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Wannabe Geek
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  Reply # 2028049 3-Jun-2018 23:35
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I'll bet that those lamps are just standard 2W 12V grain-of-wheat bulbs.
One can work it out based on the wiring diagram given which shows the 40W ballast resistors in series with 2x 2W lamps.
If you take the mains voltage of (say) 230V and multiply it by 2/44th (2+2+40W), you get a bit over 10V per bulb.
Going further, I bet that ballast resistor is likely to be either 1100 ohm or more likely 1200 ohms being a 'Prefered value'.

 

(40/44)*230V = 209V = voltage drop across ballast.

 

R = (V^2) / P
so (209*209)/40
43681 / 40
= 1092 ohms

 

The numbers work out slightly better if we assume 240V instead on 230V (closer to 1200 ohms).
It'll be interesting to see what a multimeter thinks that resistor is.

 

Anyway, as for where to get a bulb, try these Jaycar ones - SL2675 or SL2685
(Being new I can't post a link to it, so you'll have to use Jaycar's less than stellar web search function for those 2 part numbers)

 

They state they are 12V 0.1A so you'd need to wire 2 of them in parallel with each other so they don't get too much current through them and burn out.

 

The old lamps would be about 0.2A

 

I = P/V
230V / 44W = 0.191A = 191mA

 

Either that, or hunt around for some 2W, 12V grain of wheat lamps from somewhere.


645 posts

Ultimate Geek
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  Reply # 2028052 3-Jun-2018 23:50
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That's only true if the ballast is a non-active component (which could be the case looking at it, I agree) . It's also only true if the ballast is running at that full rating given in the schematic - you wouldn't normally design a circuit to run components at their max rating. Was that value given in the schematic a maximum rating or an operating value?

 

OTOH 12v light bulbs are cheap and easy to get so OP doesn't have much to lose.

 

OP could even use some car bulbs for experimentation.


3006 posts

Uber Geek
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  Reply # 2028057 4-Jun-2018 01:50
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IanIan:

I'll bet that those lamps are just standard 2W 12V grain-of-wheat bulbs.
One can work it out based on the wiring diagram given which shows the 40W ballast resistors in series with 2x 2W lamps.
If you take the mains voltage of (say) 230V and multiply it by 2/44th (2+2+40W), you get a bit over 10V per bulb.
Going further, I bet that ballast resistor is likely to be either 1100 ohm or more likely 1200 ohms being a 'Prefered value'.


(40/44)*230V = 209V = voltage drop across ballast.


R = (V^2) / P
so (209*209)/40
43681 / 40
= 1092 ohms


The numbers work out slightly better if we assume 240V instead on 230V (closer to 1200 ohms).
It'll be interesting to see what a multimeter thinks that resistor is.


Anyway, as for where to get a bulb, try these Jaycar ones - SL2675 or SL2685
(Being new I can't post a link to it, so you'll have to use Jaycar's less than stellar web search function for those 2 part numbers)


They state they are 12V 0.1A so you'd need to wire 2 of them in parallel with each other so they don't get too much current through them and burn out.


The old lamps would be about 0.2A


I = P/V
230V / 44W = 0.191A = 191mA


Either that, or hunt around for some 2W, 12V grain of wheat lamps from somewhere.



I agree. Although I used a different calculation method.

I calculated the current would be drawn by a 240V 44W load. Then the voltage drop required to get 40W at the same current to get the resistor voltage. The remaining voltage I got was 21.4V or 10.7V per bulb.

So definitely 12V 2W bulbs.

That's an awesome first post @lanlan





7192 posts

Uber Geek
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  Reply # 2028086 4-Jun-2018 08:31
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Those "ballast" resistors will have the resistance value and wattage printed on them.  No doubt it will be printed on the hidden side, but the tabs holding them to the panel will bend back.  That would also confirm that they are what they look exactly like - just a standard ceramic encapsulated wire wound resistor.

 

If they were 40 watt 1,200 ohm, it'd usually be printed "40W 1K2" (K for thousand).

 

The bulbs are not the mini neon indicator bulbs as shown in a photo above.  Those are often supplied with a tiny resistor pre-soldered on one of the wires, if you connected them directly to 230V, they'll go pop.




42 posts

Geek
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  Reply # 2028098 4-Jun-2018 09:23
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IanIan:

 

They state they are 12V 0.1A so you'd need to wire 2 of them in parallel with each other

 

 

That sounds convincing. There are actually 2 lamps within each of the two 2W light units. I forgot to check when I had the unit out yesterday whether wired in parallel or series. So 12V 0.1A might be right on the money.




42 posts

Geek
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  Reply # 2028185 4-Jun-2018 11:10
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IanIan:

 

I'll bet that those lamps are just standard 2W 12V grain-of-wheat bulbs.
One can work it out based on the wiring diagram given which shows the 40W ballast resistors in series with 2x 2W lamps.
If you take the mains voltage of (say) 230V and multiply it by 2/44th (2+2+40W), you get a bit over 10V per bulb.
Going further, I bet that ballast resistor is likely to be either 1100 ohm or more likely 1200 ohms being a 'Prefered value'.

 

(40/44)*230V = 209V = voltage drop across ballast.

 

R = (V^2) / P
so (209*209)/40
43681 / 40
= 1092 ohms

 

The numbers work out slightly better if we assume 240V instead on 230V (closer to 1200 ohms).
It'll be interesting to see what a multimeter thinks that resistor is.

 

Anyway, as for where to get a bulb, try these Jaycar ones - SL2675 or SL2685
(Being new I can't post a link to it, so you'll have to use Jaycar's less than stellar web search function for those 2 part numbers)

 

They state they are 12V 0.1A so you'd need to wire 2 of them in parallel with each other so they don't get too much current through them and burn out.

 

The old lamps would be about 0.2A

 

I = P/V
230V / 44W = 0.191A = 191mA

 

Either that, or hunt around for some 2W, 12V grain of wheat lamps from somewhere.

 

 

The "ballast: is marked as "40W 1K 2J". Measured resistance at 1210 ohms.

 

Measured voltage over both light units at 23.1V and over one light unit at 12.04V.

 

Have visited Jaycar this morning and bought a pair of those 12V lamps.

 

Thanks loads and you know your stuff!




42 posts

Geek
+1 received by user: 15


  Reply # 2028214 4-Jun-2018 13:01
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Thanks all - especially IanIan.

 

Replaced with Jaycar SL2685 12V miniature lamps as suggested (2 x $1.90).

 

Now all oven lights working...

 

Click to see full size

 

... so a successful outcome :-)

 

For backup I have ordered these bulbs for future spares. These have pigtails instead of PVC leads, which will make mounting easier.

 

https://www.ebay.com/itm/Grain-of-Wheat-Bulbs-20-pcs-3mm-Clear-12V-100mA-Mini-Bulbs-Lighting-Kits/400110358903?epid=1131301164&hash=item5d286f9177:g:ekIAAOxyGb1RFpJA&vxp=mtr

 

$NZ12 for 20 shipped.


1905 posts

Uber Geek
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  Reply # 2029371 4-Jun-2018 19:09
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Super effort by all. Well done.

 

Geekzone is such an amazing place!


1738 posts

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  Reply # 2030907 6-Jun-2018 14:33
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Pleased you got it sorted, though it's an enormously inefficient way to light a couple of light bulbs.

 

If you have both sets cooking - upper and lower, that's 80W of power being dumped to light 4 lamps.

 

Though I suppose of you're drawing kilowatts whilst cooking, 0.08KW aint going to matter much.





My thoughts are no longer my own and is probably representative of our media-controlled government


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