Sunbeam personal heating pad. With AC thru diode, power is 1/2, right ? H

[RJSV]

   Hi, I have a heating pad with little controller that has failed, so I'm checking if that can be run using straight battery power, 12 V or 24 V.

   But I'm wondering, maybe those little controllers only output during one half of the AC
input waveform (at 120 VAC input).
The power would then be 1/2, correct ?
That is because it's essentially only with output for 1/2 the time.  Usually, when voltage is halved then so is the current, so that case power is 1/4 of full.
   Am I correct assuming that ?
   And, do some consumer type products use a triac, for controlling both AC input polarities ?

   I measured approx 220 ohms in the pad heater wire, so full output is about 500 mA, and that gives a power of about 60 watts, for the personal heat-pad.
   If it's a controller that only outputs 1/2 of the AC cycle, having simpler cheaper circuitry, then that means 30 Watts on the high setting.

   Am I estimating correctly, here ?
Thanks.
   

[RJSV]

   Heating pad and connector.

[thm_w]

Do you have the controller to open up and look at?
What is the specification of the unit, it should tell you if its 30 or 60W on the box.

[RJSV]

Thanks I'll check that.  Opened up it has a controller processor and for the PC board output looks to be a pair of (3 pin) transistors for the 500 mA or whatever.  Likely PWM control but I don't know if it switches, like direct off the AC Mains, or switches after some full wave rectification.  I'm thinking maybe a cheaper circuit could just deal with switching only one polarity of the incoming AC waveform.

   I'll do some checking, thanks.

[RJSV]

   Ok, answered my own question, it's with a full-wave rectifier, so my question about AC is n/a.

   But, still curious, question is if you place a diode on AC Mains coming to the load, then that's going to deliver half of available power (in a simpler circuit without PWM or processor).

   At any rate the startling thing is, I found an assembly number, 109375N, and
DATE:  2008 - 8 - 18.        Wow, that was just sold maybe 2021....!

[bdunham7]

But, still curious, question is if you place a diode on AC Mains coming to the load, then that's going to deliver half of available power (in a simpler circuit without PWM or processor).

Yes, if you put a diode in series with a resistive load on an AC-powered circuit, the dissipated power will be almost exactly half, less the diode voltage drop.

However, the measured RMS voltage and current will both be more than half because....math.

[RJSV]

   Yes, I see what you mean, thanks, I would never thought of that ...probably would have stumped me, if asked what the RMS voltage and current has to be.
   That's a compounded deal, to make a certain power level.  Meanwhile a duty cycle approach would (almost) obviously mean a 50 % duty cycle would be delivery of 50 %.  Got it, thanks.

   Calculation would involve square root of 0.5
so roughly 70 % times 70 % or 0.7 X 0.7 gets you the value each is reduced (RMS volts and amps).

[Siwastaja]

However, the measured RMS voltage and current will both be more than half because....math.

Yes, because RMS voltage and current is not any actual physical entity, but an arbitrary data compression model made for us humans to be helpful, to turn waveform pictures into a single number which might be intuitive or useful on its own. But in many cases, especially for those past the beginner stage but still far from enlightenment, the concept hinders more than helps, and in such cases, we should avoid talking about it too far; let the math be a tool, and choose right tool for the job.

This would be one of such cases - it's pretty easy to see on time domain how power is halved when the heater is normally energized for the 50% the time and off otherwise. Compressing this information into RMS current and RMS voltage must, by definition, produce the same answer, but whether those numbers (especially RMS voltage) are of any use is a big question. RMS current can be used at least to choose a fuse or wire gauge.

Similar difficulties are caused by concepts of real, apparent and reactive power. Physics only knows power, which is rate of energy, and P = U*I simply holds at any point in time. All the difficulties are introduced when we choose not to show the waveforms and compress this information into two numbers.

[retiredfeline]

Early in my tinkering career I decided to reduce power consumption in a valve TV that used a heater chain, with a dropping resistor (because the heater voltages didn't add up to 230V). I calculated that RMS voltage if I used a half wave rectifier in series with a small wirewound resistor, instead of the large wirewound resistor, would be 230/sqrt(2). Of course this was the correct answer and the hack worked.

[RJSV]

   Here is the heating pad started to dismantle, but the thing is very securely built, with strong thread and strong base fabric!
   Had wanted to verify that it is just a single run of the thermal heating wire, and since the whole thing was useless / wornout, no harm in opening the thing, and grabbing the wire for my JUNK BOX.
   That's a 220 ohm wire run, as mentioned, and produces approx 55 watts, which is also printed on the connector.  (Mainly was curious if single run, or a couple of separate runs, as the controller has 4 heat levels.)
Yup, nope: It's all just the one (220 ohm) wire, on PWM control 55 watts max.
   The neck or body personal hot pad included a run of 'Bean Bag' material around the periphery for comfort... But the dang thing is just too securely built to have to cut almost every stitch, unless I have a year to separate out that wire.
Byebye that Xtra part.

(Coulda used that, for heating a worm can at fishing pier)

- - Rick B.

[RJSV]

Oops.
   The attachments thingy isn't working straight.