half wave rectifier...

[Fred Bloggs]

On Thursday, June 29, 2023 at 6:37:31 PM UTC-4, TTman wrote:

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 and can be driven by a smart switch so that it doesn\'t suck too
much out of the 9.5kWh battery or solar panels.. What diode? This is UK
spec.

When you do things like halve any of the quantities in the RMS estimate equation, including conduction time, the \'R\' part means the reduction is only 1/sqrt(2) which is about 0.7 in round numbers. The 3kW heater will therefore draw 2.1 kW from the voltage source.

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[Chris Jones]

On 30/06/2023 8:22 pm, Phil Allison wrote:

TTman wrote:
-------------------------
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 and can be driven by a smart switch so that it doesn\'t suck too
much out of the 9.5kWh battery or solar panels.. What diode? This is UK
spec.

Is the source the AC line, what you folks call \"mains\" ?

If not, the DC component could be a problem.

I have conjectured that a lot of DC current will slow down an electric
meter, disk or electronic.

Yes, AC line, 220/250V 50Hz.Mains


** Think the idea creates an illegal mains installation.
The DC component and harmonic currents are both way too large too comply with UK regulations.
Also , you may cook your grid tie inverter.

It\'s certainly very antisocial. It may greatly increase the losses in
all sorts of wound devices (the distribution transformer, any
transformers in his house, induction motors etc.)

Running a resistive water heater from a battery is pretty silly anyway -
the energy could much better be stored as heat in a bigger tank of water
than in a battery that will wear out due to the unnecessary cycling.

 

[TTman]

P = E^2 / R

If you string two heaters in series, you have the same voltage, but twice the resistance, so half the power.

No, you don\'t understand this english stuff. Twin heaters are rare in the UK

Not sure what you are talking about. You want to rig a diode to create half wave AC to drive a heater. That\'s not exactly \"standard\" practice anywhere.

Correct. Not standard. Read my original post that explains why.

If you have mechanical limitations, that\'s what you have. Electrically, this makes sense.
No mechanical limitations.

You could add a resistance in series. It would need to be about 30% of the resistance of the heater. The main element would dissipate half of the original heat. The heat in this added resistance would be about 22% of the heat dissipated in the original heating element. One nice thing about this is that you can tailor the heat in the main element to other than half the original heat level.

No chance. 3kW = ~18 Ohms. Where do I get 18 Ohms 1.5kW rating?


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[TTman]

On 30/06/2023 11:15, Phil Allison wrote:

TTman wrotr:
-------------------

Note that this rectifier method produces a DC component on the mains.
This may play havoc with transformers and motors on that same mains.

** But the OP is not using the mains.

Yes, it is a houshold mains application..


** Remember posting this?

\" 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 and can be driven by a smart switch so that it doesn\'t suck too
much out of the 9.5kWh battery or solar panels. \"

I see the words \"experiment \" , \" solar \" & \"battery\" but no sign of the word \"mains\".



....... Phil
Sorry, my big mistake

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[Clive Arthur]

On 29/06/2023 23:37, TTman wrote:

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 and can be driven by a smart switch so that it doesn\'t suck too
much out of the 9.5kWh battery or solar panels.. What diode? This is UK
spec.

A big fuck-off Variac is what you need, if you can find/borrow one.

[I did use a diode in a very similar though much lower power
application. From a magazine idea in the dark ages, I put a diode
across the contacts in a mains \'torpedo\' switch supplying my
uncontrolled soldering iron. Switch on for full power, off for standby.]

--
Cheers
Clive

 

[TTman]

On 30/06/2023 11:17, Jan Panteltje wrote:

On a sunny day (Fri, 30 Jun 2023 10:55:12 +0100) it happened TTman
kraken.sankey@gmail.com> wrote in <u7m8q1$2gqfg$4@dont-email.me>:

On 30/06/2023 00:04, John Larkin wrote:
On Thu, 29 Jun 2023 23:37:22 +0100, TTman <kraken.sankey@gmail.com
wrote:

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 and can be driven by a smart switch so that it doesn\'t suck too
much out of the 9.5kWh battery or solar panels.. What diode? This is UK
spec.

Is the source the AC line, what you folks call \"mains\" ?

If not, the DC component could be a problem.

I have conjectured that a lot of DC current will slow down an electric
meter, disk or electronic.

Yes, AC line, 220/250V 50Hz.Mains

I\'d prefer a triac over a diode to limit any DC (saturation inductors etc)
Google has many examples, here is one:
https://www.homemade-circuits.com/how-to-make-25-amp-1500-watts-heater/

I bought this... It works, but I thought a diode would be a simpler
solution...
https://tinyurl.com/4ndkfddw
AC 220V 4000W SCR Thyristor Digital Control Electronic Voltage Regulator

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[John Larkin]

On Fri, 30 Jun 2023 04:20:55 -0700 (PDT), Fred Bloggs
<bloggs.fredbloggs.fred@gmail.com> wrote:

On Thursday, June 29, 2023 at 6:37:31?PM UTC-4, TTman wrote:
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 and can be driven by a smart switch so that it doesn\'t suck too
much out of the 9.5kWh battery or solar panels.. What diode? This is UK
spec.

When you do things like halve any of the quantities in the RMS estimate equation, including conduction time, the \'R\' part means the reduction is only 1/sqrt(2) which is about 0.7 in round numbers. The 3kW heater will therefore draw 2.1 kW from the voltage source.

3 KW half the time is 1.5 KW.

 

[Ricky]

On Friday, June 30, 2023 at 7:21:01 AM UTC-4, Fred Bloggs wrote:

On Thursday, June 29, 2023 at 6:37:31 PM UTC-4, TTman wrote:
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 and can be driven by a smart switch so that it doesn\'t suck too
much out of the 9.5kWh battery or solar panels.. What diode? This is UK
spec.
When you do things like halve any of the quantities in the RMS estimate equation, including conduction time, the \'R\' part means the reduction is only 1/sqrt(2) which is about 0.7 in round numbers. The 3kW heater will therefore draw 2.1 kW from the voltage source.

What nonsense. If you time slice the power using a diode, you don\'t need to use any equation other than 1/2 P. Do you have some other image of what the OP wants to do?

--

Rick C.

+- Get 1,000 miles of free Supercharging
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[Don]

John Larkin wrote:

Fred Bloggs wrote:
TTman wrote:
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 and can be driven by a smart switch so that it doesn\'t suck too
much out of the 9.5kWh battery or solar panels.. What diode? This is UK
spec.

When you do things like halve any of the quantities in the RMS estimate
equation, including conduction time, the \'R\' part means the reduction is
only 1/sqrt (2) which is about 0.7 in round numbers. The 3kW heater will
therefore draw 2.1kW from the voltage source.


3 KW half the time is 1.5 KW.

OK. 3 kWh half the time is 1.5 kWh. Does the -3 dB half power point play
a role with kW (eg not kWh)?

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.

 

[Ricky]

On Friday, June 30, 2023 at 7:54:26 AM UTC-4, TTman wrote:

P = E^2 / R

If you string two heaters in series, you have the same voltage, but twice the resistance, so half the power.

No, you don\'t understand this english stuff. Twin heaters are rare in the UK

Not sure what you are talking about. You want to rig a diode to create half wave AC to drive a heater. That\'s not exactly \"standard\" practice anywhere.
Correct. Not standard. Read my original post that explains why.

If you have mechanical limitations, that\'s what you have. Electrically, this makes sense.
No mechanical limitations.

You could add a resistance in series. It would need to be about 30% of the resistance of the heater. The main element would dissipate half of the original heat. The heat in this added resistance would be about 22% of the heat dissipated in the original heating element. One nice thing about this is that you can tailor the heat in the main element to other than half the original heat level.

No chance. 3kW = ~18 Ohms. Where do I get 18 Ohms 1.5kW rating?

You don\'t design much electronics, do you? You need about 600 ohms at 700 watts. Six 1 ohm, 150 watt resistors will do the job.

https://www.digikey.com/en/products/filter/chassis-mount-resistors/54?s=N4IgjCBcoExaBjKAzAhgGwM4FMA0IB7KAbXAAYyA2ATgHYR8wywbqHyYYBWNxqsgMwAOdmDGUKosABZaYXuDBCYlSqOllqVdWKFd1QiXEZcytLmpO0Bp0ZRjTK%2BxvaHUIL6RePhly57600oaibrQqotQCDmSRQtLSAuwwzPLSyWDCIfgwYOHUPpxcwrE5Ata0liDclAJiyfZc0oVO1NIe1dYCssnmeSI55gJmvaryozRVMJXhAdOUfb2VCUuqc7SV1FMbQnnJhpTj%2BMNkQkKlIMPDWuxXPFVXThfDjpQDl2SOdOxv7T6sGh%2B1Gou3Y4VSYJSXHo%2BHC3XSsPBAQ2MHKYNmMJAGwKPlo8TAPiEAlUCJAei4TXYwLomJkzAp6g2R3AFj8oRSegydS4HVRqmJIAAuvgAA4AFygIAAymKAE4ASwAdgBzEAAXxyp300BASEgaCweEIJHAAAIAPIACwAtpghaKJZAQABVRXysXm5AAWWwqEwAFdZdh1RqQGodfKACaSnziyUyhUq9higCeIuDTr9SDVaqAA

--

Rick C.

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

 

[Martin Brown]

On 30/06/2023 10:55, TTman wrote:

On 30/06/2023 00:04, John Larkin wrote:
On Thu, 29 Jun 2023 23:37:22 +0100, TTman <kraken.sankey@gmail.com
wrote:

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 and can be driven by a smart switch so that it doesn\'t suck too
much out of the 9.5kWh battery or solar panels.. What diode? This is UK
spec.

Is the source the AC line, what you folks call \"mains\" ?

If not, the DC component could be a problem.

I have conjectured that a lot of DC current will slow down an electric
meter, disk or electronic.

Yes, AC line, 220/250V 50Hz.Mains

I don\'t think your inverter is going to like facing a resistive 3kW load
every alternate half cycle. Things may break and magic smoke comes out.
It is also probably against UK electrical code to do this with a high
power load.

It is one thing to run an electric blanket at tens of watts in
configurations R+R+diode for the lowest heat, R+R, R, R||R

But is quite another to ask it to supply a load that effectively
unbalances the local mains to the extent of 3kW. There are commercial
smart switches that divert solar panel electricity to the immersion
heater whenever the sun shines and the water is below temperature.

With UK feed in tariffs today this is generally worthwhile unless you
are an early adopter on a very favourable ancient tariff.

--
Martin Brown

 

[Eddy Lee]

On Friday, June 30, 2023 at 5:26:53 AM UTC-7, TTman wrote:

On 30/06/2023 11:17, Jan Panteltje wrote:
On a sunny day (Fri, 30 Jun 2023 10:55:12 +0100) it happened TTman
kraken...@gmail.com> wrote in <u7m8q1$2gqfg$4...@dont-email.me>:

On 30/06/2023 00:04, John Larkin wrote:
On Thu, 29 Jun 2023 23:37:22 +0100, TTman <kraken...@gmail.com
wrote:

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 and can be driven by a smart switch so that it doesn\'t suck too
much out of the 9.5kWh battery or solar panels.. What diode? This is UK
spec.

Is the source the AC line, what you folks call \"mains\" ?

If not, the DC component could be a problem.

I have conjectured that a lot of DC current will slow down an electric
meter, disk or electronic.

Yes, AC line, 220/250V 50Hz.Mains

I\'d prefer a triac over a diode to limit any DC (saturation inductors etc)
Google has many examples, here is one:
https://www.homemade-circuits.com/how-to-make-25-amp-1500-watts-heater/

I bought this... It works, but I thought a diode would be a simpler
solution...
https://tinyurl.com/4ndkfddw
AC 220V 4000W SCR Thyristor Digital Control Electronic Voltage Regulator

Wouldn\'t it be simpler to use a temperature controlled heater and turn down the knob?

 

[Martin Brown]

On 30/06/2023 16:39, Ricky wrote:

On Friday, June 30, 2023 at 7:54:26 AM UTC-4, TTman wrote:
P = E^2 / R

[snip]

You could add a resistance in series. It would need to be about
30% of the resistance of the heater. The main element would
dissipate half of the original heat. The heat in this added
resistance would be about 22% of the heat dissipated in the
original heating element. One nice thing about this is that you
can tailor the heat in the main element to other than half the
original heat level.

No chance. 3kW = ~18 Ohms. Where do I get 18 Ohms 1.5kW rating?

When you put the second 18R in series the current halves as does the
voltage drop across each one so that the power in each of the two
resistive loads is then 1/4 of the original 3kW. Total 1.5kW.

He will be wasting 50% of the heat though unless the second load
resistor is also an immersion heater inside the hot water tank.

You don\'t design much electronics, do you? You need about 600 ohms
at 700 watts. Six 1 ohm, 150 watt resistors will do the job.

600R would leave him with about 100W most of it in the ballast resistor
- I presume it was a typo for 6x 1R. BTW I get it as 7.5R 650W

https://www.digikey.com/en/products/filter/chassis-mount-resistors/54?s=N4IgjCBcoExaBjKAzAhgGwM4FMA0IB7KAbXAAYyA2ATgHYR8wywbqHyYYBWNxqsgMwAOdmDGUKosABZaYXuDBCYlSqOllqVdWKFd1QiXEZcytLmpO0Bp0ZRjTK%2BxvaHUIL6RePhly57600oaibrQqotQCDmSRQtLSAuwwzPLSyWDCIfgwYOHUPpxcwrE5Ata0liDclAJiyfZc0oVO1NIe1dYCssnmeSI55gJmvaryozRVMJXhAdOUfb2VCUuqc7SV1FMbQnnJhpTj%2BMNkQkKlIMPDWuxXPFVXThfDjpQDl2SOdOxv7T6sGh%2B1Gou3Y4VSYJSXHo%2BHC3XSsPBAQ2MHKYNmMJAGwKPlo8TAPiEAlUCJAei4TXYwLomJkzAp6g2R3AFj8oRSegydS4HVRqmJIAAuvgAA4AFygIAAymKAE4ASwAdgBzEAAXxyp300BASEgaCweEIJHAAAIAPIACwAtpghaKJZAQABVRXysXm5AAWWwqEwAFdZdh1RqQGodfKACaSnziyUyhUq9higCeIuDTr9SDVaqAA

I took his intention being to reduce the load presented to the mains by
a factor of two not the power dissipated in the water tank by half.

Only the OP knows what he actually wants to do but he could end up with
a very expensive mess by using a big power diode if the grid tie
inverter isn\'t very robustly built! He would be much better off with a
properly designed diverter given the lack of knowledge shown already.

--
Martin Brown

 

[Fred Bloggs]

On Friday, June 30, 2023 at 10:15:29 AM UTC-4, John Larkin wrote:

On Fri, 30 Jun 2023 04:20:55 -0700 (PDT), Fred Bloggs
bloggs.fred...@gmail.com> wrote:
On Thursday, June 29, 2023 at 6:37:31?PM UTC-4, TTman wrote:
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 and can be driven by a smart switch so that it doesn\'t suck too
much out of the 9.5kWh battery or solar panels.. What diode? This is UK
spec.

When you do things like halve any of the quantities in the RMS estimate equation, including conduction time, the \'R\' part means the reduction is only 1/sqrt(2) which is about 0.7 in round numbers. The 3kW heater will therefore draw 2.1 kW from the voltage source.
3 KW half the time is 1.5 KW.

The operative word there is \"time\". An inverter reacts to instantaneous current for its protection. Use 2100 Watts for purposes of sizing his loading.

 

[Fred Bloggs]

On Friday, June 30, 2023 at 8:26:53 AM UTC-4, TTman wrote:

On 30/06/2023 11:17, Jan Panteltje wrote:
On a sunny day (Fri, 30 Jun 2023 10:55:12 +0100) it happened TTman
kraken...@gmail.com> wrote in <u7m8q1$2gqfg$4...@dont-email.me>:

On 30/06/2023 00:04, John Larkin wrote:
On Thu, 29 Jun 2023 23:37:22 +0100, TTman <kraken...@gmail.com
wrote:

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 and can be driven by a smart switch so that it doesn\'t suck too
much out of the 9.5kWh battery or solar panels.. What diode? This is UK
spec.

Is the source the AC line, what you folks call \"mains\" ?

If not, the DC component could be a problem.

I have conjectured that a lot of DC current will slow down an electric
meter, disk or electronic.

Yes, AC line, 220/250V 50Hz.Mains

I\'d prefer a triac over a diode to limit any DC (saturation inductors etc)
Google has many examples, here is one:
https://www.homemade-circuits.com/how-to-make-25-amp-1500-watts-heater/

I bought this... It works, but I thought a diode would be a simpler
solution...
https://tinyurl.com/4ndkfddw
AC 220V 4000W SCR Thyristor Digital Control Electronic Voltage Regulator

https://www.aliexpress.us/item/2255800921813779.html

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[TTman]

On 30/06/2023 16:39, Ricky wrote:

On Friday, June 30, 2023 at 7:54:26 AM UTC-4, TTman wrote:
P = E^2 / R

If you string two heaters in series, you have the same voltage, but twice the resistance, so half the power.

No, you don\'t understand this english stuff. Twin heaters are rare in the UK

Not sure what you are talking about. You want to rig a diode to create half wave AC to drive a heater. That\'s not exactly \"standard\" practice anywhere.
Correct. Not standard. Read my original post that explains why.

If you have mechanical limitations, that\'s what you have. Electrically, this makes sense.
No mechanical limitations.

You could add a resistance in series. It would need to be about 30% of the resistance of the heater. The main element would dissipate half of the original heat. The heat in this added resistance would be about 22% of the heat dissipated in the original heating element. One nice thing about this is that you can tailor the heat in the main element to other than half the original heat level.

No chance. 3kW = ~18 Ohms. Where do I get 18 Ohms 1.5kW rating?

You don\'t design much electronics, do you? You need about 600 ohms at 700 watts. Six 1 ohm, 150 watt resistors will do the job.

I used to. Back in the day, 250V @3kW = 12Amps. Also R = V/I so 250/12 =
20 Ohms ish, give or take a bit. So, back in the day, to halve the
power, one had to double the resistance. Still with me ? So double 20 =
40. Correct? So what I need is a 20 ohm resistor capable of dissipating
half the power.i.e. 1500 watts in series with the 3kW heater. 1500 watts
will go in heat into the resistor, the other 1500 watts in the 3kW
heater element will go to heat the water.
Have things changed so much in modern technology that the basic
equations have been repaced with garbage ? P=ZxI and V=I/R back in the day.
So what are these new equations that tell me I need 600 Ohms at 700 watts ?

https://www.digikey.com/en/products/filter/chassis-mount-resistors/54?s=N4IgjCBcoExaBjKAzAhgGwM4FMA0IB7KAbXAAYyA2ATgHYR8wywbqHyYYBWNxqsgMwAOdmDGUKosABZaYXuDBCYlSqOllqVdWKFd1QiXEZcytLmpO0Bp0ZRjTK%2BxvaHUIL6RePhly57600oaibrQqotQCDmSRQtLSAuwwzPLSyWDCIfgwYOHUPpxcwrE5Ata0liDclAJiyfZc0oVO1NIe1dYCssnmeSI55gJmvaryozRVMJXhAdOUfb2VCUuqc7SV1FMbQnnJhpTj%2BMNkQkKlIMPDWuxXPFVXThfDjpQDl2SOdOxv7T6sGh%2B1Gou3Y4VSYJSXHo%2BHC3XSsPBAQ2MHKYNmMJAGwKPlo8TAPiEAlUCJAei4TXYwLomJkzAp6g2R3AFj8oRSegydS4HVRqmJIAAuvgAA4AFygIAAymKAE4ASwAdgBzEAAXxyp300BASEgaCweEIJHAAAIAPIACwAtpghaKJZAQABVRXysXm5AAWWwqEwAFdZdh1RqQGodfKACaSnziyUyhUq9higCeIuDTr9SDVaqAA

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[TTman]

OdmDGUKosABZaYXuDBCYlSqOllqVdWKFd1QiXEZcytLmpO0Bp0ZRjTK%2BxvaHUIL6RePhly57600oaibrQqotQCDmSRQtLSAuwwzPLSyWDCIfgwYOHUPpxcwrE5Ata0liDclAJiyfZc0oVO1NIe1dYCssnmeSI55gJmvaryozRVMJXhAdOUfb2VCUuqc7SV1FMbQnnJhpTj%2BMNkQkKlIMPDWuxXPFVXThfDjpQDl2SOdOxv7T6sGh%2B1Gou3Y4VSYJSXHo%2BHC3XSsPBAQ2MHKYNmMJAGwKPlo8TAPiEAlUCJAei4TXYwLomJkzAp6g2R3AFj8oRSegydS4HVRqmJIAAuvgAA4AFygIAAymKAE4ASwAdgBzEAAXxyp300BASEgaCweEIJHAAAIAPIACwAtpghaKJZAQABVRXysXm5AAWWwqEwAFdZdh1RqQGodfKACaSnziyUyhUq9higCeIuDTr9SDVaqAA

I took his intention being to reduce the load presented to the mains by
a factor of two not the power dissipated in the water tank by half.

Correct. So that the power taken from the grid, or solar or battery will
end up at around 1500 Watts.

Only the OP knows what he actually wants to do but he could end up with
a very expensive mess by using a big power diode if the grid tie
inverter isn\'t very robustly built! He would be much better off with a
properly designed diverter given the lack of knowledge shown already.

I know a bit about electronics, in fact a lot in one area but not mains
power stuff. My ignorance wants to know why adding an appropriate series
diode would likely damage the inverter.? Is it because the load
effectively becomes discontinuous?



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[TTman]

Yes, AC line, 220/250V 50Hz.Mains

I don\'t think your inverter is going to like facing a resistive 3kW load
every alternate half cycle. Things may break and magic smoke comes out.
It is also probably against UK electrical code to do this with a high
power load.

It is one thing to run an electric blanket at tens of watts in
configurations R+R+diode for the lowest heat, R+R, R, R||R

But is quite another to ask it to supply a load that effectively
unbalances the local mains to the extent of 3kW. There are commercial
smart switches that divert solar panel electricity to the immersion
heater whenever the sun shines and the water is below temperature.

With UK feed in tariffs today this is generally worthwhile unless you
are an early adopter on a very favourable ancient tariff.

I have one of those- I-Boost.
My mad professor brain was thinking of a cheap alternative. By all
accounts, ill founded.Thx.

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[Ricky]

On Friday, June 30, 2023 at 5:31:50 PM UTC-4, TTman wrote:

On 30/06/2023 16:39, Ricky wrote:
On Friday, June 30, 2023 at 7:54:26 AM UTC-4, TTman wrote:
P = E^2 / R

If you string two heaters in series, you have the same voltage, but twice the resistance, so half the power.

No, you don\'t understand this english stuff. Twin heaters are rare in the UK

Not sure what you are talking about. You want to rig a diode to create half wave AC to drive a heater. That\'s not exactly \"standard\" practice anywhere.
Correct. Not standard. Read my original post that explains why.

If you have mechanical limitations, that\'s what you have. Electrically, this makes sense.
No mechanical limitations.

You could add a resistance in series. It would need to be about 30% of the resistance of the heater. The main element would dissipate half of the original heat. The heat in this added resistance would be about 22% of the heat dissipated in the original heating element. One nice thing about this is that you can tailor the heat in the main element to other than half the original heat level.

No chance. 3kW = ~18 Ohms. Where do I get 18 Ohms 1.5kW rating?

You don\'t design much electronics, do you? You need about 600 ohms at 700 watts. Six 1 ohm, 150 watt resistors will do the job.
I used to. Back in the day, 250V @3kW = 12Amps. Also R = V/I so 250/12 =
20 Ohms ish, give or take a bit. So, back in the day, to halve the
power, one had to double the resistance. Still with me ? So double 20 =
40. Correct? So what I need is a 20 ohm resistor capable of dissipating
half the power.i.e. 1500 watts in series with the 3kW heater. 1500 watts
will go in heat into the resistor, the other 1500 watts in the 3kW
heater element will go to heat the water.
Have things changed so much in modern technology that the basic
equations have been repaced with garbage ? P=ZxI and V=I/R back in the day.
So what are these new equations that tell me I need 600 Ohms at 700 watts ?


https://www.digikey.com/en/products/filter/chassis-mount-resistors/54?s=N4IgjCBcoExaBjKAzAhgGwM4FMA0IB7KAbXAAYyA2ATgHYR8wywbqHyYYBWNxqsgMwAOdmDGUKosABZaYXuDBCYlSqOllqVdWKFd1QiXEZcytLmpO0Bp0ZRjTK%2BxvaHUIL6RePhly57600oaibrQqotQCDmSRQtLSAuwwzPLSyWDCIfgwYOHUPpxcwrE5Ata0liDclAJiyfZc0oVO1NIe1dYCssnmeSI55gJmvaryozRVMJXhAdOUfb2VCUuqc7SV1FMbQnnJhpTj%2BMNkQkKlIMPDWuxXPFVXThfDjpQDl2SOdOxv7T6sGh%2B1Gou3Y4VSYJSXHo%2BHC3XSsPBAQ2MHKYNmMJAGwKPlo8TAPiEAlUCJAei4TXYwLomJkzAp6g2R3AFj8oRSegydS4HVRqmJIAAuvgAA4AFygIAAymKAE4ASwAdgBzEAAXxyp300BASEgaCweEIJHAAAIAPIACwAtpghaKJZAQABVRXysXm5AAWWwqEwAFdZdh1RqQGodfKACaSnziyUyhUq9higCeIuDTr9SDVaqAA

The equations are all the same. But you have to apply them correctly. Maybe I don\'t understand your setup, but my mental image is a resistance heater built into a pot of some sort. You want to heat the pot at half power. Your calculations are figuring the power in the entire load. Only the power in the pot heater is useful. The power in the added resistors is waste heat, no?

My calculations give you half power in the pot heater. Please remember that the added resistor will cut both the voltage and the current to the pot heater. If you are trying to do something else, I have no understanding of it at this point.

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Rick C.

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[Fred Bloggs]

On Friday, June 30, 2023 at 8:26:53 AM UTC-4, TTman wrote:

On 30/06/2023 11:17, Jan Panteltje wrote:
On a sunny day (Fri, 30 Jun 2023 10:55:12 +0100) it happened TTman
kraken...@gmail.com> wrote in <u7m8q1$2gqfg$4...@dont-email.me>:

On 30/06/2023 00:04, John Larkin wrote:
On Thu, 29 Jun 2023 23:37:22 +0100, TTman <kraken...@gmail.com
wrote:

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 and can be driven by a smart switch so that it doesn\'t suck too
much out of the 9.5kWh battery or solar panels.. What diode? This is UK
spec.

Is the source the AC line, what you folks call \"mains\" ?

If not, the DC component could be a problem.

I have conjectured that a lot of DC current will slow down an electric
meter, disk or electronic.

Yes, AC line, 220/250V 50Hz.Mains

I\'d prefer a triac over a diode to limit any DC (saturation inductors etc)
Google has many examples, here is one:
https://www.homemade-circuits.com/how-to-make-25-amp-1500-watts-heater/

I bought this... It works, but I thought a diode would be a simpler
solution...
https://tinyurl.com/4ndkfddw
AC 220V 4000W SCR Thyristor Digital Control Electronic Voltage Regulator

That is probably best because it eliminates the DC bias. Looks like the input is entered as a percentage between 0-100. Enter 70% for half power. I doubt it\'s any kind of real regulator, but it doesn\'t need to be for your use.. It\'s going to put harmonics and RFI on the line, but your use is so brief, it shouldn\'t matter. You have the same problem with the diode. You can wire 220V receptacle to discrete wire breakout that is permanently wired into the lugs of the regulator. Packaging is not good. You really can\'t enclose it anything, and it doesn\'t look very spill proof.

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