Biggest problem I can see with that circuit is trip current. As drawn any short will have to get Q1 to drop 9.5 volts while fully on. With a 0.1 ohm on resistance FET that would require 95 amps, which a short couldn't draw anyway due to R2. If the 12v and 5v supplies are truly independent, any short would trigger the circuit after bringing the 12v power supply down to near 2.5 volts. If your simulation program can make graphs, see what the power supply voltage does when J1 is closed. It would also be instructive to watch the voltage of both inputs to the comparator at the same time.
My discrete component solution can be flipped over by replacing the NPN with a PNP and the N-FET with a P-FET and reversing the power supply connections.
The mosfet you mention is very close to tolerable so I would be nervous using it. I am curious as to where you are going with this? There must be a noticeably flaw in my design that you see and I don’t. I suspect it’s the oscillation as you mentioned a few days back but my mind can’t grasp this notion, yet.
Also the mosfet data sheet states "vertical mosfet", ill check into this now as I have never seen this term applied in a mosfet before. I hope this does not reflect my inexperience.
Okay, 'just wanted to make sure your motor was not going to draw more than 200 mA.
It's not clear to me how the MOSFET gets turned on if it's off to start with. If it's off, the negative input to to the comparator will be pulled low by the load and by the 0.5-ohm resistor, so it will remain off. IOW, it can only turn on if the comparator's minus input is higher than the plus input; but that can only happen if it''s already on.
Also, I'm not entirely sure what the purpose of the 0.5-ohm resistor is, unless it's just to protect the MOSFET from peak over-current in case of a short. The MOSFET's abs max peak current rating is 600 mA. 12V into 0.5 ohms is 24 A.
The instant the circuit is powered the following as I envision this should take place.
The mosfet is set to conduct current, as we know the pull up is in place to accomplish this.
The comparator is waiting for a fault condition. The fault is when the reference resistor voltage shoots up over 2.5 volts. This will then trigger the comparator to set the output to low forcing the mosfet to turn off.
As I look now the sense resistor is useless and I can remove it. Thank you for noticing.
The one thing I may be overlooking is the state of the comparators output when the circuit is powered on the first time. In fact as I type this I may be digesting what you have been cramming down my through all along. Is this it Phil, are you concerned about the comparators output when the circuit is first turned on?
Biggest problem I can see with that circuit is trip current. As drawn any short will have to get Q1 to drop 9.5 volts while fully on. With a 0.1 ohm on resistance FET that would require 95 amps, which a short couldn't draw anyway due to R2. If the 12v and 5v supplies are truly independent, any short would trigger the circuit after bringing the 12v power supply down to near 2.5 volts. If your simulation program can make graphs, see what the power supply voltage does when J1 is closed. It would also be instructive to watch the voltage of both inputs to the comparator at the same time.
My discrete component solution can be flipped over by replacing the NPN with a PNP and the N-FET with a P-FET and reversing the power supply connections.
Lawson
Oops Lawson did not see you post there. Ill check into the graphs casue if I have that feature that's a fantastic approach.
The mosfet is set to conduct current, as we know the pull up is in place to accomplish this.
The pullup prevents the pMOSFET from conducting until the gate is driven low, the same as a gate pulldown prevents an nMOSFET form conducting until driven high. (You did say you tested the circuit, right? )
Phil I am going to scream about my office now, and I haven't yet finished my Redbull! You've got me trapped again, Ill post more on this after some thinking.
Alright I think I have this fixed and yes it is bread-boarded this time. Take a look Phil / others, I am excited to know what you think. This circuit will not latch but its fine with me. You can see I added a capacitor so that on a new boot (power up) it will create a short for the mosfet to turn on. Yet the cap is small enough to let the mosfet turn off when a fault condition is true.
Attachment not found.
edited****
"This circuit will not latch but its fine with me." I guess it will latch and depends on the capacitor size.
You asked about vertical MOSFETs ... Look at the bottom of this page and here for a picture. The MOSFET is laid out along a groove etched along crystal planes and appears sort of vertical in relation to the wafer surface. As I recall, this technique allowed better control of gate construction than could otherwise be achieved at that time.
In your new circuit, what is the R1C1 time constant? That's how long it will take to respond to a short circuit, during which time the MOSFET's peak current spec will be violated. Is that what you want?
BTW, there is someplace you could place a cap where I think it would be effective.
That might work. I was thinking more of the bottom end where you've got it and the top end on V1+. It wouldn't have to be a very big one, I don't think, since the motor is an inductive load that won't sink much current right away.
I have to say that the electrolytic cap might be a small explosion ready to happen so I am now using a 0.047uF with only positive trade offs meaning the RC time will be shorter and it wont explode.
I think I got a winner, unless any one finds another flaw.
The biggest flaw I can see is that Q1 still requires 9.5 volts across it for the circuit to transition from on to off. If the motor draws 800mA, Q1 will only drop ~8 of the 9.5 needed volts. (a 5 ohm load should draw this current) In this case, Q1 is then stuck dissipating 6.4 watts which is enough to fry Q1. (assuming it's still a BSS192)
Could you please dig a little deeper in your explanation? Other words how did you come up with these numbers? I have my osilicope here and the voltage at the mosfet gate is almost a perfect square wave from 0 - 12 volts, 0 volts being normal mode and 12 volts at the gate during a short.
I did mistakenly list the mosfet as you noted. The real mosfet is an IRF5210
Attached is the circuit as I have it on my bench with all the corrected values.
- J2 is a connector to a DC pump with max current at 800ma.
Ok looking at the most recent schematic I assume all grounds are common.
First lets find the conditions are required to get this circuit to switch off. Lets start at the comparator. It's powered by an (independent?) 5 volt supply with a voltage divider supplying half that (2.5v) to the + input. In the "on" state of the circuit the - input of the comparator is at almost 12 volts. This drives activates the open collector output of the comparator and keeps Q1 on. Now to turn off the comparator and Q1 the - input must be pulled to a lower voltage than the + input of the comparator. If I assume that the 12v supply is ideal, this would require dropping (12 - 2.5) = 9.5 volts across Q1 while it is fully on, so across a 0.06 ohm resistor. This ideal 158 amp fault current would require a short with an equivalent series resistance of 0.015 ohms or less. Now your real 12 volt power supply can't supply 158 amps, so when it's rated current output is exceed the supply voltage starts to drop. In this case, the 12v supply drops down to about 5 volts before Q1 starts turning off, at which point the voltage at the - input of the comparator will rapidly drop from current voltage of the 12v supply to ground potential.
The above bad behavior will show up in two ways. First, the circuit will happily draw the full rated current of the 12 volt power supply before tripping, generally a bad idea for a circuit intended to limit fault current to a light load. If the motor only ever draws 800mA, should the current limiting circuit let it draw up to 150 amps? Second, there will be a ball-park 100uS dip in the 12v supply voltage down to about 5 volts every time the circuit trips. Hard to see when triggering the circuit manually, but easy to see if you have a sampling oscilloscope configured for one-shot operation. (or use a function generator and some spare parts to repeatedly trigger and reset the circuit) You can get away with this abuse right now, because Q1 tough enough to take the current abuse.
I appreciate your analysis but I respectfully disagree with the resistor value that you listed. I have since removed the feedback resistor and instead use the motors inductance as a feedback source to the comparators negative pin, far greater than a 0.06Ohm. Will this still create a issue in your professional opinion? I could drop the voltage to the comparators + side to help in the current rise. The mosfet I have selected would be happy dropping 30Amps and since this is a quick transition I was not concerned. All said and done I could be way off so please lets continue this discussion.
I instead use the motors inductance as a feedback source to the comparators negative pin, far greater than a 0.06Ohm.
True, the motor's resistance is far greater than 0.06ohms and that matters when Q1 is off with an equivalent resistance over 100Kohm. When Q1 is on, the equivalent resistance of Q1 will be about 0.06 ohms and this overwhelms the equivalent resistance of the motor this pulls the motor positive and U1A.2 ( - input ) up to 12 volts. You want this to happen, otherwise your motor wouldn't spin!
*snip* I could drop the voltage to the comparators + side to help in the current rise.
This would make things worse. I'd suggest a voltage of ~11.88 volts on the comparator + input to get a trip current of ~2 amps with the chosen FET. (you'll also have to run the comparator off of the 12v supply and find one with a rail to rail input. for instance a LMC6772)
A quick experiment would help show my point. Set your bench supply to it's maximum current limit, and see how many 10-30 ohm power resistors you can put in parallel with the running motor before the circuit trips.
The mosfet I have selected would be happy dropping 30Amps and since this is a quick transition I was not concerned.
I wouldn't call the IRF5210 "happy" to conduct 30 amps steady state unless it's screwed to a LARGE heat sink. I estimate ~50 watts dissipated steady state at 30 amps, a good bit more that the 2-3 watts the TO-220 package can dissipate on it's own.
This would make things worse. I'd suggest a voltage of ~11.88 volts on the comparator + input to get a trip current of ~2 amps with the chosen FET.
.
I was under the impression that the sooner the comparator seen (sensed) a voltage close to the reference, the sooner it would shut the mosfet off. Under this notion the voltage at the (+) pin on the comparator should be as low as possible, so it would trigger much sooner. I am not to sure about setting the reference voltage close to 11Volts because when the motor is running the voltage is close to zero. It would require more time to shut off the mosfet if the ref was 11Volts.
Well Ill test more tomorrow and think about what you are saying.
Comments
My discrete component solution can be flipped over by replacing the NPN with a PNP and the N-FET with a P-FET and reversing the power supply connections.
Lawson
BTW, before I comment further, the BSS192 is rated for 200mA with an RDS(on) of about 10 ohms. Is that adequate for your app?
-Phil
The mosfet you mention is very close to tolerable so I would be nervous using it. I am curious as to where you are going with this?
Also the mosfet data sheet states "vertical mosfet", ill check into this now as I have never seen this term applied in a mosfet before. I hope this does not reflect my inexperience.
It's not clear to me how the MOSFET gets turned on if it's off to start with. If it's off, the negative input to to the comparator will be pulled low by the load and by the 0.5-ohm resistor, so it will remain off. IOW, it can only turn on if the comparator's minus input is higher than the plus input; but that can only happen if it''s already on.
Also, I'm not entirely sure what the purpose of the 0.5-ohm resistor is, unless it's just to protect the MOSFET from peak over-current in case of a short. The MOSFET's abs max peak current rating is 600 mA. 12V into 0.5 ohms is 24 A.
-Phil
The mosfet is set to conduct current, as we know the pull up is in place to accomplish this.
The comparator is waiting for a fault condition. The fault is when the reference resistor voltage shoots up over 2.5 volts. This will then trigger the comparator to set the output to low forcing the mosfet to turn off.
As I look now the sense resistor is useless and I can remove it. Thank you for noticing.
The one thing I may be overlooking is the state of the comparators output when the circuit is powered on the first time. In fact as I type this I may be digesting what you have been cramming down my through all along. Is this it Phil, are you concerned about the comparators output when the circuit is first turned on?
Oops Lawson did not see you post there. Ill check into the graphs casue if I have that feature that's a fantastic approach.
-Phil
Phil I am going to scream about my office now, and I haven't yet finished my Redbull! You've got me trapped again, Ill post more on this after some thinking.
Attachment not found.
edited****
"This circuit will not latch but its fine with me." I guess it will latch and depends on the capacitor size.
In your new circuit, what is the R1C1 time constant? That's how long it will take to respond to a short circuit, during which time the MOSFET's peak current spec will be violated. Is that what you want?
BTW, there is someplace you could place a cap where I think it would be effective.
-Phil
thank you for the links I just happen to visit them last night and find it interesting. I am a nerdy girl :P
So currently I am testing with a 47K pull up and a 1uf electrolytic capacitor. And if t = R x C then it would be 47mS.
Phil,
I am interested in where you would place a capacitor. I am guessing you would place it here.
Attachment not found.
-Phil
I think I got a winner, unless any one finds another flaw.
Lawson
Could you please dig a little deeper in your explanation? Other words how did you come up with these numbers? I have my osilicope here and the voltage at the mosfet gate is almost a perfect square wave from 0 - 12 volts, 0 volts being normal mode and 12 volts at the gate during a short.
I did mistakenly list the mosfet as you noted. The real mosfet is an IRF5210
Attached is the circuit as I have it on my bench with all the corrected values.
- J2 is a connector to a DC pump with max current at 800ma.
Attachment not found.
First lets find the conditions are required to get this circuit to switch off. Lets start at the comparator. It's powered by an (independent?) 5 volt supply with a voltage divider supplying half that (2.5v) to the + input. In the "on" state of the circuit the - input of the comparator is at almost 12 volts. This drives activates the open collector output of the comparator and keeps Q1 on. Now to turn off the comparator and Q1 the - input must be pulled to a lower voltage than the + input of the comparator. If I assume that the 12v supply is ideal, this would require dropping (12 - 2.5) = 9.5 volts across Q1 while it is fully on, so across a 0.06 ohm resistor. This ideal 158 amp fault current would require a short with an equivalent series resistance of 0.015 ohms or less. Now your real 12 volt power supply can't supply 158 amps, so when it's rated current output is exceed the supply voltage starts to drop. In this case, the 12v supply drops down to about 5 volts before Q1 starts turning off, at which point the voltage at the - input of the comparator will rapidly drop from current voltage of the 12v supply to ground potential.
The above bad behavior will show up in two ways. First, the circuit will happily draw the full rated current of the 12 volt power supply before tripping, generally a bad idea for a circuit intended to limit fault current to a light load. If the motor only ever draws 800mA, should the current limiting circuit let it draw up to 150 amps? Second, there will be a ball-park 100uS dip in the 12v supply voltage down to about 5 volts every time the circuit trips. Hard to see when triggering the circuit manually, but easy to see if you have a sampling oscilloscope configured for one-shot operation. (or use a function generator and some spare parts to repeatedly trigger and reset the circuit) You can get away with this abuse right now, because Q1 tough enough to take the current abuse.
Lawson
I appreciate your analysis but I respectfully disagree with the resistor value that you listed. I have since removed the feedback resistor and instead use the motors inductance as a feedback source to the comparators negative pin, far greater than a 0.06Ohm. Will this still create a issue in your professional opinion? I could drop the voltage to the comparators + side to help in the current rise. The mosfet I have selected would be happy dropping 30Amps and since this is a quick transition I was not concerned. All said and done I could be way off so please lets continue this discussion.
True, the motor's resistance is far greater than 0.06ohms and that matters when Q1 is off with an equivalent resistance over 100Kohm. When Q1 is on, the equivalent resistance of Q1 will be about 0.06 ohms and this overwhelms the equivalent resistance of the motor this pulls the motor positive and U1A.2 ( - input ) up to 12 volts. You want this to happen, otherwise your motor wouldn't spin!
This would make things worse. I'd suggest a voltage of ~11.88 volts on the comparator + input to get a trip current of ~2 amps with the chosen FET. (you'll also have to run the comparator off of the 12v supply and find one with a rail to rail input. for instance a LMC6772)
A quick experiment would help show my point. Set your bench supply to it's maximum current limit, and see how many 10-30 ohm power resistors you can put in parallel with the running motor before the circuit trips.
I wouldn't call the IRF5210 "happy" to conduct 30 amps steady state unless it's screwed to a LARGE heat sink. I estimate ~50 watts dissipated steady state at 30 amps, a good bit more that the 2-3 watts the TO-220 package can dissipate on it's own.
Marty
.
I was under the impression that the sooner the comparator seen (sensed) a voltage close to the reference, the sooner it would shut the mosfet off. Under this notion the voltage at the (+) pin on the comparator should be as low as possible, so it would trigger much sooner. I am not to sure about setting the reference voltage close to 11Volts because when the motor is running the voltage is close to zero. It would require more time to shut off the mosfet if the ref was 11Volts.
Well Ill test more tomorrow and think about what you are saying.