High power pulses killing low power resistors

[CasualKilla]

So I am currently struggling with resistors blowing out on a PCB. They are 1/4 Watt 1206 resistors, The resistors where designed for steady state, but there is a power spike when the device is turned on (Power spike shown below), also, the power spike is shared by two resistors. What wattage resistor do I need to handle this type of spike. The spike will occur worst case average every couple minutes but generally only every couple hours.

 

Draft2.bmp

 

I am currently testing with 10W resistors, which don't even get mildly hot with these spikes so I know these would be overkill.

Edited December 11, 2015 by CasualKilla

[Carrock]

So I am currently struggling with resistors blowing out on a PCB. They are 1/4 Watt 1206 resistors, The resistors where designed for steady state, but there is a power spike when the device is turned on (Power spike shown below), also, the power spike is shared by two resistors. What wattage resistor do I need to handle this type of spike. The spike will occur worst case average every couple minutes but generally only every couple hours.

 

Draft2.bmp

You're relying on the non repetitive surge rating of the resistor, which is a function of peak power and energy of the surge. You may also be exceeding the smaller resistor's voltage rating.

These things depend on the resistor design; you'd need the manufactuter's data.

I am currently testing with 10W resistors, which don't even get mildly hot with these spikes so I know these would be overkill.

The whole resistor doesn't have to warm up for the resistive part to burn out; a fuse burnt out by a surge typically remains cold while it may be warm if it blows through a continuous slight overload.

If you're using a 10W wire wound resistor I'd expect the thermal inertia of the wire to cope with a 3 1/2 joule 50W peak pulse.

 

Otherwise, without more information the only reliable solution is a 50W resistor.

[edit] maybe 25W if two resistors[/edit]

Edited December 11, 2015 by Carrock

[CasualKilla]

You're relying on the non repetitive surge rating of the resistor, which is a function of peak power and energy of the surge. You may also be exceeding the smaller resistor's voltage rating.

These things depend on the resistor design; you'd need the manufactuter's data.

The whole resistor doesn't have to warm up for the resistive part to burn out; a fuse burnt out by a surge typically remains cold while it may be warm if it blows through a continuous slight overload.

If you're using a 10W wire wound resistor I'd expect the thermal inertia of the wire to cope with a 3 1/2 joule 50W peak pulse.

 

Otherwise, without more information the only reliable solution is a 50W resistor.

[edit] maybe 25W if two resistors[/edit]

 

Thankyou for your response. I made a mistake in the power integration and it turns out it is only a 0.76J peak. I will have a look at the datasheets for more info in the meantime.

I just need some advice about the pulse duration, since the pulse dies down almost asymptotically slowly it doesn't make sense to take the full duration of 100ms, does one approximate the pulse as by integrating to find the total joules then take joules/Pmax to find the pulse time?

[Carrock]

Thankyou for your response. I made a mistake in the power integration and it turns out it is only a 0.76J peak. I will have a look at the datasheets for more info in the meantime.

 

I just need some advice about the pulse duration, since the pulse dies down almost asymptotically slowly it doesn't make sense to take the full duration of 100ms, does one approximate the pulse as by integrating to find the total joules then take joules/Pmax to find the pulse time?

This could be good, depending what you want to use the pulse duration length for.

 

I suspect your application has many similar solutions by other people and you could adapt your circuit in that light. eg is it really necessary to deliver 0.76J at 50Watts?

[Sensei]

Have you tried using zener diode parallel to circuit prior your current PCB?

This will make sure no high voltage reach PCB. It will pass through zener branch.

 

The spike will occur worst case average every couple minutes but generally only every couple hours.

 

That does not sound like "power spike when the device is turned on"...

Edited December 12, 2015 by Sensei

[CasualKilla]

Have you tried using zener diode parallel to circuit prior your current PCB?

This will make sure no high voltage reach PCB. It will pass through zener branch.

diode24.gif

 

 

That does not sound like "power spike when the device is turned on"...

 

So what seems to be happening is that the 110Vdc to 24Vdc converters input pin is taking time to charge up from 0V to 110V. This means that during startup the equivalent load behaves like a capacitor with negligible inductance and resistance. (see below of the circuit and simplified circuit).

 

 

This capacitive load during startup also means there is a voltage pulse of 110V across the resistors which dies down asymptotically (shown in first post).

 

The problem with the zener solution seems to be that we require an input voltage range of around 70V to 150V. So the steady-state losses across the resistor will be too large in the 150V case.

 

This could be good, depending what you want to use the pulse duration length for.

 

I suspect your application has many similar solutions by other people and you could adapt your circuit in that light. eg is it really necessary to deliver 0.76J at 50Watts?

 

I plan on using the pulse duration to select for appropriate resistors using the pulse power graphs in certain resistor datasheets.

 

 

In light of the circuit diagram I provided, would you say the design is flawed? If so, I am open to recommendations, but the problem is the PCB have already been printed and only allow for 1206 package chips, so this is a last resort.

 

EDIT: I should add that we can't make the resistors too large since that would deteriorate the steady-state efficiency too much and may lead to too much heat generation. The steady state output power is only around 1.6W, meaning input current on the 110V is around 15mA. Meaning that a resistance of 1k would lead to a 1000*0.015 = 15V steady state voltage drop and efficiency drop of 100%*((0.015^2)*1000)/1.6 = 14%.

Edited December 12, 2015 by CasualKilla

[Carrock]

A few points.

If the converter is switch mode I doubt there's significant equivalent input capacitance.

Is the dc input fairly constant? eg rectifier+capacitor? (But rectifier+capacitor drifts up to peak input voltage - 150V - if there's no load eg prior to closing DC power switch.)

If so C8 isn't needed or can be replaced by a smaller value.

It's probably taking nearly all the surge current rather than the 110Vdc to 24Vdc converter.

It may then be better to switch the power source for the 110V D.C. input if that is possible rather than the D.C. output.

If the input is rectified unfiltered A.C. C8,R25 and R26 should be moved.

 

I just need some advice about the pulse duration, since the pulse dies down almost asymptotically slowly it doesn't make sense to take the full duration of 100ms, does one approximate the pulse as by integrating to find the total joules then take joules/Pmax to find the pulse time?

I think that would be a safe approximation.

[CasualKilla]

A few points.

If the converter is switch mode I doubt there's significant equivalent input capacitance.

Is the dc input fairly constant? eg rectifier+capacitor? (But rectifier+capacitor drifts up to peak input voltage - 150V - if there's no load eg prior to closing DC power switch.)

If so C8 isn't needed or can be replaced by a smaller value.

It's probably taking nearly all the surge current rather than the 110Vdc to 24Vdc converter.

It may then be better to switch the power source for the 110V D.C. input if that is possible rather than the D.C. output.

If the input is rectified unfiltered A.C. C8,R25 and R26 should be moved.

 

I think that would be a safe approximation.

 

Thanks reponse, I have a few comments on your points.

 

1) I am not sure, but the datasheet mentions switching frequency and isolation.

2) The 110Vdc rail is produced on a public transport train and is not part of the design, however the specification says it can vary significantly from 70 to 150V. There will most likly already be mainly loads on the 110V rail when connecting the switch.

3) -

4) -

5) I am not sure what you mean by this

6) The input is 110Vdc

 

I think the protection circuitry too much. The datasheet DC to DC boasts that this device "virtually eliminates the needs for protection circuitry", but is hard to determine what is actually needed and what is not. The datasheet is very sparse (http://dtsheet.com/doc/238188/murata-ruw15).

 

Question with regard to removing components:

 

If R25 and R26 is removed, but not the capacitor, I assume this will cause the PCB traces to burn out? Spice is telling be the peak power in the traces with be in the kW range XD

Edited December 12, 2015 by CasualKilla

[Carrock]

Thanks reponse, I have a few comments on your points.

 

1) I am not sure, but the datasheet mentions switching frequency and isolation.

2) The 110Vdc rail is produced on a public transport train and is not part of the design, however the specification says it can vary significantly from 70 to 150V. There will most likly already be mainly loads on the 110V rail when connecting the switch.

3) -

4) -

5) I am not sure what you mean by this

6) The input is 110Vdc

 

I think the protection circuitry too much. The datasheet DC to DC boasts that this device "virtually eliminates the needs for protection circuitry", but is hard to determine what is actually needed and what is not. The datasheet is very sparse (http://dtsheet.com/doc/238188/murata-ruw15).

 

Question with regard to removing components:

 

If R25 and R26 is removed, but not the capacitor, I assume this will cause the PCB traces to burn out? Spice is telling be the peak power in the traces with be in the kW range XD

If R25 and R26 is removed, but not the capacitor, I assume this will cause the PCB traces to burn out?

No, but I said moved, which your latest post indicates is impossible, not removed.

 

Overvoltage can destroy resistors though I'd not expect it if the 110V rail is always between 70 and 150V (which I suspect it isn't).

C8 ensures that most of the A.C. voltage transients appear across R25/R26.

C8 isn't needed to protect the inverter but it may be needed for reasons you know about but I don't.

Based on the limited information you've provided I'd suggest:

 

Remove C8 and replace R25/R26 with either one 110 ohm 1 watt resistor ( for its higher voltage rating) or with two one watt 220 ohm resistors in parallel.

 

There are many possible reasons this won't work but I'm done with guesswork and this topic.

 

I hope this helped.

[CasualKilla]

No, but I said moved, which your latest post indicates is impossible, not removed.

 

Overvoltage can destroy resistors though I'd not expect it if the 110V rail is always between 70 and 150V (which I suspect it isn't).

C8 ensures that most of the A.C. voltage transients appear across R25/R26.

C8 isn't needed to protect the inverter but it may be needed for reasons you know about but I don't.

Based on the limited information you've provided I'd suggest:

 

Remove C8 and replace R25/R26 with either one 110 ohm 1 watt resistor ( for its higher voltage rating) or with two one watt 220 ohm resistors in parallel.

 

There are many possible reasons this won't work but I'm done with guesswork and this topic.

 

I hope this helped.

Yes you helped did alot. Sorry I could not provide better information. I try to do my self study, but am still a novice at many aspects of power electronics. I am doing a course next year.

[Externet]

What I would do is removing the failing surface mount resistor and replace it with a stack of other surface mount value ones.

 

Say it is a 10 Ohm, replace with four 40 Ohm soldered one on top of the other. That multiplies the power handling capability by 4 and there is no need to redesign pads and traces on the board, and barely makes it bulkier.

[MountainGuardian]

I am not an electronic engineer, but I am a great technician. Without knowing more about the application or being able to actually physiscally test the circuit it is a bit hard for me to wrap my head around the problem. You are describing damage to your transistor on start up due to voltage spike, then later mention that this happens so many times per x minutes of operation?

.

An initial spike upon powering up would imply that you may not have adequate power supply, if your voltage drops enough initially it can cause damage the same as too much voltage can. Whether it is too much voltage or too little voltage when the circuit is powered, this can be remedied by bigger capacitors, the more capacitance the better it will smooth out the signal when you power the circuit.

.

As you later describe, the circuit also has spikes while in operation, if this is true then it is not just on powering the circuit. It is apparently imtermittent from what you describe, I would look for sources of static within the circuit, irregularities within the voltage supply and I would look for any kind of light or thermal protection within the circuit. I had a brand new electronic dart board that worked perfectly until you actually played, it would simply lock up at what seemed completely random times.We went through power supply, line voltage, static discharge, everything we could think of and could find nothing wrong. In the end it turned out someone at the factory had put the wrong light bulbs in it and it was drawing too much power when it would go through the light flashing mode which was just enough to drop voltage and lock up the processor. A thermally protected component could also slowly heat up and potentially cause an intermittent problem.

.

I would also look for the potential of a short within the circuit or the potential of a bad trace or solder joint, both of these can potentially cause this effect. I would also take a close look and measure of components within the circuit, because it is supposed to tolerate x amount of voltage or heat does not always mean that it does. Think about the flasher unit in your car, it flashes due to the heating of a small metal strip which makes and breaks the circuit, the same thing can happen within your circuit if a part is not up too par, it may say it is and may measure within tolerance, but that does not always mean that it is working properly under power.

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For me, I generally work on equiptment that has seen a lot of use, I often go and replace all high voltage caps right from the start, if they aren't the cause yet eventually they will be and you save yourself a few service calls for the cost of a few dollars in caps.

Edited January 16, 2016 by MountainGuardian