Halving AC Voltage with a single diode. What might go wrong? - Page 1
Forum section: electronics / beginners

author: Watth
author_rank: Frequent Contributor
timestamp: May 13, 2016, 06:24:43 pm
content: Hi there,I use a heating resistance to warm etching solution (ammonium persulfate).The thing is it's a bit over powered : it runs on 230V RMS (mains) and has resistance of 367?. I didn't measure the current, but if my calculations are correct, it develops 144 W for 0.627 A.That's a bit too much: the solution boils, and I don't like the idea of breathing ammonium persulfate fumes.I wonder if it's a safe solution to use a diode to halve the voltage, thus the power. I have 1A 800V rated diodes: 1N4006RLG. I RTFMed (http://docs-europe.electrocomponents.com/webdocs/1221/0900766b81221291.pdf), and concluded these diodes are fit for this task.I could go with a trial and error method, although I'm not a big fan of learning from my mistakes when mains power is involved; so I prefer to ask your opinion(s) first.TL:DR:Is it safe to use a single diode to lower AC voltage for a heating resistor?Thanks for you attention.

author: DmitryL
author_rank: Regular Contributor
timestamp: May 13, 2016, 06:32:18 pm
content: With a single diode you will halve the power, not the voltage.Amplitude voltage will not change at all (ok, it will be 0.7V less), RMS voltage will be Sqrt2(your AC voltage)

author: RTFM
author_rank: Newbie
timestamp: May 13, 2016, 06:41:31 pm
content: Transformer is what u search to lower the voltage,also u may use potentiometer to get exact the voltage u want. Diodes just lower the current,and resistors lower the voltage but create a lot of heat and aren't really solution u want... Diodes are also used to transform ac to dc with bridge rectifier simply 4 diodes connected...anyway. Hope u find the best solution for your project!

author: joun
author_rank: Contributor
timestamp: May 13, 2016, 06:47:33 pm
content: Of course it will work I have done it many times..By adding a diode in series you will cut the ac wave in half and so the rms voltage will be half.The voltage will drop by 0.7V only if you add a diode in series in DC voltage..And of course you cand do that with a simple potensiometer at 140W you will need a high wattage one.

author: DmitryL
author_rank: Regular Contributor
timestamp: May 13, 2016, 06:55:54 pm
content: Quote from: joun on May 13, 2016, 06:47:33 pm...Cura te ipsum...

author: IanB
author_rank: Super Contributor
timestamp: May 13, 2016, 07:01:06 pm
content: A much better solution is to use a lamp dimmer. It is fully adjustable and is designed for the purpose.For a professional job you can mount one in an enclosure with an output socket and a fuse.

author: Watth
author_rank: Frequent Contributor
timestamp: May 13, 2016, 07:02:02 pm
content: Quote from: DmitryL on May 13, 2016, 06:32:18 pmWell a single diode will disable half of the sine, i.e. from 0 to 325V to 0 (minus the direct voltage, witch we can be neglected here) will go through. The second half (0V to -325V to 0) will be blocked.Quote from: RTFM on May 13, 2016, 06:41:31 pmThe resistor is not a mean to lower the voltage. It's the load. I specified it's a heating resistor so that it's clear that the waveform doesn't matter.

author: DmitryL
author_rank: Regular Contributor
timestamp: May 13, 2016, 07:09:08 pm
content: Quote from: Watth on May 13, 2016, 07:02:02 pmThank you for noticing obvious thing. Do you know how amplitude of a sine wave is defined ?

author: Watth
author_rank: Frequent Contributor
timestamp: May 13, 2016, 07:12:10 pm
content: Quote from: IanB on May 13, 2016, 07:01:06 pmOoh, I didn't think of the lamp dimmer. And with halogen lamps getting obselete, I might find some old ones (the kind that lays on the floor).It's not a professional job, but indeed, if I used one I'd add a female plug.Quote from: joun on May 13, 2016, 06:47:33 pmI was worried I missed something that made it unsafe/unpractical.Should I worry about heat ? I'd like to inclose it in adhesive shrink tubing for strain relief. At less than .7A I guess it's OK but I'd rather have it confirmed.

author: Watth
author_rank: Frequent Contributor
timestamp: May 13, 2016, 07:18:40 pm
content: Quote from: DmitryL on May 13, 2016, 07:09:08 pmI'm not sure if that is a passive agressive answer... Anyway if this is a genuine question, if I'm correct, for a sine waveform :Peak Voltage = VRMS1/2Peak-to-peak Voltage = 2 x Peak Voltage = 2.VRMS1/2Edit:I tried to copy/paste the square root symbol into my answer, and it failed. I retyped square root of x as x1/2

author: suicidaleggroll
author_rank: Super Contributor
timestamp: May 13, 2016, 07:26:37 pm
content: Quote from: Watth on May 13, 2016, 07:18:40 pmYes, but when you cut half of the sine wave off, it's no longer a sine wave. The RMS voltage of half of a sine wave is not half of the RMS voltage of a full sine wave, it's 70.7% of it.For a sine wave:Vrms = Vpeak*0.707For a sine wave that has the bottom half cut off:Vrms = Vpeak*0.5Since power is proportional to V^2, this means you're cutting the RMS voltage by 70.7%, and cutting the power in half, as DmitryL said.

author: Watth
author_rank: Frequent Contributor
timestamp: May 13, 2016, 08:21:33 pm
content: Quote from: suicidaleggroll on May 13, 2016, 07:26:37 pmSorry, I didn't check if my copy/paste of the square root symbol worked, and it didn't! My previous (unedited) message was indeed weird.Now, you mention the .707 figure, which equals 1/(21/2) a.k.a. the reciprocal of the square root of 2. So, VRMS= VPEAKx 2-1/2. I get that, no prob.But you say that if you cut the lower half (or any half actually, once it's squared we get absolute values), VRMS= VPEAK/2.I don't get it. If you cut a sine wave in half, both have the same RMS value, since as soon as you use the square function, both are positive. Therefore the square roots of the means of these two halves are equal. So, I don't see why the RMS half of the sine for a full period is not half the RMS value of a full sie wave, therefore VPEAK/2 x 2-1/2.Since the load can be reduced to a simple resistance (no power factor), RMS power is directly proportionnal to VRMS. Therefore halving voltage halves power.

author: IanB
author_rank: Super Contributor
timestamp: May 13, 2016, 08:32:20 pm
content: Quote from: Watth on May 13, 2016, 08:21:33 pmNo!Since the load is a resistance, halving the voltage also halves the current. And since power is voltage times current, you end up with a quarter of the power, not half.

author: Watth
author_rank: Frequent Contributor
timestamp: May 13, 2016, 08:40:10 pm
content: Quote from: IanB on May 13, 2016, 08:32:20 pmI guess I mixed up current's with power's relation to Voltage. P = U²/R. So yeah. A bit counter-intuitive to me!

author: LaurentR
author_rank: Frequent Contributor
timestamp: May 13, 2016, 09:01:43 pm
content: Quote from: Watth on May 13, 2016, 07:12:10 pmThe dimmer is a great idea, but the diode is a pretty common way of doing this too. I have seen many 3-position lamp switched that have a small diode built into the switch.Here is a vacuum cleaner (I think) 3-position switch:http://smile.amazon.com/dp/B00Y35LWGY/

author: suicidaleggroll
author_rank: Super Contributor
timestamp: May 13, 2016, 09:23:55 pm
content: Quote from: Watth on May 13, 2016, 08:21:33 pmYou could derive it analytically, but for simple things like this I prefer to do it discretely in [pick a language].eg:Code:IDL> rad = dindgen(10000)/10000 * 2 * !dpiIDL> volts = sin(rad)IDL> print,sqrt(mean(volts^2))0.70710678IDL> print,sqrt(mean((volts>0)^2))0.50000000In IDL, var>0 is equivalent to "the greater of each element or 0", which essentially just wipes out the negatives and replaces them with zero, like you'd get from a diode.edit: You could also think of it in practical/experimental terms.Imagine taking a single power resistor and placing it between L and N. When L > N, current flows one way through the power resistor and burns off power, when L < N, current flows the other way and burns off the same amount of power per cycle. If you eliminate one of those half cycles, so it only burns off power when L > N for example, and block the other with a diode, then the power dissipation will be cut in half. If power has dropped by half, then RMS voltage must have dropped by 70.7%, since P ~= V^2.

author: Zero999
author_rank: Super Contributor
timestamp: May 13, 2016, 09:51:48 pm
content: Quote from: Watth on May 13, 2016, 06:24:43 pmWhy not use a thermostat to control temperature properly?

author: Watth
author_rank: Frequent Contributor
timestamp: May 13, 2016, 11:30:05 pm
content: Quote from: Hero999 on May 13, 2016, 09:51:48 pmI do use one, but the heating thingy gets really hot, the solution boils almost immediately, even if I make it circulate or even if the thermostat's probe is right by the heating piece.It happens because the heating resistor is quite a small cylinder (D 10mm, L 50mm), another solution would be to add a heat sink like soldering aluminium winglets to the thing.

author: sokoloff
author_rank: Super Contributor
timestamp: May 13, 2016, 11:31:53 pm
content: Quote from: IanB on May 13, 2016, 08:32:20 pmWhich is not what's happening in the proposed single-diode circuit. It's chopping off half the wave (in the time domain) and (VERY) slightly reducing the power on the half of the wave that's left. Ignore the second effect as it's not really material to the discussion. Same amount of power on the forward bias; zero power on the reverse bias. Half the total power.

author: IanB
author_rank: Super Contributor
timestamp: May 13, 2016, 11:57:47 pm
content: Quote from: sokoloff on May 13, 2016, 11:31:53 pmIt is of course material to the discussion, if you would follow along from the beginning and read each post:1. When we insert a diode, we halve the power supplied2. Given that we have the relationshipP=VRMS² /R, we have halvedVRMS²3. Ergo, we have multipliedVRMSby sqrt(0.5) = 0.7074. Hence,VRMS(half wave) = 0.707VRMS(full wave), which was the point in question.

author: sokoloff
author_rank: Super Contributor
timestamp: May 14, 2016, 12:39:52 am
content: The "not really material to the discussion" was clearly specified to apply to the second effect: the voltage drop across the diode during the conducting phase.I can't see any way that could have been misinterpreted.

author: ovnr
author_rank: Frequent Contributor
timestamp: May 14, 2016, 01:36:41 am
content: Oh, what a heap of nonsense for such a simple question.Yes, a diode will work perfectly well. It's suboptimal from a power factor standpoint, butnoone gives a shit.A lamp dimmer is more flexible, and is probably a better tool for the job. But if you have a diode on hand and half power is fine, go for it!

author: Dave
author_rank: Super Contributor
timestamp: May 14, 2016, 05:17:38 am
content: So much arguing over math.\$ V_{RMS} = \sqrt{\frac{1}{T} \int_{0}^{T}{(V_{pk}\cdot sin(2\pi\frac{t}{T}))}^{2} dt} = \frac{V_{pk}}{\sqrt{2}} \$Since you have the second half-wave of the sine blocked by the diode, you can simply halve the upper limit of integration.\$ V_{RMS\, diode} = \sqrt{\frac{1}{T} \int_{0}^{\frac{T}{2}}{(V_{pk}\cdot sin(2\pi\frac{t}{T}))}^{2} dt} = \frac{V_{pk}}{2} \$\$ \frac{V_{RMS\, diode}}{V_{RMS}} = \frac{1}{\sqrt{2}} \$When in doubt, go back to the definition.

author: Watth
author_rank: Frequent Contributor
timestamp: May 14, 2016, 06:08:13 am
content: Actually my question was not so much about the ratio (although the thread subject mentions halving Voltage) ; as it was about the specific model of diode (see first post) and its ability to do the job without firy and/or harmful outcome.

author: joun
author_rank: Contributor
timestamp: May 14, 2016, 06:31:00 am
content: In theory you won't have a problem..But if the diode can't handle the power it will most likelly short so your resistor gets full voltage and works as before so not a real problem ,right?And if diode opens again no problem at all..

author: Dave
author_rank: Super Contributor
timestamp: May 14, 2016, 06:34:06 am
content: You can use the diode you linked in your first post, just make sure it will get enough cooling.