# Is DC heating faster than AC heating?

If I have a 120V @ 50Hz AC heater rated at 750W, and I run it using 120V DC, will the DC heat it faster, and why? (Assuming I could get a clean 120V DC power source).

• Since the duty cycle of power is 50% and assuming that the heating element is somthing like nichrome, the inertia of this element can not respond that fast. Using proper adjustable PWM is different. In any case you can calculate the thermal time constant of the material – GR Tech Aug 6 '19 at 23:32
• Note that while the heating element will probably not care, the fan motor might not work with DC. In any case, it's probably not a good idea to run these things out of spec due to the fire hazard… – wrtlprnft Aug 7 '19 at 5:16
• As well as any Thyristor that may be trying to control the circuit. It won't switch off if powered with DC. – user77232 Aug 7 '19 at 10:57
• Please also remember that if a 120 DC load is rated for 6.25 A and has no design margin it may exceed instantaneous currents to over 8.8A when driven with 120VRMS AC at the peak of the cycle. Fortunately it is rare for a heater or fuse to fail with 41% over current in practice at limited duty cycle. Moving from AC rating to DC increases your current design safety margin by 29%. – KalleMP Aug 9 '19 at 6:20

When you say 120V @ 50Hz AC you are implicitly saying 120Vrms.

The RMS voltage is qualitatively defined as the voltage which will give the same resistive heating (averaged out over time) as a DC voltage of the same number. Therefore, by the definition of RMS, the heating will be the same because the RMS voltages are the same.

If you said 120Vpeak or something different, then things would be different.

This is in reference to a heater modeled only as a resistor. No extra real-world components like motors for fans.

• So the utility supply is rated at 120Vrms which is actually 170V peak? – user77232 Aug 6 '19 at 21:00
• @user77232 Indeed. – DKNguyen Aug 6 '19 at 21:00
• Correct answer, but you should incorporate what wrtlprnft already commented: if the heater includes a fan, then this may not work with DC, and that could potentially be quite dangerous! – leftaroundabout Aug 7 '19 at 9:19
• There is also the issue of any inductive component of the heater element. Most likely this is negligible, but I'm sure there are exceptions. – Hot Licks Aug 7 '19 at 21:39
• @user77232 Indeed. If ever working on a typical home/small business triphase setup, you should also not forget that it's nominal voltage (230V here) from phase to neutral, but 400V phase to phase. (Also RMS.) The peak phase to phase voltage is over 560V here in europe. (disclaimer: I'm not sure how common triphase is in the US.) – Demonblack Aug 8 '19 at 10:09

No, because 120V RMS is the AC voltage that produces the same heat as 120 VDC. Theoretically it should be no difference at all.

For heating, the only thing that matters is active power. If the load is a resistor, active power is $$\ V^2/R \$$ with V being the RMS voltage.

The RMS value of 120V DC is 120V RMS, which will give the same active power in a resistor than 120V AC RMS.

AC power will be pulsed, but a 750W resistor will have enough thermal mass to smooth it out, so there is no difference.

No...

On a purely theoretical level, the resistor should get to working temperature a (really) tiny bit faster with DC power. The reason is the Stefan–Boltzmann law which state that the radiative output is proportional to the fourth power of the temperature.

During the beginning of the heating phase, during the power peak of the AC current, the resistor will get hotter than it's DC counterpart at the same time. But it will loose much more energy during the "low power" part of the AC cycle. Fourth power is a really step function.

The effect will be really small, because the ripple in the temperature will be microscopic 1/100th of second is infinitesimally small compared to the time constant of the resistor (typically 10th of seconds), but that doesn't means that it doesn't exist.

But... this is only about the temperature of the resistor itself. In fact, the thermal energy dissipated in the room during the heating time will be higher. Two reason to this :

The first one is that, as stated earlier, instantaneous power dissipation will be higher on average.

The second is that real resistors are not perfect and generally act as Negative Temperature Coefficient resistors. So, by being slightly cooler, the AC one will have lower resistance and drain more current, so more power.

But keep in mind that this calculation is purely theoretical and that in real live, even with perfectly equivalent power source, the effect will certainly not be measurable.

• I'm not sure I buy this argument, it would be much more helpful if you could show this effect graphically or at least derive the relevant equations. – pipe Aug 7 '19 at 18:59
• "Dissipate" ? English is not my main language. And you are right that the difference will be dwarfed by other effect, I just wanted to include that answer for completeness, and to show that difference might exist, they just don't have a practical application. – Kevin FONTAINE Aug 8 '19 at 8:34
• I like this answer a lot, the others are correct in an ideal circuit, but I think this gets deeper into the details. I agree also it would be nice to see the graph showing why this is true. I think an important point to be made is that if the radiative losses are higher, the total heat output must also be higher, which would be explained because the resistance of the element drops with the temperature. – Joel Keene Aug 8 '19 at 15:11
• @user77232 No, because at 0V it heats up slower than 120VDC so it balances out over time even without thermal inertia. with thermal inertia it is even more reduced. But if your frequency is very low and your mass is very small then yes it is different. – DKNguyen Aug 8 '19 at 15:56
• +1, I like this because it touches on the thermal inertia of a non ideal resistor and how this will make a gentle mockery of the RMS equivalence. The effects will be slight and only of theoretical interest. Lay person will struggle to measure the effect. – KalleMP Aug 9 '19 at 6:15