Continue to Site

# Want to reduce power of portable heater

Status
Not open for further replies.

#### LC123

##### Newbie level 3
I have a portable space heater which is this model but in a cream color:
https://www.amazon.com/Comfort-CZ792BK-Milkhouse-Electric-Utility/dp/B002Q8HDJK/

When I bought it, I didn't notice that the two settings for wattage are LOW: 1300 watts and HIGH: 1500 watts.

This is not that much of a difference between the low and the high setting. I actually measured the wattage and it is really 1200 watts and 1400 watts. But even at 1200 watts, it is too much because I worry about the old wiring in the house and would rather run the heater at a lower temperature and less wattage.

I want to reduce the wattage this heater uses to approximately 750-800 watts. I don't care about the high level.

So I opened the heater up and it seems fairly simple inside. The main power line has the ground to the case. A red wire goes first to the tip over switch and then to the main switch with off/fan/1300w/1500. That switch goes to one side of the heating coils.

The other AC power blue wire goes to the temperature knob switch and that connects to the other side of the coils. The fan and an LED is also spliced into the circuit.

Can I simply solder a resistor between the temperature switch and the heating element connector side?

I've tried a few calculations:

Currently:
1500 watts / 120 volts = 12.5 amps
120 volts / 12.5 amps = 9.6 ohms (The amount of resistance of the heating elements)

Desired:
750 watts / 120 volts = 6.25 amps
120 volts / 6.25 amps = 19.2 ohms

So I need an additional resistance to stop the heating element from receiving as much power:
19.2 ohms - 9.6 ohms (from heater element) = 9.6 ohms (needed resistance to reduce the power)

I realize this is all approximate. But if I get a 10 ohm resistor, will this reduce the overall current draw of the heater to about 750 watts?

The next question is, exactly what kind of resistor will work? I'm used to tiny electronics resistors for DC type circuits. I assume that a resistor attached to the AC mains is going to be very different.

If anyone can show me an example of the kind of resistor that I can buy from Radio Shack or an electronics store, that would really help.

There is a lot of spare room in the metal case of the heater, so I can get a large-ish resistor if needed. I realize the resistor will get warm but will it be okay to put shrink wrap around it to seal it from the metal case. I assume the resistor will not get as warm as the heating element or it would melt everything. Is the resistor wasting 750 watts in this example? The whole point is to reduce the total wattage of the unit.

I have enough soldering experience so that's not a problem. I just want this to be done safely.

Thanks.

Add a diode. Halves the power and is very cheap.

Keith

Add a diode. Halves the power and is very cheap.

Keith

Okay, this sounds like a great idea! I never thought of that. So you are saying with AC power, since the current only goes in one direction, it will cut the wattage in half.

I just need a bit more help. I tried to describe the simple circuit.

1. Does it sound like the place I want to put it is correct?

2. Will the Diode heat up? Can i just put shrink wrap around the whole diode/soldered connections?

3. What kind of Diode do I need for this? Again, I'm use to tiny components for 5v DC type applications. Does the diode need to handle 1500 watts or only 750 watts? Where to buy?

I'm looking here:

This is a rectifier and I don't know if it would handle the current - seems like the voltage is still too low.

I also tried here:
https://www.mouser.com/Semiconductors/Discrete-Semiconductors/Diodes-Rectifiers/_/N-ax1ma/

and I basically have no idea with thousands of possible diodes. Most of them are rated for 5v or less it seems.
Maybe this:
https://www.mouser.com/ProductDetai...GAEpiMZZMtQ8nqTKtFS/E6cG%2brOIQfRIe3TI9O2I9Q=

Any input to confirm I'm on the right track.

Last edited:

For 120 VAC your diode should be rated 200V reverse.

It needs to handle at least 6 amperes (although conventional wisdom advises a rating of 2x the maximum expected amount).

You can shrink-wrap the exposed leads, however as much area as possible ought to be exposed to the air, in order to cool it.

-------------------

For the sake of completeness in examining other options (although these are not better than using a diode)...

You can use a coil's inductive reactance to drop half the AC volts. A value of 92 mH will do it for a 20 ohm load. The coil must carry 6 A.

Or 77 uF of capacitance will create the necessary reactance. You'll need to assemble a bank of them (non-polarized) in parallel, to handle 6 A.

I would say that probably the easiest method of reducing power is to rewire the heating elements in series, from the power settings at the moment they are probably in parallel.

The diode soluton will work but bear in mind it will drop probably one volt at those currents so will dissipate 1 Watt per Amp. It will need heatsinking and in view of the nature of the appliance, that may not be easy!

Forget using resistors. They would be bigger than the heater!

Brian.

Don't look for Schottky diodes. You need a rectifier diode, normal recovery, then select current and voltage. I would suggest you need more than 6A to be safe.

Keith

I really do appreciate everyone's help. I think I figured it out.

I would say that probably the easiest method of reducing power is to rewire the heating elements in series, from the power settings at the moment they are probably in parallel.

Rewiring in series doesn't quite work. There are actually two sets of coils. One set has 6 coils which are all just one long piece strung up and down vertically. There is one other coil by itself placed horizontally crossing behind the other coils. It turns out that when the switch is on 1500w, they are run in parallel. But when they switch is set to 1300w, it seems the coils are serial. I rewired it to make sure they were all in serial and it gave the same reading as when the switch is set to 1300w.

I actually have two of these heaters. So I opened up the second one and ran the 2nd minor coil in series with the other two. With the extra resistance of the third coil, the heater pulled 1030 watts. So it went from 1180w -> 1030w. I don't need the second heater so I'm just going to pull the single coil from the second heater and place it with the coils in the first heater and then wire it up serially. There is plenty of room to do this and I just need to drill two holes and solder two connections. I'm hoping the extra 150w savings will be good enough.

I actually had a couple diodes picked out already from Mouser. If this is still too much wattage, I might have to do that.

Thanks again!

Last edited:

Status
Not open for further replies.