Hi

I have a commercial power supply which was pre-jumpered to 120 volts when I received it. I plugged it into 240 volts without checking. The fusible resistor blew … as it should have. Very pleased.

I searched the iNet and found out that to change from 120v to 240v, I just needed to change an internal jumper. Too late!

The fusible resistor is 6.8 ohms 1.6 watts, ceramic, wire-wound (I think), radial.

So first question. Can anyone direct me to anywhere in the world where I can source a replacement, please

Second question. If I can’t find a direct replacement, what should be the approach to choosing something close?

it strikes me that the fusible resistor is only there to overheat and melt (fuse) when the voltage exceeds the jumpered value. By and large, voltage fluctuations not withstanding, that will only happen when it is set for 120v and you plug it into 240v. So I’m guessing if I’ve corrected the jumper to 240 volts, I cld actually replace the fusible resistor with a dead short at worst, and an ordinary (non-fusible) resistor of 6.8 ohms at best.

I can get a fusible resistor at 6.8 ohms and 1 watt or at 2 watts. 2 watts would risk damage to the power supply by tolerating a higher current before blowing while 1 watt might blow too often/easily.

Would appreciate any advice, please.

Regards

 

I think it will be rated at more than 1.6 watts. Here is a link to a similar resistor but with 10 watt rating. there is a link to the manufacturers datasheet on this range of resistors with the dimensions so you can use this to see which ones match the physical size of your resistor. It would be helpful if you gave some idea of your location in your profile when asking about suppliers.

Les.

 

It may also figure highly in the inrush current the device will experience. This fusible resistor has already proven that it can be very useful, and besides that it might save your hour from burning down someday, so I would advise you to find a replacement fusible resistor. I had a similar problem with a tiny fuse that went into one of my floor fans. I took the dead part to a place where appliances are repaired and put the dead part in the technician's hand and a minute later I was paying the equivalent of U.S. 60¢ and then on my may to install the part.

Get the correct fusible resistor and then you won't have to worry about it again.

 

Why did it blow? I suspect something is shorted (input diodes? filter capacitors?) and without a fix
I'd think it likely to blow the new resistor (or something, fuse?).

story: New computer device installed in machine room. Hardwired power connection
even though inside it's like a PC. All install tests run successfully. Then it blew up.
It was set for 120 VAC and powered with 208 VAC (normal 3 phase line to line voltage).
After some repairs (swap in new power supply?) and matching setting to input voltage worked fine.

 

Ok. Good point, buy a spare!

 

Thank you all for your responses to my problem.

The power supply is an Icom PS-125 (used). It shipped with the internal jumper set to 120 VAC. In New Zealand the mains voltage is 240 VAC. As a consequence the fusible resistor blew pretty much instantly, protecting the power supply.

As I said earlier, I have searched the world and I can't source a replacement.

Here is some documentation on changing the jumper from 120 VAC to 240 VAC -

https://www.ab4oj.com/icom/ps125/ps125v.html

Here is the circuit diagram of the PS-125 -

https://www.ab4oj.com/dl/ps125v.pdf

You will see the fusible link in the top left corner of the circuit. It is called R2 and its value on the circuit is 6.8 ohms 1.6 watts. It is ceramic, wire-wound, radial/vertical.

I'm guessing that the 6.8 ohms is not critical but the low wattage is critical.

I sourced part number A142652CT-ND from Digi-Key: 6.8 ohms 5% 1/2 watt axial. I deliberately went low on the wattage. The outcome was fantastic: within milliseconds this fusible resistor blew up, great flash of flame and sparks. The power supply closed down. The temperature was so hot that a nearby wire link on the PCB melted.

I have located a 2 watt ceramic 6.8 ohm fusible resister but it can't ship to New Zealand. Why is that?

I have sourced a 33 ohm 5% 1 watt axial but haven't tried it.

I have also sourced a ceramic, radial/vertical, 6.8 ohms 5 watts, identical looking component to the original but different wattage.

If my math is correct i = sqrt(p/r) and 1.6 watts will tolerate 485 mA while 5 watts will tolerate 857 mA before it blows. The 1/2 watt resister would only cope with 271 mA.

Logic tells me that since the original component at 1.6 watts tolerated a max of 485 mA, then under normal stable utility supply at 240 VAC, and the correct jumper setting, the in-rush current will be approx 1/2 Amp. So assuming there are no surges in the mains voltage, the 5 watt resistor that I have sourced should allow the 1/2 Amp working current thro without any problems, ie it will act like a gate, but protection wont be there.

If there is a line voltage surge, the current will peak at 857 mA before the resistor blows, by which time severe damage could have been done to the power supply, I'm guessing. In simple terms, the power supply will work (presupposing no other damage has been sustained by the power supply to date), but it will have no voltage surge protection

I wonder if you can give me advice as to which one I should try. Have to say I find it pretty amazing that I cannot source an exact replacement part anywhere in the world.

Thank you for your assistance

Kind regards

 

That 120/240 voltage circuit is the standard voltage double for 120, full wave for 240 used in many power supplies.

The 6.8 ohm resistor (& related triac) form a slow start. They limit the current at start up. After that the triac is supposed
to be turned on bypassing the 6.8 ohm resistor.

When the supply was originally powered from 240 volts in 120 volt voltage doubler mode the result was
240*1.4*2 -> 672 volts. The main filter capacitors are (from the schematic/parts list) are only rated at 200 volts
so even in series the maximum voltage should be limited to 400. In my mind it's very unlikely that they would
survive that and are currently shorted. When you power it now (even in 240 volt mode) the short circuit current
is killing your replaced R2 6.2 ohm resistor.

 

More on slow start. At power up the main capacitors C9/C10 are at 0 volts and look like a short.

With a 240 volt feed correctly connected (peak voltage 240*1.4 -> 336 volts) and only the 6.8 ohm resistor limiting the current,
the maximum initial current is about 50 A into the capacitors. This depends on when/where in the input sine wave the power is connected (336 is the input peak). As the capacitors charge the input current quickly drops. The capacitors likely charge in less than 1/2 cycle of the power line.

So while the peak power in R2 is very high, it's for a very short duration (likely less than .01 seconds, possibly less than .005
seconds).

With really shorted main capacitors the current isn't limited to a short duration and the resistor is going to die.

I can get a fusible resistor at 6.8 ohms and 1 watt or at 2 watts.

A 2 watt resistor should be fine. I'd guess the resistor choice was to avoid a resistor which would explode
(and or burn the place down) if the current continued rather than to protect other components in the supply
from excessive electrical power. There is the F1 fuse for real (slower) overcurrent protection.

 

I agree with micael8 that more components than the 6R8 resistor are likely to have been destroyed. Did you not find a similar physical size resistor in the data sheet from the link I provided in post #2 ? The physical size should give a good idea of the wattage rating. (I would guess that the 10 on the original resistor was the wattage rating.)
When you next try to power it up again I suggest limiting the current with a 100 watt old fashioned light bulb in series with the mains supply. (If you can still obtain one.) I found some similar looking 5 watt resistors on Ebay (UK) so you might find some on the part of Ebay for your part of the world.

Les.

 

With really shorted main capacitors the current isn't limited to a short duration and the resistor is going to die.

I think you know that you can test the capacitors to see whether they really are shorted by using a ohmmeter.

 

Thank you for your responses. It reminded me of Electrical Engineering 101 Circuits. Switching power supplied weren't around in those days.

I have put a digital ohmmeter across each of the two capacitors. It behaves exactly as you would expect - the digital reading keeps on increases steadily. I didn't take the capacitors off the PCB.

Regarding the wattage of the original component, the circuit says its 1.6 watts. However, if you think it is 10 watts, then I do have a 5 watt 6.8 ohm ceramic radial resistor.

I can get 2 watts, 6.8 ohms ceramic from China.

what do you think? Thanks Regards.

 

Sorry, meant to say regarding the link that was posted. I can get 2 watts from there.

 

Here is the service manual. R2 is specified here.

 

I feel I'm bombarding you. Sorry.

However, the service manual describes R2 as Panasonic ERWU5TAK6R8. This is pretty easy to find on the net. The pdf attached deciphers the part number. So, the part number is described as 5 watt while the circuit specifies 1.6 watt and the actual installed component appears to be 5 watt. The thermal cutoff for the installed component is not available in the data sheet, however, the service manual says it is 2 Amp 133 degrees C.

I have a ceramic 6.8 ohm 5 watts, so I think I'll give that a go. Thanks. Regards,