Hi all,

Before I start, I am not an electronics engineer, at best a hobbyist who is working on an Arduino project and is looking to add some protection to the circuit when designing a PCB to put all the bits on. There should be an attachment of the schematic. I apologize if this is a stupid question.

I've been doing a lot of reading and perhaps unnecessarily want to add reverse voltage and over voltage protection the my circuit. The two circuit designs that I had come across were the P-Channel Mosfet for the reverse voltage, and the Crowbar for the over voltage. Now below is my first schematic. It doesn't have all the values for each component at this stage. What I am interested in knowing is if the design would actually work?

From what I understand the M1 PMOS (P Channel Mosfet) will only activate when it has the correct polarity. The Zener D1 will clamp the voltage to 15V. The resistor R1 is there to limit the current that will flow through the zener D1? If the polarity is reversed the circuit will never be complete so no current flows. Am I correct so far?

Next... if the polarity is correct, M1 PMOS allows current to flow. Capacitors C1, C4, C5 I have as bypass caps as the voltage supply will be coming via a rotating slip ring so there could be moments where a good contact is not available so these caps fill in the gaps.

Next I found the LTC1696 which is an over voltage device that monitors 2 voltages. So my theory is when voltage flows, D2 clamps the voltage to 15V, R6 limits the current through D2 (actually I'm thinking I dont need D1 if I have D2?).

The LTC1696 has a threshold voltage of 0.88V on the FB1 and FB2 pins. If more voltage than 0.88 is received on those pins it fires the Out pin. Voltage dividers (R4-R5, R2-R3) are used to set the actual voltage value that you want to trip FB1/FB2.

If FB2 goes over its threshold it fires the Out pin which latches Q1 transister which latches U1 SCR to ground which takes all the current and blows the fuse. This is protecting against the actual supply voltage being to high.

Assuming none of that is triggered, voltage flows to the v7805-2000R non isolated switching regulators. If FB1 goes over its threshold it fires the Out pin which latches Q1 transister which latches U1 SCR to ground which takes all the current and blows the fuse.

I would need to figure out the values of resistors based on what voltage value I am protecting against (say protect up to 50V on the supply, and protect against 5.5V or higher on the switching regulator out side).

Excuse my ignorance, but am I on the right track with this design?

 

Resistors R2,R3,R4,R5 can be any values, as long as you maintain 0.88V at the junction and 50v or 5v at the supply side.

So for the 50V, R5 can be 1K, R4 55.8K.
for 5V R3 1K, R2 4.68K

I would omit Q1,R7,R7 and connect the out to the scr Gate direct.

 

Resistors R2,R3,R4,R5 can be any values, as long as you maintain 0.88V at the junction and 50v or 5v at the supply side.

So for the 50V, R5 can be 1K, R4 55.8K.
for 5V R3 1K, R2 4.68K

I would omit Q1,R7,R7 and connect the out to the scr Gate direct.

Awesome, so I take it that this circuit fundamentally will work as desired. To explain the 50V side a little more. The V7805 switching regulators work in the 7-18V (max) range. This is the reason I chose 15V zeners for D1 and D2 to clamp the input to those at 15V, and would tweak R4/R5 to go above 0.88V at say 16V from the supply. The 50V is the arbitrary value that I would size all the other components for so the supply side of the circuit would protect up to 50V. So for example the P-MOSFET has a maximum voltage of 60V, the SCR 600V (not an issue ), the LTC1696 27V. So if I understand this correctly, the zener D1 and D2 have a power rating, say 1/2 watt and the resistor R1/R6 again we'll say 1/2 watt. So I would have to calculate the R1/R6 resistor value based on 50V so that would be 35V left for the resister to handle. So I=V / R, I = 35V x 5000Ω, I = 0.007 or 7mA. Then P = I x V, P = 0.007A x 35V, P = 0.245 watts. So I could use a 5K resistor to handle the 35V. Am I right so far?

Do I need D1 and D2, or just D1? I'm thinking I just need D1. Am I correct in thinking that just D1 will clamp the voltage to all components to 15V after the P-MOSFET and before the LTC1696 and V7805 regulators?

Thanks for your help, it is very much appreciated.

 

Why not omit the zeners, and set the resistors to trip the thyristor at 5.5V and 16V??

 

Why not omit the zeners, and set the resistors to trip the thyristor at 5V and 16V??

The LTC1696 is a 27v MAX on the Vcc. I'm guessing as long as my supply voltage doesn't spike over 27v then I'm good as the LTC1696 wont get fried, and would protect the regulators, supply side and 5v loads, but if I got 50v on the supply the p-mosfet would cope, but the LTC1696 would fry and then potentially not perform its function and not crowbar the circuit which would mean 50V to the switching regulators which would fry those and potentially the 5v loads also. So I'm thinking the zener and its resistor give some insurance that the LTC 1696 will always be available to do its job?? Or at least with my limited knowledge I think that is what would happen?? Is my reasoning correct??

As this is a personal project I'm proposing this from two angles, a) to learn something, and b) to protect the circuit's components.

 

Not related to OVP, but

Connecting voltage regulators of any sort, with a very few exceptions, will typically not result in anything close to equal sharing of the load current. The regulator with the highest output voltage, even if by just millivolts, will supply all of the load current until it reaches its current limit and its voltage begins to decline. The second regulator will then start to supply current. If the "load regulation" (decrease in output voltage with increase in output current) isn't very good, you may get current sharing below the current limit of the higher-voltage device.

This is not necessarily a big problem provided the regulators do limit current at a level that protects them from damage. It does mean shorter overall lifetime for the system since temperature is a big player in failure rate and the hotter device will fail first. If they shared equally, this would be less of an issue. In some cases the device carrying the higher current will shut down due to over-temperature. This can lead to complete shut-down if the remaining device(s) overheat before the one(s) that shut down cool sufficiently to re-start. Again, as long as the device can tolerate this, no damage should result, but the system will turn off.

There are methods that can enhance sharing, such as using low-value "ballast" resistors in series with the output of each individual regulator, but this degrades overall voltage regulation accuracy and reduces efficiency. Usually the best solution , where possible, is to use a single regulator that can safely handle the full load current.

===
If you want to protect the fault detection IC, the zener on its supply should go directly to ground and the resistor should be in series between the input supply and the zener cathode. If you change the circuit so the IC drives the gate of the SCR directly, be sure to have some decoupling local to the IC so that it can provide a good high-current pulse to drive the SCR. If the SCR isn't driven with lots of current, it will tend to turn on in a localized area at first and almost certainly fail short-circuit - which of course protects your circuit. If you can get the whole SCR to turn on fast enough there is some chance it will survive the current required to blow the fuse.

 

Not related to OVP, but

Connecting voltage regulators of any sort, with a very few exceptions, will typically not result in anything close to equal sharing of the load current. The regulator with the highest output voltage, even if by just millivolts, will supply all of the load current until it reaches its current limit and its voltage begins to decline. The second regulator will then start to supply current. If the "load regulation" (decrease in output voltage with increase in output current) isn't very good, you may get current sharing below the current limit of the higher-voltage device.

This is not necessarily a big problem provided the regulators do limit current at a level that protects them from damage. It does mean shorter overall lifetime for the system since temperature is a big player in failure rate and the hotter device will fail first. If they shared equally, this would be less of an issue. In some cases the device carrying the higher current will shut down due to over-temperature. This can lead to complete shut-down if the remaining device(s) overheat before the one(s) that shut down cool sufficiently to re-start. Again, as long as the device can tolerate this, no damage should result, but the system will turn off.

There are methods that can enhance sharing, such as using low-value "ballast" resistors in series with the output of each individual regulator, but this degrades overall voltage regulation accuracy and reduces efficiency. Usually the best solution , where possible, is to use a single regulator that can safely handle the full load current.

===
If you want to protect the fault detection IC, the zener on its supply should go directly to ground and the resistor should be in series between the input supply and the zener cathode. If you change the circuit so the IC drives the gate of the SCR directly, be sure to have some decoupling local to the IC so that it can provide a good high-current pulse to drive the SCR. If the SCR isn't driven with lots of current, it will tend to turn on in a localized area at first and almost certainly fail short-circuit - which of course protects your circuit. If you can get the whole SCR to turn on fast enough there is some chance it will survive the current required to blow the fuse.

Thanks for the info, didn't know that about the connected regulators. I can probably split my loads. Load 1 would be 36 dotstar LEDs. 20mA per color, so 60mA per LED if showing white. 36 x 0.06 = 2.16A which is a smidge over the max output, my code doesn't set any of the LED's white at the same time, so this would only be an issue in the case of a malfunction. The V7805-2000 has a Thermal shutdown protection (150° at internal IC Junction). Am I correct in thinking that this would protect the regulator somewhat - drawing to much current would "overheat" the regulator and it would shutdown?? If not I guess I need to consider some sort of current limiter which cant be a resistor (limit current to 2A, P = 2A * 5v, P = 10w). If I need to go down this path I guess there will some voltage drop to the load, the dotstar's are 5v devices, so perhaps I need a 6.5v regulator to give me some headroom.

The other load would be a Teensy, Wireless board, TFT screen which could run off the 2nd regulator. But now that I have said that I have 3 voltages to monitor.... So I need an IC that will monitor 3 voltages?

Sorry for the 20 questions, but I am very appreciative of the help.

 

Thanks for the info, didn't know that about the connected regulators. I can probably split my loads. Load 1 would be 36 dotstar LEDs. 20mA per color, so 60mA per LED if showing white. 36 x 0.06 = 2.16A which is a smidge over the max output, my code doesn't set any of the LED's white at the same time, so this would only be an issue in the case of a malfunction. The V7805-2000 has a Thermal shutdown protection (150° at internal IC Junction). Am I correct in thinking that this would protect the regulator somewhat - drawing to much current would "overheat" the regulator and it would shutdown?? If not I guess I need to consider some sort of current limiter which cant be a resistor (limit current to 2A, P = 2A * 5v, P = 10w). If I need to go down this path I guess there will some voltage drop to the load, the dotstar's are 5v devices, so perhaps I need a 6.5v regulator to give me some headroom.

The other load would be a Teensy, Wireless board, TFT screen which could run off the 2nd regulator. But now that I have said that I have 3 voltages to monitor.... So I need an IC that will monitor 3 voltages?

Sorry for the 20 questions, but I am very appreciative of the help.

Ok so been thinking after having some feedback, and have done some more googling for IC's that might reduce the complexity of this circuit. I'll endeavour to post up a revised schematic for some comment.

 

Ok so been thinking after having some feedback, and have done some more googling for IC's that might reduce the complexity of this circuit. I'll endeavour to post up a revised schematic for some comment.

Ok spent to much time surfing. But thanks to ebp's great advice going to ditch the V7805 regulators and swap in just one OKY-T/3,T/5-D12 Series, Adjustable Output 3 and 5-Amp DOSA-SMT DC/DC Converter (5 Amp version). Keeping the LTC1696 to monitor the supply voltage and the regulated voltage for over-voltage, firing the SCR to crowbar the circuit blowing the fuse.

I've attached an LTSpice (v15) schematic file for comment. LTC1696 still works as described above, voltage dividers to set the voltage threshold to fire an over voltage condition, have included the resistor values. Still have the P-MOSFET to protect against reverse voltage at beginning of circuit.

The max voltage of the DC/DC converter is now 14v. The max voltage of LTC1696 is 27v.

I have some questions I am hoping someone can school me on. I'm a little confused with the zener/resistor combo if I need to use it and the values to use.

1. Do I need to clamp the voltage to the LTC1696 with a zener to protect it from voltages over 27v?
2. If I do need to clamp voltage to the LTC1696 what needs to be added to the circuit?
3. Do I need to clamp the voltage to the DC/DC converter to protect it from voltages over 14v, or will the LTC1696 work fast enough to crowbar the circuit and protect it?
4. If I do need to clamp the voltage to the DC/DC converter what needs to be added to the circuit?

Thanks in advance.

 

You could stick with the two regulators if you can partition your load suitably. You could also do a pretty good job of overvoltage protection by diode-ORing the two outputs together. Two 1N4148 diodes could be used, one from each regulator output to the "top" end of R2. The diodes would drop about 0.6 to 0.7 V, so you would adjust the value of R2 accordingly. You do lose a bit of accuracy this way, but usually there is sufficient margin between normal operating voltage and damaging overvoltage that it shouldn't be a problem. Most things that normally operate on 5 V will survive up to 6 V without damage. If you used the diodes and set the resistors for say 5.7 V at the regulator outputs (so about 5 V at the top end of R2), you should get good protection. You would need to check the spec's for "absolute maximum" supply voltage for the devices running from the regulator.

It is extremely rare for the output of a voltage regulator to go just a little above what it should be, though time and temperature do cause small shifts. A 5 volt regulator, for example, will typically be specified for some specific maximum under worst-case conditions. If it goes higher than that, it is nearly invariably due to something in the regulator short-circuiting which results in the input voltage appearing at the output. That tends to be a very fast process so you don't need to worry about something like LEDs getting excessive current for a long time before the crowbar would fire. They'll get a large but very short overcurrent which they will most likely survive with no problem. Most likely.

If you program your crowbar circuit for 14 volts, it should very adequately protect the LTC1696 as long as there is some capacitance on the input supply (which slows the rate of change of voltage), as long as you don't set the delay to be too long. The regulator will likely survive quite well.

One thing I overlooked before is the slip ring issue. When you have a lot of capacitance on the "output" side of the slip ring, you can get very high transient currents if brush breaks then remakes contact. The caps will discharge a bit during break and then recharge on make, with the current limited only by what the supply can deliver and the circuit impedance. If the supply has capacitors at the output (always the case, for anything expect batteries), it may be able to deliver a transient current well above the normal full-load rating. Some means of preventing such transient currents may be necessary to protect the slip rings. Depending on the rings rating, a small resistance might be adequate, but of course it will waste some power in normal operation. Inductance can be used, but that introduces problems with voltage transients. An active current limiting circuit could be used, but like the resistor that will come at a cost in overall efficiency and add some complexity.

 

You could stick with the two regulators if you can partition your load suitably. You could also do a pretty good job of overvoltage protection by diode-ORing the two outputs together. Two 1N4148 diodes could be used, one from each regulator output to the "top" end of R2. The diodes would drop about 0.6 to 0.7 V, so you would adjust the value of R2 accordingly. You do lose a bit of accuracy this way, but usually there is sufficient margin between normal operating voltage and damaging overvoltage that it shouldn't be a problem. Most things that normally operate on 5 V will survive up to 6 V without damage. If you used the diodes and set the resistors for say 5.7 V at the regulator outputs (so about 5 V at the top end of R2), you should get good protection. You would need to check the spec's for "absolute maximum" supply voltage for the devices running from the regulator.

It is extremely rare for the output of a voltage regulator to go just a little above what it should be, though time and temperature do cause small shifts. A 5 volt regulator, for example, will typically be specified for some specific maximum under worst-case conditions. If it goes higher than that, it is nearly invariably due to something in the regulator short-circuiting which results in the input voltage appearing at the output. That tends to be a very fast process so you don't need to worry about something like LEDs getting excessive current for a long time before the crowbar would fire. They'll get a large but very short overcurrent which they will most likely survive with no problem. Most likely.

If you program your crowbar circuit for 14 volts, it should very adequately protect the LTC1696 as long as there is some capacitance on the input supply (which slows the rate of change of voltage), as long as you don't set the delay to be too long. The regulator will likely survive quite well.

One thing I overlooked before is the slip ring issue. When you have a lot of capacitance on the "output" side of the slip ring, you can get very high transient currents if brush breaks then remakes contact. The caps will discharge a bit during break and then recharge on make, with the current limited only by what the supply can deliver and the circuit impedance. If the supply has capacitors at the output (always the case, for anything expect batteries), it may be able to deliver a transient current well above the normal full-load rating. Some means of preventing such transient currents may be necessary to protect the slip rings. Depending on the rings rating, a small resistance might be adequate, but of course it will waste some power in normal operation. Inductance can be used, but that introduces problems with voltage transients. An active current limiting circuit could be used, but like the resistor that will come at a cost in overall efficiency and add some complexity.

Thanks for your reply ebp . Ok so I think I'll stick with the one DC/DC converter as shown below. I am reasonably comfortable that an over voltage up to 27v will be covered as the LTC1696 is rated to that voltage, so it should still be operational and crowbar the circuit. If I understand you correctly you are saying to have a capacitor local to the LTC1696 to ensure it has a voltage, but not to make this to big because potentially it will delay the LTC1696 starting (I am assuming this situation would be when first turning the circuit on and an over voltage exists immediately, the cap at the LTC1696 will take some moments to fully charge before current flows to the LCT1696 and during that time the circuit will experience the over voltage - correct??)

I should probably stop there... but my curiosity has the better of me. What to do if I wanted to protect against a higher voltage? I had originally shown/thought of placing a zener at the LTC1696 to limit the voltage on the supply rail so it doesn't get fried but I couldn't figure the correct resistor/zener combo. But I've looked some more at the LTC1696 datasheet (snips shown below). Am I correct in reading the graphs that the LTC1696 draws ~170uA at idle and when active will draw 1.2mA from around 8-9v supply (I am assuming that active means it is responding to an over voltage situation and firing the OUT pin).

If that is the case can I do something like shown below with a resistor and zener rated at 1 watt?



This is from the LTC1696 data sheet.
VCC (Pin 3): PowerSupply. The pin is connected separately from the power supply output that the chip is monitoring. Its input range is from 2.7V to 27V. The quiescent current is typically 100µA in standby mode when the device is operating at 5V. The quiescent current increases to 170µA when operating at 12V.

OUT (Pin 4): Output Current Limit Driver. Capable of delivering continuous current, typically 80mA, at high supplies.





A note on the slip ring I mentioned. This is something that I will design and 3d print. It will consist of 2 machined metal rings that will sit (insulated) on a servo motor shaft. A "carrier" will house two roller ball bearings that will be spring loaded onto their respective rings.