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Basic electronics question

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NutmegCT

NutmegCT

Great Pumpkin
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I have an electric heater (resistance), 120vac, 1500 watts/750 watts.

The 750 watt side died, when the resistor (?) in that circuit blew out. The 1500 watt side still runs fine.

What resistor do I put in, to re-enable the 750 watt part of the circuit?

The one that burned out was only about 1/2 inch long. That seems pretty "small" to handle the current involved.

Thanks.
Tom
a/k/a "Easily Befuddled"
 
That may have been a diode instead of a resistor.
 
Thanks Bay. Any idea of what diode I should buy to replace it?

I just want to get the 750 watt side of the heater working again.

Thanks.
Tom
 
Here's a picture of what burned out. Looks like two items in parallel; one has "evaporated". The unit was in a power feed line to the heating element.
 
What Brand and what model heater is this? Perhaps an electrical diagram is online somewhere? That will tell you what should be there.

Mark
 
The actual heating coils *are* the resistors.

That sort of looks like a fusible link (a one-shot fuse).

If it's blown, it may mean there's other trouble with the unit.

Maybe just turn on the 1500 watt side of the unit for half the time! :jester:
 
It's a Holmes HQH307, two quartz crystal tubes. Can't find a wiring diagram to save my life. I called Customer Service and was told they "no longer sell that model". And "there are no parts available". And "I don't know which part it is that exploded sir."

But ...

"The heating coils *are* the resistors"

As I never use the High setting ... maybe I'll just disconnect the feed wire to one of the two quartz columns. 1500/2 = 750, if memory serves.

Yeah!

Tom
 
NutmegCT said:
... maybe I'll just disconnect the feed wire to one of the two quartz columns. 1500/2 = 750, if memory serves.

Yeah!

Tom
Click to expand...

or 0 if they are wired in series...

m
 
<span style="font-style: italic">0 if wired in series ..</span>

Hey - 100% energy savings! Another green tip!

(actually they were in parallel, so I dis-wired (?) half the circuit. Now one quartz is cranking away full blast, but it's only using about 700w.)

Progress!

T.
 
Th' gizmo in th' pic ~looks~ like a diode. Pea brane can't pull th' number out, "144?" popped in first.
 
No, not a diode. if it was, current would only flow for 1/2 cycle and the output in watts would be approx. 50 percent. They are probably fusible links that open if the quartz elements short out. This is a two element heater. When both elements are healthy and for max. heat they are connected in parallel at 120-volts for 1500 watt output.

Power = V squared/R = 1500 = 120 x 120/R

Solve for R = 1.04 ohms.

Solve for current = 1500/120 = 12.5 amps.

With one element out (open circuit)

Power = 750 watts or 1/2 of rated

You do not need to , nor should you replace the defective element with any other sort of resistor. This would be dangerous!!!

If you want more heat, buy a replacement element.

Good luck and please be more careful!
 
Those are most certainly diodes. the silver stripe around the top of the unexploded one is to indicate the + side (if I recall correctly. When the diodes are in the circuit, it cuts out 1/2 of the current flow (one direction of of the AC) and cuts the output from 1500W to 750W. Now at P=V*I so 750=120*I or the current is 6.25A The company was cheapish and used two diodes so that each diode would only handle 1/2 of the current. Obviously they didn;t quite get the calculation right!

This should do the job, considering how cheap they are I would tie 3 of them together. Make sure they are all in the same direction.
https://www.radioshack.com/product/index.jsp?productId=2062579
 
angelfj said:
No, not a diode. if it was, current would only flow for 1/2 cycle and the output in watts would be approx. 50 percent.
Click to expand...

Which is exactly what is happening in this circuit. Full power =1500W To get low power, the diodes are switched into the circuit and the power goes down to 750W since current is only flowing 1/2 the time.
 
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