half wave rectifier

On Saturday, July 1, 2023 at 6:22:09 PM UTC-4, TTman wrote:
> > 
> > The communication hasn't worked so far, perhaps it's been a bit 
> > unclear. I'll try a bit differently. The key point is that doubling 
> > the resistance does NOT halve the power. 
> > 
> > Take the resistance of your heater (HR), which is about 20.8 ohms, 
> > and figure what voltage needs to be applied to it to run it at 1500 
> > watts.  Here's the math: 
> > Solve for heater resistance (HR): 
> > P=E^2/HR so HR=E^2/P so HR=250*250/3000 so HR=62500/3000 = 20.83 
> > 
> > Now what voltage must be applied to HR to equal 1500 watts? 
> > 1500 = E*E/20.83 so 20.83 * 1500 = E^E so 31250 = E*E so 
> > E = sqroot 31250 so E = 176.77volts 
> > 
> > Ok, once you know that voltage you can solve for the rest. 
> > We need to add a series resistance (AR) to drop the voltage 
> > across HR from 250 to 176.77volts 
> > 
> > Compute the current through the series string of HR plus AR 
> > 
> > Solve for I through HR at 1500 watts: 
> > P=E*I so 1500 = 176.77 * I so I = 1500/176.77 = 8.485amps 
> > 
> > The total voltage (250) minus VHR (176.77) equals VAR, the 
> > voltage across the (AddedResistance) AR 
> > 
> > 250-176.77 = 73.223volts 
> > 
> > Solve for power dissipated in AR 
> > 
> > P=I*E= 8.486*73.223=621.32 watts in AR 
> > 
> > I hope this makes it a bit clearer. 
> > Ed
> Too complicated...HR= 20 Ohms. Add another resistor in series = 20 Ohms 
> making Rt - 40 Ohms. 
> If I apply 250 V across 40 Ohms, I get 125V across each resistor.

and... you get half the current, so each resistor will dissipate a quarter of the original power.  The total power will be half of the original power, but if I understand your limitations on the design, the series resistor can not be used to heat the pot.  

I'm not going to continue to beat you over the heat to get you to pay attention to the equations.  But here are equations for power, one more time. 

P = E * I
P = E^2 / R
P = I^2 * R

Apply them to the appropriate devices, rather than just saying, "twice the resistance, half the power".  Also, try solving for the actual current and the voltage on each device.  This isn't hard, but sometimes it takes a bit of work to get used to it. 

-- 

  Rick C.

  ---- Get 1,000 miles of free Supercharging
  ---- Tesla referral code - https://ts.la/richard11209

Ricky wrote:
> TTman wrote:
>> Ed wrote:
>> >
>> > The communication hasn't worked so far, perhaps it's been a bit
>> > unclear. I'll try a bit differently. The key point is that doubling
>> > the resistance does NOT halve the power.
>> >
>> > Take the resistance of your heater (HR), which is about 20.8 ohms,
>> > and figure what voltage needs to be applied to it to run it at 1500
>> > watts.  Here's the math:
>> > Solve for heater resistance (HR):
>> > P=E^2/HR so HR=E^2/P so HR=250*250/3000 so HR=62500/3000 = 20.83
>> >
>> > Now what voltage must be applied to HR to equal 1500 watts?
>> > 1500 = E*E/20.83 so 20.83 * 1500 = E^E so 31250 = E*E so
>> > E = sqroot 31250 so E = 176.77volts
>> >
>> > Ok, once you know that voltage you can solve for the rest.
>> > We need to add a series resistance (AR) to drop the voltage
>> > across HR from 250 to 176.77volts
>> >
>> > Compute the current through the series string of HR plus AR
>> >
>> > Solve for I through HR at 1500 watts:
>> > P=E*I so 1500 = 176.77 * I so I = 1500/176.77 = 8.485amps
>> >
>> > The total voltage (250) minus VHR (176.77) equals VAR, the
>> > voltage across the (AddedResistance) AR
>> >
>> > 250-176.77 = 73.223volts
>> >
>> > Solve for power dissipated in AR
>> >
>> > P=I*E= 8.486*73.223=621.32 watts in AR
>> >
>> > I hope this makes it a bit clearer.
>> Too complicated...HR= 20 Ohms. Add another resistor in series = 20 Ohms
>> making Rt - 40 Ohms.
>> If I apply 250 V across 40 Ohms, I get 125V across each resistor.
>
> and... you get half the current, so each resistor will dissipate a quarter
> of the original power.  The total power will be half of the original power,
> but if I understand your limitations on the design, the series resistor can
> not be used to heat the pot.
>
> I'm not going to continue to beat you over the heat to get you to pay
> attentn to the equations.  But here are equations for power, one more time.
> 
> P = E * I
> P = E^2 / R
> P = I^2 * R
>
> Apply them to the appropriate devices, rather than just saying, "twice the
> resistance, half the power".  Also, try solving for the actual current and
> the voltage on each device.  This isn't hard, but sometimes it takes a bit
> of work to get used to it.

Thanks guys. You helped me put it all together. If we keep 250VAC 
constant then V^2 (eg 250VAC^2) is a constant. Initially, my mind 
misrepresents the V^2 constant as a slope in a linear equation.
    But V^2 is not a slope because P=V^2/R is not linear. The 
multiplicative inverse term, 1/R, makes it non-linear. 
    Here's a plot: <https://crcomp.net/misc/pr.png> [1]. As expected, it
looks similar to a 1/x plot. 

This (so far unanswered) question was previously asked by me:

    Does the -3 dB half power point play a role with kW (eg not kWh)?

Ordinarily the half power point pertains to bandwidth. It's a little
eccentric to apply it to a constant frequency. But it can be done. By
inspection, the half power point of 3000 W yields 40 Ohms.

Note.

[1] octave code:

    R=linspace(10,50,40);
    P=250^2 ./ R;
    plot(R, P);
    xlabel('R (Ohms)');
    ylabel('P (Watts)');

Danke,

-- 
Don, KB7RPU, https://www.qsl.net/kb7rpu
There was a young lady named Bright Whose speed was far faster than light;
She set out one day In a relative way And returned on the previous night.

Don wrote:
--------------------
>
> This (so far unanswered) question was previously asked by me: 
> 
> Does the -3 dB half power point play a role with kW (eg not kWh)? 
>

**  I suggested adding a series capacitor so the load voltage falls to half power.
     The capacitive reactance then needs to be the same as the load resistance.
     Same calculation as the -3dB point.
      C= 1/2.pi.f.R

......   Phil 

On Saturday, July 1, 2023 at 11:20:23 AM UTC-7, Ricky wrote:
> On Saturday, July 1, 2023 at 1:45:10 PM UTC-4, whit3rd wrote: 
> > On Saturday, July 1, 2023 at 5:33:42 AM UTC-7, TTman wrote: 
> > 
> > > It's absolutely clear to me, and no doubt to many others on here that 
> > > you have no idea. Which part of 3kW heater running off 250V don't you 
> > > understand ? I want to halve the power in the 3kW heater. Show me the 
> > > maths... 
> > If you want to halve the power in a heater, use an electric oven control 
> > knob. Those are available as repair parts, and they cycle the heating 
> > element they control, to give proportional heat control. 
> > 
> > Every broken down cooktop has four of those, and three of 'em will work.

> I've never seen an oven or stove control that gave proportional control. Every one I've seen was an on/off switch with some hysteresis, around 10-15 degrees. What am I missing? 

If you want to 'halve the power in a heater', that's the way an electric resistance stovetop
acts with the knob set on 'medium'; it's a time-proportioning switch, with an internal heater
run by a rheostat; 'high' is always on, because heat pushes the bimetallic strip together.
The hysteresis is why the contacts don't make/break on a fast cycle (which would
wear out the high-current contacts).  It's electrically on/off, but time proportioning.

Only a few fancy stovetops have thermostats, though ovens often do.  Those haven't
half-heat-like settings.

On Sunday, July 2, 2023 at 12:31:18 AM UTC-4, Don wrote:
> Ricky wrote: 
> > TTman wrote:
> >> Ed wrote: 
> >> > 
> >> > The communication hasn't worked so far, perhaps it's been a bit 
> >> > unclear. I'll try a bit differently. The key point is that doubling 
> >> > the resistance does NOT halve the power. 
> >> > 
> >> > Take the resistance of your heater (HR), which is about 20.8 ohms, 
> >> > and figure what voltage needs to be applied to it to run it at 1500 
> >> > watts. Here's the math: 
> >> > Solve for heater resistance (HR): 
> >> > P=E^2/HR so HR=E^2/P so HR=250*250/3000 so HR=62500/3000 = 20.83 
> >> > 
> >> > Now what voltage must be applied to HR to equal 1500 watts? 
> >> > 1500 = E*E/20.83 so 20.83 * 1500 = E^E so 31250 = E*E so 
> >> > E = sqroot 31250 so E = 176.77volts 
> >> > 
> >> > Ok, once you know that voltage you can solve for the rest. 
> >> > We need to add a series resistance (AR) to drop the voltage 
> >> > across HR from 250 to 176.77volts 
> >> > 
> >> > Compute the current through the series string of HR plus AR 
> >> > 
> >> > Solve for I through HR at 1500 watts: 
> >> > P=E*I so 1500 = 176.77 * I so I = 1500/176.77 = 8.485amps 
> >> > 
> >> > The total voltage (250) minus VHR (176.77) equals VAR, the 
> >> > voltage across the (AddedResistance) AR 
> >> > 
> >> > 250-176.77 = 73.223volts 
> >> > 
> >> > Solve for power dissipated in AR 
> >> > 
> >> > P=I*E= 8.486*73.223=621.32 watts in AR 
> >> > 
> >> > I hope this makes it a bit clearer.
> >> Too complicated...HR= 20 Ohms. Add another resistor in series = 20 Ohms 
> >> making Rt - 40 Ohms. 
> >> If I apply 250 V across 40 Ohms, I get 125V across each resistor. 
> > 
> > and... you get half the current, so each resistor will dissipate a quarter 
> > of the original power. The total power will be half of the original power, 
> > but if I understand your limitations on the design, the series resistor can 
> > not be used to heat the pot. 
> > 
> > I'm not going to continue to beat you over the heat to get you to pay
> > attentn to the equations. But here are equations for power, one more time.
> > 
> > P = E * I 
> > P = E^2 / R 
> > P = I^2 * R 
> > 
> > Apply them to the appropriate devices, rather than just saying, "twice the 
> > resistance, half the power". Also, try solving for the actual current and 
> > the voltage on each device. This isn't hard, but sometimes it takes a bit 
> > of work to get used to it.
> Thanks guys. You helped me put it all together. If we keep 250VAC 
> constant then V^2 (eg 250VAC^2) is a constant. Initially, my mind 
> misrepresents the V^2 constant as a slope in a linear equation. 
> But V^2 is not a slope because P=V^2/R is not linear. The 
> multiplicative inverse term, 1/R, makes it non-linear. 
> Here's a plot: <https://crcomp.net/misc/pr.png> [1]. As expected, it 
> looks similar to a 1/x plot. 

Of course it does, that's the way you've constructed it.  Your graph is not particularly relevant to adding a series resistor.  The pot heater is the only resistance which provides useful heat.  The added series resistance is just to lower the power in the pot heater.  To figure the power in the pot heater, the reduced voltage on the pot heater needs to be taken into account.  


> This (so far unanswered) question was previously asked by me: 
> 
> Does the -3 dB half power point play a role with kW (eg not kWh)?  

Not sure what role you are referring to.  -3 dB, is a way of expressing half power.  What more are you thinking of? 


> Ordinarily the half power point pertains to bandwidth. 

It has nothing to do with bandwidth, other than it is the point chosen for specifying an attenuation.  


> It's a little 
> eccentric to apply it to a constant frequency. 

Why?  Because you've not see this before?  


> But it can be done. By 
> inspection, the half power point of 3000 W yields 40 Ohms. 

You are solving an irrelevant problem.  No one cares about reducing the total power to half.  That's the mistake that TTman is making.  I originally said he could add an equivalent heater in series with the one he's already using and the total power would be half.  He said that won't work because only the pot heater is producing useful heat. 

If the pot heater (at 18 or 20 ohms, pick one) is dissipating 3,000 watts, how much resistance do you need to add in series to make the pot heater dissipate 1,500 watts? 

-- 

  Rick C.

  ---+ Get 1,000 miles of free Supercharging
  ---+ Tesla referral code - https://ts.la/richard11209

On Sunday, July 2, 2023 at 1:32:34 AM UTC-4, whit3rd wrote:
> On Saturday, July 1, 2023 at 11:20:23 AM UTC-7, Ricky wrote: 
> > On Saturday, July 1, 2023 at 1:45:10 PM UTC-4, whit3rd wrote: 
> > > On Saturday, July 1, 2023 at 5:33:42 AM UTC-7, TTman wrote: 
> > > 
> > > > It's absolutely clear to me, and no doubt to many others on here that 
> > > > you have no idea. Which part of 3kW heater running off 250V don't you 
> > > > understand ? I want to halve the power in the 3kW heater. Show me the 
> > > > maths... 
> > > If you want to halve the power in a heater, use an electric oven control 
> > > knob. Those are available as repair parts, and they cycle the heating 
> > > element they control, to give proportional heat control. 
> > > 
> > > Every broken down cooktop has four of those, and three of 'em will work. 
> 
> > I've never seen an oven or stove control that gave proportional control. Every one I've seen was an on/off switch with some hysteresis, around 10-15 degrees. What am I missing?
> If you want to 'halve the power in a heater', that's the way an electric resistance stovetop 
> acts with the knob set on 'medium'; it's a time-proportioning switch, with an internal heater 

What you are calling "Time proportioning" is called Bang-Bang. It's not proportional, it's on or off. 

-- 

  Rick C.

  --+- Get 1,000 miles of free Supercharging
  --+- Tesla referral code - https://ts.la/richard11209

On 01/07/2023 23:09, TTman wrote:
> 
>>
>>
>>> Which part of 3kW heater running off 250V don't you
>>> understand ? I want to halve the power in the 3kW heater. Show me the
>>> maths...
>>
>> I already did, but here it is again.
>>
>> P = E^2 / Rh, you want to reduce the voltage on the heater by adding a 
>> resistor in series, i.e. replace Rh with Rh + Rs, where Rs is the 
>> added resistor.  So, how large does Rs need to be for the power in Rh 
>> to be half the original value (1,500 W vs. 3,000 W)?
>>
>> So, now we have Ph = Eh^2 / Rh.  Since Ph is 1/2 * P from above, 2 * 
>> Eh^2 / Rh = E^2 / Rh.
>>
>> Multiply by Rh / Eh^2 to get, √ 2 = E / Eh = 1.414, Eh = E * 0.707.
>>
>> So now, we know the voltage needed on the heater for it to dissipate 
>> half the original power.
>>
>> Easy to get the required added resistance, Es = E * Rs / (Rs + Rh), E 
>> * 0.293 = E * Rs / (Rs + Rh), 0.293 = Rs / (Rs + Rh).  0.293 Rs + 
>> 0.293 Rh = Rs, 0.293 Rh = 0.707 Rs, Rs = Rh * 0.293 / 0.707 = Rh * 0.414.
>>
>> Rh can be found from P = E^2 / Rh, Rh = E^2 / P = 16.133 ohms, so Rh = 
>> 6.6861 ohms.
>>
>> Was there any part you didn't follow?
> 
> None of it. You're over thinking it. Rh = Rs. Rh = ~18 Ohms. therefore 
> Rs = 18 ohms.Half the power = twice the resistance.Simple.

Whilst I don't like the heavy handed obfuscated way he has done it or 
that fact that his solution sets power in the heater load to be 1.5kW 
and leaves about 630W dissipated as waste heat in the dropper resistor.

So the total mains load is 2.1kW and it is one possible interpretation 
of what you said you wanted to do at the outset.

> 1500W into Rh 
> and 1500W into Rs.

No. If you double the resistance you halve the current flowing in each 
resistor and you also halve the voltage drop across each of them.

Power is V^2/R = 3kW in the original configuration
and V^2/2R = 1.5kW in the series configuration 750W in each.

The total dissipation it is then 1500W in *total* with 750W delivered in 
each resistor. IOW you waste 50% of the energy supplied in the dropper.

The only way to use all the heat would be to have your additional series 
resistor as two immersion heaters in series in the hot water tank.

>> I would point out that anyone familiar with using dB to compare power 
>> levels has likely discovered that -3 dB is half power, but 0.707 of 
>> the voltage and current.  Most designers simply know this off the top 
>> of their heads.  I take it you don't use such calculations very often.

His objective was to reduce the load presented to his mains supply from 
3kW to 1.5kW. An older >3kW industrial filament lighting or infrared 
heater controller might be one option for the OP. Something like this 
from Amazon depends a bit on whether it is really a thyristor (which 
would be no better than his original power diode) or a triac. eg.

www.amazon.co.uk/dp/B0BY96W111/0TmFtZT1zcF9kZXRhaWwy

Spec says thyristor but it is anybody's guess if it is or not. Heaven 
knows how much RFI a cheap and nasty Chinese one might throw out.

Better quality ones might be available scrap from old theatres that are 
replacing their incandescent lights with much lower power LED systems.
(although most have already done so)

-- 
Martin Brown

On 02/07/2023 06:17, Phil Allison wrote:
> Don wrote:
> --------------------
>>
>> This (so far unanswered) question was previously asked by me:
>>
>> Does the -3 dB half power point play a role with kW (eg not kWh)?
>>
> 
> **  I suggested adding a series capacitor so the load voltage falls to half power.
>       The capacitive reactance then needs to be the same as the load resistance.
>       Same calculation as the -3dB point.
>        C= 1/2.pi.f.R
> 
> ......   Phil

A non-polar 160uF capacitor rated for peak voltage 450v 50Hz and 10A 
continuous ripple current isn't going to be cheap and cheerful.

You run up against the maximum permitted ripple current PDQ. Of Rapid's 
stock today 3x 470uF electrolytics in parallel would get you ~160uF and 
7A ripple. Obviously need 6 in total and a couple of low drop diodes.

Or a pair of non-polar 80uF 450v motor start capacitors in parallel - 
can't figure out their ripple current rating (Italian datasheet).

Phase shifting is OK for a 12W filament bulb nightlight but not very 
realistic or cost effective for a 3kW load!

-- 
Martin Brown

On 02/07/2023 01:11, Ricky wrote:
> On Saturday, July 1, 2023 at 6:09:30 PM UTC-4, TTman wrote:
>>>
>>>
>>>> Which part of 3kW heater running off 250V don't you
>>>> understand ? I want to halve the power in the 3kW heater. Show me the
>>>> maths...
>>>
>>> I already did, but here it is again.
>>>
>>> P = E^2 / Rh, you want to reduce the voltage on the heater by adding a resistor in series, i.e. replace Rh with Rh + Rs, where Rs is the added resistor. So, how large does Rs need to be for the power in Rh to be half the original value (1,500 W vs. 3,000 W)?
>>>
>>> So, now we have Ph = Eh^2 / Rh. Since Ph is 1/2 * P from above, 2 * Eh^2 / Rh = E^2 / Rh.
>>>
>>> Multiply by Rh / Eh^2 to get, √ 2 = E / Eh = 1.414, Eh = E * 0.707.
>>>
>>> So now, we know the voltage needed on the heater for it to dissipate half the original power.
>>>
>>> Easy to get the required added resistance, Es = E * Rs / (Rs + Rh), E * 0.293 = E * Rs / (Rs + Rh), 0.293 = Rs / (Rs + Rh). 0.293 Rs + 0.293 Rh = Rs, 0.293 Rh = 0.707 Rs, Rs = Rh * 0.293 / 0.707 = Rh * 0.414.
>>>
>>> Rh can be found from P = E^2 / Rh, Rh = E^2 / P = 16.133 ohms, so Rh = 6.6861 ohms.
>>>
>>> Was there any part you didn't follow?
>> None of it.
> 
> Yes, that's what I figured.
> 
> 
>> You're over thinking it. Rh = Rs. Rh = ~18 Ohms. therefore
>> Rs = 18 ohms.Half the power = twice the resistance.Simple. 1500W into Rh
>> and 1500W into Rs.
> 
> I'm really sorry you were not willing to try to understand the math.  Unfortunately, such simplistic talking about the problem, is not remotely the same as math.  Once you learn how to do the math, come back and I'll be happy to discuss this with you.  Meanwhile, you will find others agree with me, and not you.  Try to recognize when there is something you need to learn.
> 
Why do I need to consider all the RMS stuff? How is that relevant ? P= V x I


-- 
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Martin Brown wrote:
-----------------------------
>Phil Allison wrote: 
> > Don wrote: 
> 
> >> 
> >> This (so far unanswered) question was previously asked by me: 
> >> 
> >> Does the -3 dB half power point play a role with kW (eg not kWh)? 
> >> 
> > 
> > ** I suggested adding a series capacitor so the load voltage falls to half power. 
> > The capacitive reactance then needs to be the same as the load resistance. 
> > Same calculation as the -3dB point. 
> > C= 1/2.pi.f.R 
>
>
> A non-polar 160uF capacitor rated for peak voltage 450v 50Hz and 10A 
> continuous ripple current isn't going to be cheap and cheerful. 
> 

 **  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