A flyback diode across the resistor? I used to do switching for stepper motors. Fried a bunch of mosfets though the supply voltage was well within the absolute maximum ratings specified.

 

You could also use two resistors in series of about 1/2 the total resistance each.
That would double both your power and voltage limits.

 

A flyback diode across the resistor? I used to do switching for stepper motors. Fried a bunch of mosfets though the supply voltage was well within the absolute maximum ratings specified.

How can it help?

 

A flyback diode across the resistor? I used to do switching for stepper motors. Fried a bunch of mosfets though the supply voltage was well within the absolute maximum ratings specified.

I don't see how that can help the op's problem.
There is no flyback voltage or current.

 

How can it help?

I was thinking, potential energy is acquired slowly during close of the first switch and is dumped rapidly right afterwards along the whole length of the wire. The potential at the other end decreases just as rapid hence the diode will provide the other path for the dump. Since R2 is dying, the capacitor does not help dissipating or has high impedence. Hence the diode must have lower impedence than R2.

 

I don't see how that can help the op's problem.
There is no flyback voltage or current.

I'm sorry, then those resistors are not like stepper motors.

 

Where is the input voltage coming from?

Is this a naughty , naughty?

 

Below is my LTspice simulation of the circuit.
It shows an energy dissipation of 101.6J for R1 and 132.5J for R2.
An explanation for the difference is left to the reader.
Hint: It's not the dissipation in the diodes, which is a little over 1J total for the 4 diodes.

View attachment 108472View attachment 108473

I work by Orcad Pspice and I could not find the average power and Energy. would you find it from 0.5sec to 1.5sec for R1 and 4sec to 4.5sec for R2.

 

I work by Orcad Pspice and I could not find the average power and Energy. would you find it from 0.5sec to 1.5sec for R1 and 4sec to 4.5sec for R2.

You don't need further simulations. It's a simple calculation based upon the energy.
Do you not know the relation between energy and power (1 J/s = 1 W)?
Based upon my simulation the average power for R1 is thus 101.6J / 1s = 101.6W and 132.5J / 0.5s = 265W for R2.

 

"I work by Orcad Pspice and I could not find the average power and Energy. would you find it from 0.5sec to 1.5sec for R1 and 4sec to 4.5sec for R2."

He gave you the total energy; and it is dissipated within the bounds you set so you can calculate the average power in those intervals yourself.

 

I expect that two resistor (R1 and R2) dissipate the same power but why R1 can withstand and R2 fails?

The capacitor is charged up by rectified-sine-wave pulses, so energy is not flowing through R1 continuously. 100 times each second the current drops to zero. The capacitor is discharged directly into R2 in one continuous blast than discharges the cap in much less time. According to the simulation in post #8, it takes about seven times longer for the cap to charge than to discharge. That spreads out the heat over time, letting R1 run cooler.

Also, because the input voltage is sinusoidal, the initial current surge after turn-on is not instantaneous. It increases as the voltage increases up the first half-sine. The discharge current spike is near-infinite, limited only by the ESR of the capacitor. R2 basically blows like a fuse.

ak

 

The data sheet/page you linked to had a max 3W resistor, and searching for "SWA06" on the site brings up nothing. Maybe it's "WA06"? They have that.

 

The capacitor is charged up by rectified-sine-wave pulses, so energy is not flowing through R1 continuously. 100 times each second the current drops to zero. The capacitor is discharged directly into R2 in one continuous blast than discharges the cap in much less time. According to the simulation in post #8, it takes about seven times longer for the cap to charge than to discharge. That spreads out the heat over time, letting R1 run cooler.

Also, because the input voltage is sinusoidal, the initial current surge after turn-on is not instantaneous. It increases as the voltage increases up the first half-sine. The discharge current spike is near-infinite, limited only by the ESR of the capacitor. R2 basically blows like a fuse.

But the interesting thing is that the total energy dissipated in R1 is also less than R2, neither of which is a function of time.

The discharge current spike is very large but not near infinite, it's obviously limited by the resistor resistance.

 

But the interesting thing is that the total energy dissipated in R1 is also less than R2, neither of which is a function of time.

The discharge current spike is very large but not near infinite, it's obviously limited by the resistor resistance.

LTspice sure is handy. I never had this problem before. But first on paper and know your components.

 

the average power for R1 is thus 101.6J / 1s = 101.6W and 132.5J / 0.5s = 265W for R2.

Shhh! Don't tell the over-unity guys. Just 101.6J to charge up a cap and you get 132.5J out !!

 

I don't know what the purpose of this circuit is beyond wasting energy, but using a smaller capacitor would decrease the necessity of a higher power rating for R2. Half the capacitance would require half the R2 power rating, but due surge power wouldn't be reduced as much.

Or better, use a stepdown transformer (there ought to be an isolation transformer in there anyway for safety reasons) and a larger capacitor to maintain the energy to R2 at a lower voltage and lower surge power. For example a 269V peak output from the secondary into a 3.8mF capacitor would still deliver 137J to R2, but would reduce the surge power by half due to doubling the RC time constant. Step down to 220V and you might be able to use the 6W 47 ohm for R2.

 

Looking at the circuit, it appears to be a transformerless power supply.

 

I don't know what the purpose of this circuit is beyond wasting energy, but using a smaller capacitor would decrease the necessity of a higher power rating for R2. Half the capacitance would require half the R2 power rating, but due surge power wouldn't be reduced as much.

Or better, use a stepdown transformer (there ought to be an isolation transformer in there anyway for safety reasons) and a larger capacitor to maintain the energy to R2 at a lower voltage and lower surge power. For example a 269V peak output from the secondary into a 3.8mF capacitor would still deliver 137J to R2, but would reduce the surge power by half due to doubling the RC time constant. Step down to 220V and you might be able to use the 6W 47 ohm for R2.

Hi guys and big thanks to all of you for your halpfull replies although my English is really awful ,
This circuit is not a part of my real circuit and is a simulator. Only R1 is in real circuit and it works as Inrush Current Limiter in the input stage of SMPS. after a while the capacitors charge and the circuit is in a stable situation a relay bypasses this resistor.
Actually I have only R1 in my real circuit and I want to do a strict life test for a wirewound 47ohm/6w resistor. I think in worst situation (when input switch exactly switches in peak of input ac voltage) the stress of R1 may be equal to R2. I put my resistor in R2 position to see if it can withstand in worst input situation.
Please tell me (ensure me ) If you think that even in worst situation of input (when input switch exactly switches in peak of input ac voltage) my resistor R1 can withstand and it's not damaged. Suppose that my resistor is a really high quality 47ohm/6W wirewound resistor and I want switch the input every 15sec while the capacitors are completely discharged.

 

It is a charge-transfer circuit: the capacitor gets charged up by the dc (=rectified ac) and then transfer the charge to the resistor on the right.

Because a rc circuit is always 50% efficient, ideally the energy dissipated on the R2 is the same as the energy dissipated on R1.

It is very difficult to envision a situation where R1 is burned out when R2 isn't, assuming that the circuit is correct.

 

The instantaneous power dissipation on those resistors can be quite significant. When the capacitor is being charged up from a fully-depleted state, the current going through the resistor is about 380v/47ohm=8amp. The instantaneous power dissipation over the resistor is 8a*380v=3Kw.

Obviously that doesn't last very long, as the capacitor is being charged up quickly. But it gives you some sense as to what "average" power dissipation over a cycle should be.