Halving AC Voltage with a single diode. What might go wrong?

[Watth]

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.

[DmitryL]

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)

[RTFM]

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!

[joun]

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.

[DmitryL]

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.

...Cura te ipsum...

[IanB]

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.

[Watth]

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)

Well 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.

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!

The 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.

[DmitryL]

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)

Well 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.

Thank you for noticing obvious thing. Do you know how amplitude of a sine wave is defined ?

[Watth]

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.

Ooh, 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.

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.

I 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.

[Watth]

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)

Well 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.

Thank you for noticing obvious thing. Do you know how amplitude of a sine wave is defined ?

I'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 = V_RMS^1/2
Peak-to-peak Voltage = 2 x Peak Voltage = 2.V_RMS^1/2

Edit: I tried to copy/paste the square root symbol into my answer, and it failed. I retyped square root of x as  x^1/2

[suicidaleggroll]

I'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 = ?2.V_RMS
Peak-to-peak Voltage = 2.Peak Voltage = 2.?2.V_RMS

Yes, 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*0.707

For a sine wave that has the bottom half cut off:
Vrms = Vpeak*0.5

Since 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.

[Watth]

I'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 = ?2.V_RMS
Peak-to-peak Voltage = 2.Peak Voltage = 2.?2.V_RMS

Yes, 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*0.707

For a sine wave that has the bottom half cut off:
Vrms = Vpeak*0.5

Since 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.

Sorry, 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/(2^1/2) a.k.a. the reciprocal of the square root of 2. So, V_RMS = V_PEAK x 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), V_RMS = V_PEAK/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 V_PEAK/2 x 2^-1/2.
Since the load can be reduced to a simple resistance (no power factor), RMS power is directly proportionnal to V_RMS. Therefore halving voltage halves power.

[IanB]

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.

No! 

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.

[Watth]

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.

No! 

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.

I guess I mixed up current's with power's relation to Voltage. P = U²/R. So yeah. A bit counter-intuitive to me!

[LaurentR]

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.

Ooh, 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.

The 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/

[suicidaleggroll]

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 V_PEAK/2 x 2^-1/2.

You could derive it analytically, but for simple things like this I prefer to do it discretely in [pick a language].
eg:

Code: [Select]

IDL> rad = dindgen(10000)/10000 * 2 * !dpi
IDL> volts = sin(rad)
IDL> print,sqrt(mean(volts^2))
      0.70710678
IDL> 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.

[Zero999]

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.

Why not use a thermostat to control temperature  properly?

[Watth]

[Snip]

Why not use a thermostat to control temperature  properly?

I 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.

[sokoloff]

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.

No! 

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.

Which 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.

[IanB]

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.

No! 

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.

Which 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.

It 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 supplied
2. Given that we have the relationship P = V_RMS² / R, we have halved V_RMS²
3. Ergo, we have multiplied V_RMS by sqrt(0.5) = 0.707
4. Hence, V_RMS (half wave) = 0.707 V_RMS (full wave), which was the point in question.

[sokoloff]

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.

[ovnr]

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, but noone 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!

[Dave]

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.

[Watth]

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.

[joun]

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..