to kick start it.

I can't picture the MOSFET requiring a "kick start", whatever that is. Its gate draws no current, so why do you think you need a push-pull stage to drive its gate? Where did you get the idea of needing a push-pull stage here?

yes the output of the comparator is to be connected to pin 8 and 4, if its a comparator the output signal should be strong enough to connect it directly

If the 555 sits there inoperational during normal use of the circuit, what will be the signal level at its pin 3 for driving the next stage?

The surge at the inverting input( is going to be replaced with the feedback from the source of the mosfet)

I can only repeat what I said earlier. The voltage at the source of the MOSFET switch is not going to show any significant change when there is a spike on the 15V supply. Shouldn't you be sampling the 15V supply instead?

 

[knowledgeseeki]

I can't picture the MOSFET requiring a "kick start", whatever that is. Its gate draws no current, so why do you think you need a push-pull stage to drive its gate? Where did you get the idea of needing a push-pull stage here?

If the 555 sits there inoperational during normal use of the circuit, what will be the signal level at its pin 3 for driving the next stage?

I can only repeat what I said earlier. The voltage at the source of the MOSFET switch is not going to show any significant change when there is a spike on the 15V supply. Shouldn't you be sampling the 15V supply instead?

you are right, the 15V should be sampled instead. that is my reference voltage. i was confused.
the push pull should be removed as well, really don't remember why i put that in there. more confusion am guessing. This is what my current circuit looks like, its almost all good except for one thing, the feedback from the mosfet source (input voltage) is much lower than the reference voltage. How do i boost the input voltage??

also i have introduced an Ac current in the load, causing the fault current condition.
the 555 is not inoperational during normal use, my mistake if i implied such. However during normal use, the 555 would be at a low, i.e sending a low signal to the mosfet thus turning it off.

 

[NascentOxygen]

Sampling the voltage across that 100 Ohm resistor is actually sampling the load current. I doubt that is what you would ever want to do. Load current is [in general] no indicator of load voltage.

You have again got the 15V supply for the control circuit intertwined with the load supply voltage. I assume you want these to be kept separate: the source powering the load should connect only to the load, nowhere else.

the 555 is not inoperational during normal use, my mistake if i implied such. However during normal use, the 555 would be at a low, i.e sending a low signal to the mosfet thus turning it off.[/color]

Have you got this wrong, or am I misunderstanding what you are trying to achieve? Surely the FET is ON during normal operation, and it turns OFF to interrupt current to the load upon detection of abnormally high voltage?

You want to interrupt the load current for a short preset period, and then reconnect the load, is that the principle of operation? Are you sure that is going to be acceptable? What is the nature of the load are you planning to power?

 

[knowledgeseeki]

Sampling the voltage across that 100 Ohm resistor is actually sampling the load current. I doubt that is what you would ever want to do. Load current is [in general] no indicator of load voltage.

You have again got the 15V supply for the control circuit intertwined with the load supply voltage. I assume you want these to be kept separate: the source powering the load should connect only to the load, nowhere else.

Have you got this wrong, or am I misunderstanding what you are trying to achieve? Surely the FET is ON during normal operation, and it turns OFF to interrupt current to the load upon detection of abnormally high voltage?

You want to interrupt the load current for a short preset period, and then reconnect the load, is that the principle of operation? Are you sure that is going to be acceptable? What is the nature of the load are you planning to power?

I am confused though, if the mosfet turns off to interrupt the current to the load, that leaves an open circuit( high resistance) won't the current just be exerted through out the load.
Shouldn't the mosfet turn on to interrupt the current, i.e allowing it to flow through it, and into ground to be drained?

My load is simply the LED connected to the resistor ( i want the LED)
to be a signaling light when there is a fault

this is what i thought about when i designed the circuit: The input voltage is the feedback voltage from the Mosfet source, when the feedback voltage is above the reference, the output at the comparator would be at a low. This low output from the comparator leads to a low output in the 555 timer, which leads to a low voltage turning the Mosfet off. (input voltage into the mosfet must be > than threshold, the low voltage would be below this value, which prompts the mosfet to turn off.)
The Mosfet would only be turned off when the current in the load is no longer at a fault. The mosfet turning off causes an open circuit, current in the load will now only be the current being supplied from the power source. The LED should be in its off state.

 

[NascentOxygen]

At this stage, don't worry about including a LED to indicate a fault condition. That's merely a decoration which can be added in later. You are confusing me and yourself by discussing two LEDs. Well, I think you are.

In your original sketch you showed an inductor and resistor representing the load. I suggested that in the preliminary stages you should substitute a LED and its current-limiting resistor for the L-R load as a visual indicator of load current status. The LED will glow when load current is flowing because in this simplified arrangement the LED current IS the load current; the LED is the load.

The MOSFET is in series with the load. When the MOSFET conducts (i.e., is ON), current flows through the load and through the MOSFET to ground, so the load gets powered. If an over-voltage is detected, I believe you want the MOSFET to become non-conducting (i.e., turn OFF), so no current flows through the load or the MOSFET.

Why are you reluctant to indicate details of the load you are wanting to protect?

 

[knowledgeseeki]

At this stage, don't worry about including a LED to indicate a fault condition. That's merely a decoration which can be added in later. You are confusing me and yourself by discussing two LEDs. Well, I think you are.

In your original sketch you showed an inductor and resistor representing the load. I suggested that in the preliminary stages you should substitute a LED and its current-limiting resistor for the L-R load as a visual indicator of load current status. The LED will glow when load current is flowing because in this simplified arrangement the LED current IS the load current; the LED is the load.

The MOSFET is in series with the load. When the MOSFET conducts (i.e., is ON), current flows through the load and through the MOSFET to ground, so the load gets powered. If an over-voltage is detected, I believe you want the MOSFET to become non-conducting (i.e., turn OFF), so no current flows through the load or the MOSFET.

Why are you reluctant to indicate details of the load you are wanting to protect?

lol, sorry i don't know the load yet. Havent decided. not reluctant at all, you've been so helpful. Any suggestions?

There is only 1 LED, represented by the diode in series with the 2.2kohm resistor

My issue is my if the mosfet conducts, i.e is on how does that interrupt the current flowing through the load? won't the mosfet being on, simply let the current flow through the mosfet?? there is no break in the circuit if the mosfet turns on to protect the load( at least i don't think there is)

 

[NascentOxygen]

My issue is my if the mosfet conducts, i.e is on how does that interrupt the current flowing through the load?

When you want to interrupt the current, you turn the MOSFET OFF. MOSFET current stops, so does load current because the load is in series with the MOSFET.

You will probably end up connecting multiple MOSFETs in parallel to produce a switch with sufficiently high current rating. I think there should be no issues with doing so.

 

[knowledgeseeki]

When you want to interrupt the current, you turn the MOSFET OFF. MOSFET current stops, so does load current because the load is in series with the MOSFET.

You will probably end up connecting multiple MOSFETs in parallel to produce a switch with sufficiently high current rating. I think there should be no issues with doing so.

well my design is evidently wrong i built part of it today and i kept on burning the 100ohm resistor as well as other components.
could you please show me how to properly connect a comparator to a 555 timer, so that the comparator output triggers the timer to turn on and off.
I think i got that quite wrong, therefore everything else connected to it has been done wrongly.

1 question though: if the mosfet is off and current can't flow through it, how does that current get to the source( that is what i am using as feedback for my input)
also if the mosfet turns off when there is a fault, won't the fault just remain where the 2.2kohm+LED is?

 

[knowledgeseeki]

so here i have reconnected the comparator to the 555 timer, properly i believe.
I tested it by introducing a pulse voltage at R8 and R7, introduced into the inverting feedback.
The timer and the comparator responded the way i wanted them to.
However pulse voltage won't be supplied to it when i am building it.
I want to introduce a feedback from my timer output into the inverting input of the comparator(in this package, the opamp)
What can i do to fix it?? it is just giving me a flat line and the timer doesn't seem to be triggering once i try and feed it back to the comparator.

 

[NascentOxygen]

I want to introduce a feedback from my timer output into the inverting input of the comparator(in this package, the opamp)

Why would you want to?

1 question though: if the mosfet is off and current can't flow through it, how does that current get to the source( that is what i am using as feedback for my input)
also if the mosfet turns off when there is a fault, won't the fault just remain where the 2.2kohm+LED is?

Current doesn't get to the FET's source terminal when the FET is OFF. So that's obviously not a suitable place to sample the voltage. You must sample the +15V supply at the +15V terminal.

sorry i don't know the load yet. Havent decided.

I see. That explains the cavalier regard for the load you are "protecting" then:

When the LED remains ON till it burns out this means, that the fault condition could not be resolved.

For the arrangement you are using, your 555 output goes HIGH for a set period during which the MOSFET requires a LOW to turn it OFF. So it looks like you need an inverter between those two stages (or else invent a 555 one-shot where the output is normally HIGH but falls LOW during the timing period).

 

[knowledgeseeki]

Why would you want to?

Current doesn't get to the FET's source terminal when the FET is OFF. So that's obviously not a suitable place to sample the voltage. You must sample the +15V supply at the +15V terminal.

I see. That explains the cavalier regard for the load you are "protecting" then:


For the arrangement you are using, your 555 output goes HIGH for a set period during which the MOSFET requires a LOW to turn it OFF. So it looks like you need an inverter between those two stages (or else invent a 555 one-shot where the output is normally HIGH but falls LOW during the timing period).

yes i have changed somethings, currently a monostable design. So my trigger for the 555 requires a low to output a high.
I want to feedback the timer output to the inverting input because i want the time period for my RC circuit to decide the switching speed. Not some pulse i have to keep introducing. (I don't think that's how it would work when i start building it) the only thing i would be introducing is extra current in the load to see if my circuit can break the circuit to protect said load.

Well my load could be anything as simple as a resistor. or a small motor. What is more important is to see if my circuit can protect it.

:'( am really confused now.

my whole circuit requires a comparator, a 555 timer and my mosfet in series with my LED+2.2Kohm resistor.

What should should my input be? what am i comparing to my reference if it isn't the voltage at the source of the FET? :'(

also if the FET is off to protect the circuit i don't see how that works. ( definitely designed it wrong then)

This load i have yet to decide upon( we will call it a 2K ohm resistor) is it supposed to be in series with the FET?

Where would my extra current be introduced?

 

[NascentOxygen]

my trigger for the 555 requires a low to output a high.

Which needs an inverter to change this to a low to turn off the MOSFET.

I want to feedback the timer output to the inverting input because i want the time period for my RC circuit to decide the switching speed. Not some pulse i have to keep introducing.

I don't understand this, but I do know that you don't want that feedback.

What should should my input be? what am i comparing to my reference if it isn't the voltage at the source of the FET? :'(

The input will be the +15V, because you indicated it may jump to a high level and imperil the load. Remember, you are building this circuit to disconnect the power in the event that the supply jumps higher than +15V.

also if the FET is off to protect the circuit i don't see how that works. ( definitely designed it wrong then)

The FET turns OFF and halts current flow. If there is no current flowing, the load can't be damaged by excessive current due to excessive voltage.

This load i have yet to decide upon( we will call it a 2K ohm resistor) is it supposed to be in series with the FET?

Where would my extra current be introduced?

The excess current would come from the + supply (when it jumps from 15V to a higher voltage).

 

[knowledgeseeki]

Which needs an inverter to change this to a low to turn off the MOSFET.

I don't understand this, but I do know that you don't want that feedback.

The input will be the +15V, because you indicated it may jump to a high level and imperil the load. Remember, you are building this circuit to disconnect the power in the event that the supply jumps higher than +15V.

The FET turns OFF and halts current flow. If there is no current flowing, the load can't be damaged by excessive current due to excessive voltage.

The excess current would come from the + supply (when it jumps from 15V to a higher voltage).

hi, so i have finally decided on what my load should be.
A small motor.
coil 1= 100mH
coil 2= 100mH
coupling= 0.5

Does this design make more sense? had to change the whole thing based on our discussion. Now i believe the mosfet should appropriately turn off to prevent current flow.

the push pull drive the switching mosfet

however i am not getting the results i expect.

I expect to get a pulse wave from my comparator.
When the input > reference the comparator output is at a low.

I expect to get a pulse wave from my 555 timer, where when comparator output is low timer output is high. This information should now be sent into the mosfet which will switch on when the timer output is high. [ this is when the input voltage> ref voltage, i want it to switch off during this instance] so to do that i should include an inverter?

 

[knowledgeseeki]

having included a NAND gate everything works just fine but 1 little glitch.
When i replace the voltage pulse with a 3K ohm load and a mosfet then feedback the output of my push pull config to the gate of the mosfet i get nothing?

am i sure this is how its done this time, because mosfet turns on when VGS>Voltage threshold. and off when VGS is less.
I used the NAND gate to inverse the timer output which was a high.

so far this is what i have

please let me know if i am doing something wrong

 

[NascentOxygen]

hi, so i have finally decided on what my load should be.
A small motor.
coil 1= 100mH
coil 2= 100mH
coupling= 0.5

That's a motor??

the push pull drive the switching mosfet

What has that reapppeared? I can't see it serves any purpose. The 555 can drive the MOSFET directly.

I expect to get a pulse wave from my 555 timer, where when comparator output is low timer output is high. This information should now be sent into the mosfet which will switch on when the timer output is high. [ this is when the input voltage> ref voltage, i want it to switch off during this instance] so to do that i should include an inverter?

You could use a spare FET (plus a resistor) to form an inverter.

 

[NascentOxygen]

There are issues with a 555 monostable when the trigger has not returned to HI before the 555's pulse ends and output returns to LO. (See its data sheet). A fix for this is probably to include a differentiator between the comparator and the 555; this will produce a narrow pulse on each transition of the comparator's output. It's just a capacitor and resistor, and possibly a diode.

 

[NascentOxygen]

There are issues with a 555 monostable when the trigger has not returned to HI before the 555's pulse ends and output returns to LO. (See its data sheet). A fix for this is probably to include a differentiator between the comparator and the 555; this will produce a narrow pulse on each transition of the comparator's output. It's just a capacitor and resistor, and possibly a diode.

Oh, I see it's precisely what you had on the 555's trigger input back here:

So you just need to restore that. The 2.2μF may be bigger than is needed, try 0.1μF and see how that works out. Have a look at the waveform on pin 2 and see how healthy it looks. It should show pulses that quickly return to baseline level well before the 555 gets triggered again.