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I am controlling a 1500W heater with an SSR at 5Hz. The heater's voltage is 110V freq is 60Hz and my breaker is rated for 15A. I have other loads on this circuit consuming about 2A. This means the maximum current draw is (1500/110) + 2A = 15.63A, so I will trip the breaker without changing something.

The heater is on a temperature control loop that I can adjust the duty cycle and frequency. If I limit it to 90% duty cycle will I be likely to trip a generic 15A breaker? This would bring my maximum current to 14.3A. Could the frequency be optimized to prevent smoother current draw or is 5Hz acceptable?

Edit: Heater espresso boiler heating element

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    \$\begingroup\$ there are two parallel heating elements: one measures 12.3ohms and other is 25.7 giving an equiv resistance of 8.3ohms. between 1590 W and 0 W, these two independent heating elements allow 930 W (12.3 Ω) and 470 (25.7 Ω) switching one end, and about 320 W (series) before even starting burst control. Harper's interjection about the effect of duty cycle on "thermal" tripping of circuit breakers looks quite consequential. \$\endgroup\$
    – greybeard
    Commented Feb 20, 2023 at 7:35

5 Answers 5

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Breakers have what is known as a trip curve. Trip curve (Taken from Understanding Trip Curves)

Modern MCCBs have two trip elements:

  1. Instant magnetic trip, usually in the 5-10 times nominal current area. There's no time delay on this.
  2. Bimetallic trip, up to the instant trip. This is a bimetallic strip heating (or a electronic integral emulating this behavior). This has a time delay and will sustain a higher current for some time.

The crip curves tells us what it is. As we can see from the one above, the breaker will last at least 10000 seconds at 1.13 nominal - for a 15A it will essentially not trip if you load it with 17A for an hour.

Your PWM works far faster than the slow thermal trip. Even though the current is above 15A for periods, the average is below - and as long as you're moving in the thermal regime, average is what matters.

You can think of the thermal trip as a series resistance that will trip if you go above n watts. When you PWM your load, you also PWM the series resistance, thus changing the average power it sees. As long as the PWM is fast enough, and average low enough, it will not trip.

Note that this looks at how things work technically , not whetever this is a good idea or allowed by code.

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Unless you have done actual experiments and measurements and had breaker trips, I suspect the situation is a little different.

The "standard" continuous use device for a US/Canada residential standard 15A circuit is 1500W. This is why nearly all portable electric space heaters sold in the US/Canada are 1500W. It doesn't matter whether they say "big room" or "small room" or specify an area. It doesn't matter whether they have a fan or not, ceramic, oil filled, radiant, etc. 1500W is the magic number.

Similarly, the "standard" short term use device for a US/Canada residential standard 15A circuit is 1875W. This is why nearly every hair dryer (whether $15 or $100) sold at retail for the US/Canada is rated 1875W.

Why?

Because the baseline voltage is actually 125V, not 110V. (It was 110V and then 120V and then 125V, though the typical actual voltage in most areas is still 120V.) 125V x 15A = 1875W - that's your short term usage (a.k.a., hair dryer). Continuous usage requires an 80% derate, and 125V x 15A x 0.8 = 1500W - that's your space heaters.

Which means that while you shouldn't actually use more than 1500W on a continuous basis on a 15A circuit, you won't actually trip any breakers until some point past 1875W. You have 3A of wiggle room. In reality, you can typically go 10% to 20% over the rating for an extended period of time without a likely breaker trip. But you shouldn't plan that way.

Depending on your situation, it may be possible to run 12 AWG wire and use a 20A breaker. In fact, the NEC (electrical code for the US) has required 20A circuits for kitchen (toasters, coffee makers) and bathroom (hair dryers) for many years, even though in residential usage most 20A circuits legitimately have 15A receptacles (this specific exception to breaker size = receptacle size is part of the NEC). That allows for a full 15A usage (which is either a true 15A hair dryer or a derated 12A space heater) to have 5A (or 3A continuous) of room to spare. Meaning that if you have a 15A circuit with a 1500W space heater you really shouldn't have much else running on the circuit at the same time, but if you have a 20A circuit then running lights and TV, etc. on the same circuit won't be a problem at all.

A similar calculation should apply to your circuit, unless you are using a specialized very sensitive/accurate circuit breaker. But with standard items (typical residential/commercial breaker panels from Eaton, GE, Siemens, etc.) this is how it works.

There is also a little extra room provided by Ohm's Law. A typical space heater, not counting fan or other extras, is a simple resistive circuit. For 12A @ 125V = 1500W, that is 10.42 Ohms. Which means that at 120V (typical for US/Canada) the actual usage will be 11.52A and 1382W. (Well, some slight variance from that is possible due to resistance varying slightly with temperature and other factors.) So in a typical setup you actually have a little extra current available in the circuit for other stuff before you run into circuit breaker trips.

For more about how circuit breakers work and why a 15A breaker doesn't magically trip at 16A, search for trip curve.

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  • \$\begingroup\$ I have added a link to the heating element. It specifically lists 110V as the baseline. Likely because this is a part to an older piece of equipment. In light of the info you posted, my application may be in the minority using 110V as a baseline. If my supply is 120V does this mean I will be pulling more current than I originally posted? For example heater resistance must be 1500/(110^2) = 8ohms and 120V/8ohms = 15A \$\endgroup\$
    – Feynman137
    Commented Feb 19, 2023 at 17:43
  • \$\begingroup\$ Interesting. Yes, that is a possibility. Or it may be somewhere in between. Only way to tell for sure is to measure it. Either measure resistance directly or measure voltage and current when running. Note that measuring high current is not, as I understand, so easy/safe with a typical meter except using a clamp meter. \$\endgroup\$ Commented Feb 19, 2023 at 17:50
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    \$\begingroup\$ I don't have a current meter capable of this. But there are two parallel heating elements: one measures 12.3ohms and other is 25.7 giving an equiv resistance of 8.3ohms. \$\endgroup\$
    – Feynman137
    Commented Feb 19, 2023 at 18:04
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    \$\begingroup\$ As far as a meter capable of the task, just get a Kill-A-Watt, they're like $25. That'll give you voltage and current, from which resistance is a simple divide. \$\endgroup\$ Commented Feb 20, 2023 at 0:53
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Circuit breakers are not precision devices, and consumer types have a long response time to overloads (as opposed to short circuits). By the way, they also tend to trip earlier if they have been caused to trip many times (at least in my experience). They also trip on overload at lower current and faster if they are in a hot environment.

Chances are good that if you limit the average over 30 or 60 seconds to less than the rated capacity of the breaker you'll be fine, particularly if your other load does not include any significant surge component (like a compressor start surge). 5Hz seems ridiculously fast to me, especially given the mains frequency and likely zero-cross switching.

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  • \$\begingroup\$ This is what I was looking for, I should not exceed 15A but for fractions of a second. And the amount it is exceeded by will be less than an amp. And so the average current pull with a 90% duty cycle should not trip the breaker. \$\endgroup\$
    – Feynman137
    Commented Feb 19, 2023 at 17:46
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If you use 90% duty cycle, you will still be drawing excess current of 15.63A for 90% of the time, even if the long term average current is 14.3A.

It will always exceed 15A for whatever time duration you decide to keep the heater on. Using any duty cycle does not change that.

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  • \$\begingroup\$ You are correct I will be exceeding 15A a fraction of the time. But breakers don't trip the instant a circuit exceeds the rated load, there is some time delay, and I am wondering if I will trip a breaker. \$\endgroup\$
    – Feynman137
    Commented Feb 19, 2023 at 17:37
  • \$\begingroup\$ @Feynman137 The heat will accumulate on the breaker thermal element eventually causing it to trip. The wiring the breaker are protecting will experience similar heating so you want the breaker to trip. If you want X% duty cycle to actually result in X% of the current rather than full current X% of the time you need to add smoothing elements like inductors (also known as line reactance) which will be rather large and expensive and introduce other things that would need to be accommodated (flyback voltage spikes from PWMing an inductive element). \$\endgroup\$
    – DKNguyen
    Commented Feb 20, 2023 at 1:27
  • \$\begingroup\$ This is a valid point but if you could show that this marginal overload would cause a trip due to thermal effects it would be more convincing. \$\endgroup\$
    – Feynman137
    Commented Feb 20, 2023 at 4:45
  • \$\begingroup\$ @Feynman137 See Vidarlo's answer. \$\endgroup\$
    – DKNguyen
    Commented Feb 20, 2023 at 7:01
  • \$\begingroup\$ As noted in another answer, the 13.63A calculation is based on the wrong assumption that the heater consumes 1500W at 110V. It's only 12A @ 125V (15A derated by 20%). \$\endgroup\$
    – MSalters
    Commented Feb 20, 2023 at 11:55
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Assuming 125V supply voltage, which is what UL would prefer you do, the 12.3 ohm element will pull 9.8 amps. The 25.7 ohm element will pull 4.67 amps.

North American practical voltage is somewhat lower than that, so current will be also.

UL requires that a plug-in appliance with a NEMA 5-15 plug limit itself to 1500 watts @ 125V.

For your application, the simplest solution is sell the appliance with a NEMA 5-20 plug. This has a rotated neutral pin so it can only fit in 20A sockets. Now you are entitled to draw 16A or 2000W in UL's thinking. "That was easy"

By Code all kitchen receptacle circuits are supposed to be 20 amps. 20A sockets are allowed but not mandatory on 20A circuits.

enter image description here

Your idea of using PWM to bring average power below 1500W does not work. That is because overload is not linear. Imagine you took a 3000W heating element and cycled it 50% of the time. Well that's 1500W yes? True, but we care about thermal heating of the wires. What happens with that? What is the thermal heating at 1500 watts? (use any arbitrary/assumed wire resistance). OK what is the thermal heating at 3000 watts? What's the proportion? HAHA, now you see why 200% overload 50% of the time is not the same as 100%.

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    \$\begingroup\$ The first portion of this answer is clear. The second portion regarding thermal heating is not.. Without the inclusion of calculations it is difficult to follow this logic and draw meaningful conclusions. \$\endgroup\$
    – Feynman137
    Commented Feb 20, 2023 at 4:34
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    \$\begingroup\$ @Feynman137 it's basic I^2R heating - the power dissipated in the supply wires increases quadratically with respect to load current, while the load current itself increases linearly with load power. So running a 3000W load at 25% duty cycle causes the same amount of heating as running a 1500W load at 100% duty cycle. You might say well 25% is too low to be useful but that's how the math works. Note that any load/current averaging assumptions you make assume sufficiently high switching frequency - overloading for 1min every 4 mins is a whole different game than overloading for 30min every 2 hrs \$\endgroup\$
    – cat40
    Commented Feb 20, 2023 at 6:50
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    \$\begingroup\$ @Feynman137 I got the I^2R thing from Ohm's law and the definition of electric power by the way. I'm assuming you know Ohm's law because if you don't you shouldn't be messing with mains at all (not that knowing it means you should, but I'll leave that decision to you). Do note I am not saying doing this is a good idea; in fact it is rarely a good idea to deliberate overload any system for any amount of time, but I'll leave it to Harper and others who are better versed in mains safety than I to chime in on exactly how unsafe this is. Do also pay attention to SSR quality-they like to fail short \$\endgroup\$
    – cat40
    Commented Feb 20, 2023 at 6:54
  • \$\begingroup\$ I don't follow the math either. We've got two resistors in series here: the device and the wires. Ohm doesn't make a difference between the two resistors. The distribution of heat production is proportional to the resistances. If we have a 1500W heater and the wire resistance is 1%, then the heat produced by the wire is 15W. If we use a 50% duty cycle, that drops to 750W and 7.5W respective. Reality is more complicated - the heater will have a temperature-dependent resistance. That 3kW heater cycled at 50% will be more than 1500W. \$\endgroup\$
    – MSalters
    Commented Feb 20, 2023 at 10:10
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    \$\begingroup\$ @MSalters You're looking at the desired heat, not loss in the wires. We're saying 18A on a wire half the time is not the same as 9A all the time. At 9A say wire is 0.1 ohm so voltage drop is 0.9V, so watts of heat in the wire is 9 x 0.9 = 8.1 watts. Agreed? Recompute at 18A. \$\endgroup\$ Commented Feb 20, 2023 at 18:24

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