18
\$\begingroup\$

Inexperienced electrical person here, but trying to save significant cost to repair an oven. The oven I have has a MR754 diode off the main convection heating element, that has failed. Unfortunately, since it is part of the wiring, the manufacturer only offers up the entire wiring harness as a replacement part at $500+.

Pending all other things checking out with a multimeter, I would like to simply splice in a replacement diode (~$2 part). However I cannot locate an MR754 in my area and it's multiple business days to get one delivered via online order. Local vendors do have an MR756 available. Both are 6 amp current, but the available replacement would be 600 V vs 400 V. Is this safe to use in this application? What is the risk, should I try this? Do I put the other components in the oven (i.e. element or control board) at risk?

inside of oven

schematic diagram

\$\endgroup\$
3
  • 14
    \$\begingroup\$ I see no reason why not. 600 V is only the reverse block rating. \$\endgroup\$
    – skvery
    Commented Apr 3, 2020 at 6:18
  • 8
    \$\begingroup\$ I think it would even be safer. More tolerance if anything goes wrong. \$\endgroup\$
    – Aloha
    Commented Apr 3, 2020 at 10:55
  • 2
    \$\begingroup\$ I there is a VTC-repair against this question, but I don't think it's applicable. This is a very specific question about one type of component, that can be easily answered. A "repair question" is for "my widget broke, here's a blurry photo of half the circuit board, where should I put my $4 multimeter?" \$\endgroup\$
    – pipe
    Commented Apr 4, 2020 at 10:34

4 Answers 4

Reset to default
35
\$\begingroup\$

Yes, higher voltage rating is fine. It just means more safety margin against mains transients (which is good). The only difference between the -4 and the -6 is the voltage rating (see datasheet).

\$\endgroup\$
4
  • 1
    \$\begingroup\$ The diode in itself is certainly fine, as you say very adequatly - but the question also asks if there are any other risk factors in changing it. You seem to have a good grasp (judging by tenure and rep) - are there none worth mentioning? \$\endgroup\$
    – Stian
    Commented Apr 3, 2020 at 11:01
  • \$\begingroup\$ A larger diode could have large leakage currents but that usually isn't a concern on mains stuff. \$\endgroup\$
    – ACD
    Commented Apr 3, 2020 at 12:48
  • 4
    \$\begingroup\$ @StianYttervik there may be risks for OP that would exist working on any mains powered appliance. You could write an answer highlighting those and I will upvote it if it is useful. I don’t see any special risks here in replacing one part with another OEM equivalent. \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented Apr 3, 2020 at 12:51
  • 3
    \$\begingroup\$ @ACD higher rated voltage is usually the exact same die. Bigger diodes are those rated for more current, not voltage. If anything a higher rated diode of the same type would likely have less leakage. \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented Apr 3, 2020 at 12:52
24
\$\begingroup\$

Generally as the comments and answers already state it's ok to use a diode of higher voltage rating when the amp rating is sufficient. It would even be ok to have a slightly higher amp rating normally.

In your case however i see no fuse in the wiring diagram and it looks like the diode could have blown due to a defective heating element.

A short circuit within the heating coil will cause the current to go up and this will as a result destroy the diode when it's going over 4 amps. Since it is more likely for a heater coil to wear off and fail than the diode you should check the heating element first. You might see your replacement diode fail very soon otherwise.

In Summary: A higher voltage diode than specified is not critical but a higher amp rating may pose a risk if the diode is working as a fuse in case of a coil short circuit.

\$\endgroup\$
4
  • 1
    \$\begingroup\$ Yes, thank you. I did check resistance across the element itself and it holds steady in the mid-40's. Seems that it checks out? \$\endgroup\$
    – LTB
    Commented Apr 3, 2020 at 15:30
  • 1
    \$\begingroup\$ @LTB I don't see right now what voltage the oven runs at, but 40Ohm is probably too little - on the other hand you are measuring cold and I assume the heating element's resistance grows with temperature (but you could still have peak currents higher than 6A, depending on voltage). \$\endgroup\$
    – Nobody
    Commented Apr 3, 2020 at 17:04
  • 1
    \$\begingroup\$ @LTB: 40 Ohm sounds reasonable for a 120V country; I measured the stovetop elements in my stove (Canada: 120V) out of curiosity when replacing a damaged one (ceramic coating chipped), and IIRC the smaller ones were like 50 ohms. 120^2 / 40 is 360W for 120Vrms across it, which sounds totally reasonable for an element. And 120V / 40 ohm is 3 amps RMS, so peak still comfortably less than the 6A rating. \$\endgroup\$
    – Peter Cordes
    Commented Apr 3, 2020 at 18:22
  • 2
    \$\begingroup\$ @Nobody: the OP is probably in a 120V country; 3A RMS would nicely explain a 6A diode rating. See my previous comment. I'd guess North America judging by the 3 languages in the manual: English, Spanish (I think), French. \$\endgroup\$
    – Peter Cordes
    Commented Apr 3, 2020 at 18:26
7
\$\begingroup\$

replying instead of answering due to lacking reputation

The question has been answered, but Stian Yttervik asked if there are any risks at all. In almost all cases, no. Most diodes will be at around 0.7 to 1 volt of forward voltage, (almost) regardless of current and breakdown voltage.

However, higher voltage ratings often come with a higher forward voltage drop and (at equal current into the load) higher power losses. If the circuit used a very-low-drop Schottky diode with a voltage drop of maybe 0.5 volts and you replace with a high-voltage diode with a voltage drop of 1.5 volts, then the replacement diode will become much hotter than the original diode since it has to dissipate (three times the voltage) * (same current). This might become a problem inside a tight enclosure.

In addition, some small-signal circuits might rely that a diode has a certain voltage drop. In that case you should also try to match the forward voltage, or at least the diode family (replace Schottky with Schottky, silicon with silicon, high voltage with high voltage).

And finally, please don't replace a Zener diode with another Zener diode with higher breakdown voltage - since for Zener diodes the breakdown voltage is the most important parameter!

\$\endgroup\$
3
  • 2
    \$\begingroup\$ But which of those concerns actually apply to the oven? The temperature one maybe I guess, but I doubt it. \$\endgroup\$
    – Nobody
    Commented Apr 3, 2020 at 17:06
  • 2
    \$\begingroup\$ This answer is a complete non-sequitur and adds nothing. \$\endgroup\$
    – TomServo
    Commented Apr 3, 2020 at 22:06
  • 1
    \$\begingroup\$ No. Diodes with reverse voltage rating in range 400V-1000V and lowest (suitable for mains rectification) speed rating do have very similar voltage drop. Replacing 400V part with 600V or 100V will keep it's thermal dissipation at the very same (or +-5% due to manufacturer tolerance) level \$\endgroup\$
    – Volodymyr Kalinyak
    Commented Apr 4, 2020 at 6:30
-1
\$\begingroup\$

For safety it is perfectly fine. But the 600V rated diode has different is made of different chemical elements than the 400V rated diode so you may get a small variation in current.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.