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.
half wave rectifier
Started by ●June 29, 2023
Reply by ●June 30, 20232023-06-30
Reply by ●June 30, 20232023-06-30
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 +- Tesla referral code - https://ts.la/richard11209
Reply by ●June 30, 20232023-06-30
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
Reply by ●June 30, 20232023-06-30
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 RegulatorWouldn't it be simpler to use a temperature controlled heater and turn down the knob?
Reply by ●June 30, 20232023-06-30
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%2BHC3XSsPBAQ2MHKYNmMJAGwKPlo8TAPiEAlUCJAei4TXYwLomJkzAp6g2R3AFj8oRSegydS4HVRqmJIAAuvgAA4AFygIAAymKAE4ASwAdgBzEAAXxyp300BASEgaCweEIJHAAAIAPIACwAtpghaKJZAQABVRXysXm5AAWWwqEwAFdZdh1RqQGodfKACaSnziyUyhUq9higCeIuDTr9SDVaqAAI 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
Reply by ●June 30, 20232023-06-30
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 Regulatorhttps://www.aliexpress.us/item/2255800921813779.html> -- > This email has been checked for viruses by Avast antivirus software. > www.avast.com
Reply by ●June 30, 20232023-06-30
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 >-- This email has been checked for viruses by Avast antivirus software. www.avast.com
Reply by ●June 30, 20232023-06-30
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? -- This email has been checked for viruses by Avast antivirus software. www.avast.com
Reply by ●June 30, 20232023-06-30
>>> >> 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. -- This email has been checked for viruses by Avast antivirus software. www.avast.com
Reply by ●June 30, 20232023-06-30
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%2BHC3XSsPBAQ2MHKYNmMJAGwKPlo8TAPiEAlUCJAei4TXYwLomJkzAp6g2R3AFj8oRSegydS4HVRqmJIAAuvgAA4AFygIAAymKAE4ASwAdgBzEAAXxyp300BASEgaCweEIJHAAAIAPIACwAtpghaKJZAQABVRXysXm5AAWWwqEwAFdZdh1RqQGodfKACaSnziyUyhUq9higCeIuDTr9SDVaqAAThe 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. -- Rick C. --- Get 1,000 miles of free Supercharging --- Tesla referral code - https://ts.la/richard11209
