half wave rectifier

Jun 29, 2023 Last reply: 2 years ago 81 Replies

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** Why not a bank of 4 x 40uF popypropylenes rated at 250VAC ( continuous ) . In use, each would have 165VAC applied and carry a tad over 2 amps - so no problem. Likely cost $80 or so. Lot safer than the live diode idea.

..... Phil

That still isn't very cheap though. I take your point that it is safer.

By inspection, from my plot, you need to add 0 Ohms in series in order to dissipate 3000 W from 250 VAC. On the other hand, when you add 20 additional Ohms, in series to the original 20 Ohms, you obtain 1500 W. Again, by inspection. As stated above, 250 VAC^2 is a constant. Let's call it V'. When you double the original resistance, Ro, you obtain:

P = V'(1/(2*Ro)) = V'(1/2)(1/Ro) = Po / 2 where Po represents the original power, 3000 W.

My post pertains to plot theory. It taught me how a slope only appears in a linear equation, where a first order independent variable is simply multiplied by a coefficient - a multiplicative inverse doesn't qualify. My newly acquired analytic insight enables me to know exactly what to expect from the 1/R term in a power equation.

Don't let my plot's implication of doubling the resistance to obtain half the power spook you.

Danke,

And, a half-wave rectifier is 'not proportional', it's also on or off. Is that relevant? Either achieves the half-power goal.

I was simply pointing out a misuse of the term "proportional control".

In control circuits, "proportional" has a specific meaning of the control being able to set the thing being controlled over a range, rather than just on or off.

No. You make a resistive voltage divider when you add a series resistor. The voltage across the original resistor is reduced (except if the added R is 0 ohms). Whatever leads one to think the voltage is constant is erroneous or not properly understood.

Ed

Let's call it V'. When you

No. ...HR= 20 Ohms. Add another resistor in series = 20 Ohms

Correct. Now for goodness sake do the next step in the math.

125 volts across 20 ohms produces 781.25 watts in the pot heater and 781.25 watts in the added resistor. But you said you want 1500 watts in the pot heater.

Ed

<snipped>

"If we keep 250VAC constant then V^2 (eg 250VAC^2) is a constant" (as stated above) is my premise. This 250VAC is treated as Vmains by me. My mind mechanically treats Vmains as a constant - because it is. Most electric outlets in my world source about 120 VAC regardless of the load connected. When Vmains is held constant, a doubling of Rmains halves Pmains. This happens because if the voltage in P=V^2/R is held constant then the power must proportionally change with resistance in order to keep the equation balanced.

Allow me to share some graphical interpretations. Here's a plot of Pmains versus Rmains:

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[1] Use conductance to linearize the resistive power plot:
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[2]

The key insight you offer above is to drop voltage Vmains down to Vheater in order to lower Pheater to 1500 W. In graphical form, lower Vheater to 176.77 VAC to bring Pheater down to 1500 W:

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[3] where the dotted vertical line shows conductance Gheater (1 / Rheater) equal to 0.048 siemens.

Note.

[1] octave code:

R=linspace(20,70); P=250^2 ./ R; plot(R, P); xlabel('Rmains (Ohms)'); ylabel('Pmains (Watts)'); ht = text(40, 2000, '250 VAC'); set(ht, "color", "blue"); grid;

[2] octave code:

R=linspace(20,70); G= 1 ./ R; P=250^2 .* G; plot(G, P); xlabel('Gmains (Siemens)'); ylabel('Pmains (Watts)'); ht = text(0.03, 1500, '250 VAC'); set(ht, "color", "blue"); grid;

[3] octave code:

R=linspace(20,70); G= 1 ./ R; P=250^2 .* G; Ph=176.77^2 .* G; plot(G, P, '', G, Ph, '', [0.048, 0.048], [0, 4000], 'k:'); xlabel('G (Siemens)'); ylabel('P (Watts)'); mt = text(0.015, 1500, '250 VAC'); set(mt, "color", "blue"); ht = text(0.035, 1000, '176.77 VAC'); set(ht, "color", "red"); hc = text(0.038, 500, '0.048 siemens ->'); set(hc, "color", "black"); grid;

Danke,

Hmmm... I am talking about the sort of burners or ovens in the home used for cooking. Why would the contacts on the heat control need to be shielded? What do you think people bake? Heck, every wall switch throws sparks!

1.44 kW, yes. The space heaters that draw the maximum contain exactly the bimetal "thermostat" you are talking about. There are zero issues with operation of such contacts. My water heater has a similar type of thermostat with similar contacts, that draws 4.5 kW. I know, because I've had it apart before.

So, when you talk about "throwing sparks", you are talking about the normal operation of the contacts. I should have known. What you don't know fills volumes.

So, you agree that stove and oven controls are in enclosures and so, can be used for more than 2,800 W? Glad we got that settled.

LOL! I've never seen you to not be happy wasting everyone's time.

I never said this was a good idea. I said it is not proportional. whit3rd referred to the on/off thermostat control as "proportional" which I corrected. You then jumped in blabbing about limited power. You continue to backpedal until we are finally in agreement. But now you are still talking about using this control. You should bring that up with someone else. It's not a point I was ever discussing.

Correct. He'll get half power out of the combination of HR (the heater resistance) plus AR (the added resistance). The original power was 3000 watts. Half power is 1500 watts. So he'll get 750 watts out of AR and

750 watts out of HR.

BUT THE OP WANTS 1500 WATTS OUT OF HR.

Your math solves the wrong problem. From the op: "As part of an experiment with solar, I want to drive my

220V 3kW immersion heater with a suitable diode so that the effective power is ~1.5kW..."

Ed

<snip>

That's the point. You don't understand this and seem to be unwilling to learn anything new. You say you don't understand what others post, then repeat the wrongthink that you have in your head. Maybe now you will pay a bit more attention to what others have posted and learn something about the problem.

The OP's problem has been solved many times in this thread, in different ways. But the OP doesn't understand any of them. Now, people are just trying to educate the OP.

<snip>

There's actually a couple of problems. One problem is people who try to shout down wrong thought of how doubling a resistor halves the power. My math solves this problem by illustrating an exception. If you keep reading my previous followup (the part you snipped) you'll discover how my math eventually illustrates a solution to the OP's 1500 W problem. But, you need to read my followup in its entirety to see it.

In the end, sloppy nomenclature caused this thread's miscommunication.

"When I use [the word volt]," Humpty Dumpty said in rather a scornful tone, "it means just what I choose it to mean - neither more nor less."

Danke,

Yes, I see it, but it defies apparent simplistic logic.I see it's V xV or IxI for power, hence the quarter/sq root. No wonder nothing I designed/built worked...

Yes, that's what we've tried to tell you. You can't just say, double resistance, so half power in the pot.

Yeah, if you ignore the math, things don't go well.

What made the lightbulb come on?

I went back to basics. I admit I just didn't think about this properly at all. I'm 72 and clearly my brain is screwed and incapable of clear thinking. Never mind, I won't be long for this world and I won't have to think about things like this.

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