# Changing wattage of a heat pad

I'm working on setting up a heater bed for my 3D printer, and looking for some help / sanity check on adjusting the heater input. I've got a Kat's automotive heater pad (150 W, 4" x 5") attached to an aluminum print bed.

My problem is that the 150 W heater is too hot. I'd like to have a 50 W heater, but haven't found one the right size (plus I have this one on hand). I'd like to add a resistor inline on the cord to make this approximately a 50 W heater.

I believe that I need to add a ~200 Ω resistor to the circuit. My reasoning is it's a 150 W, 120 V device. My electronics knowledge is very rusty (20 yrs since I've had an electronics class...), but if I recall, watts = volts * amps, so I'm expecting 1.25 A across the device. Using those values and V = IR, that means there's 96 Ω in the system, which I'd expect to be the heater pad itself. If I want 1/3 of the wattage across the heater, I need 188 (~200) Ω added in. That should reduce the current through the circuit to ~0.4 A through all components.

So, my plan is to hop down to the local friendly Radio Shack, purchase a 200 Ω resistor, cut my wall outlet cord, and solder the resistor inline. Problem is, what I see online seems to be mostly 1/2 W rated resistors. I should have ~32 W across this resistor. Does that mean I need to find a 200 Ω, 50 W resistor?

• You'd be better off hopping down to your local friendly home improvement store and buying a light dimmer. Nov 14, 2016 at 1:37
• You can knock it down to 75W with a cheap diode, and use a temperature controller and sensor to maintain a constant temperature. Open loop is not hoping to be as good. Nov 14, 2016 at 1:40
• Thanks for all the input! KISS, the dimmer seems like the best options, and I'm pricing them out at the same cost as the 50W resistors... Nov 14, 2016 at 13:26

You were generally correct in your figurings BUT calculated for new power loss of 50 Watts total. The heating pad would get 1/3 of that or about 16 Watts and the resistor about 34 Watts. As Rseries is added the current decreases (so total power drops) AND the portion available to the eatong pad drops - so power is related to I^2.

Throughout "=" === "~=" :-)

Assume 150 W , 120V heating pad. Power = V x I so I = Power/V
I = 150/120 ~= 1.25A = 1250 mA.

How big is existing R?
Power = V^2/R so R = V^2/Power
So Rpad = 120^2/150 = 96 Ohms.

For 50W in pad you want new I to be smaller by a facti\or of 3^0.5 as power is to current squared.

So new Ipad = 1250mA/3^0.5 = 1250/1.732 = 720 mA.

As I = V/R you want total new R to be 1.732 x as large as before so added series R is 0.732 X existing R Rseries = 0.732 x 96 ~= 70 Ohms.

Sanity check.
I = V/R = 120/(96+70) = 720 mA
Power = I^2 x R = (0.72)^2 x 96 = 50 Watts in heating pad. Power in 70 Ohm resistor = 36 Watts.

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Diode + Resistor. As Spehro say - adding a series diode is an easy way to halve the power to about 75 Watts. You can then if wanted add a smaller series R to lower the power further.
From above (75/50)^0.5 = 1.22
Extra series R = Rpad x 0.22
= 96 R *0.22 = 21 Ohms. (eg 20, 22, ...)
Power in 20 Ohms = 20/96 x 50W = 10W.
Rate resistors at ~= 50% of max rated power for good lifetimes.
2 x 10 Ohm 10W resistors on series or 2 x ~= 40R 10W resistors on parallel would be "about right".

As a 3D printer owner, I think I think you misunderstoodd some points.

1. Power is not the same as temperature. Power is related to the rate of change of temperature. Power still relates to the equilibrium or final temperature.
2. Higher power is good thing. More power is faster to reach setpoint temperature.

If you have high power heated bed, what you really needed is temperature controller. You will not be able to get precise bed temperature with a shunt resistor.