Running 115v device on 230v

You are assuming that he isn't already using lard oil.

Mark Rand RTFM

collection.

JG, Kind offer but probably slightly too far off track, but many thanks anyway

AWEM

A diode won't drop it by root 2, but will drop it by 2 to 115V RMS as it doesn't create a new sine wave, but simply a sign wave with half of the cycle missing.

May be intuitively the right answer, but unfortunately wrong...see:

There's a factor of root-2 difference between the RMS voltage values of the two waveforms, not a factor of 2.

Regards, Tony

voltage,

values

....mmm... I really must dig out a load resistor, diode, Avo and 12v transformer and do a simple bench test but I keep being pestered to put up lights and find large tablecloths and tinsel!

AWEM

I should know this stuff but I admit that it's a long time since I was taught it. I'd expected that Google would reveal a formula for the RMS value of a half-wave rectified sine wave along the lines of Vpk/2SQRT2 but a quick look didn't turn anything up. The power delivered to the load is related to the area under the curve (ie the RMS value of the waveform). The area under a half wave rectified sine wave equals half that of a sine wave, therefore half the energy is delivered. If the load had negligible thermal mass, or the supply frequency was very much lower, then the heater would have a problem but because the thermal mass is large the thermal time constant is effectively infinite compared to 20mS.

Have a Good Xmas. Dave

I'm afraid it doesn't quite work like this Andrew. The voltages across the resistive heater and the capacitor are in quadrature. To get 115 volts across the heater, with a 230 volt supply, you need 199 volts across the capacitor. { from SQRT( 230^2-115^2) } requiring a reactance of 15.3 Ohms in series with the 8.8 Ohms of the heater , which at 50 Hz, requires a Capacitance of 208 MicroFarads.

A 9 ohm reactance capacitor in series with a a 9 ohm heater would actually give 230/SQRT(2) volts across the heater, and across the capacitor.

Interestingly this gives the same heating effect at the heater as the other suggestion of single wave rectification. Both suggestions give an increase in the RMS voltage at the heater of SQRT(2) (over its rating), and double the rated power dissipation in the heater.

rgds Ian

voltage,

OK I've just dug out the appropriate bits: initially a VARIAC driving a 230 -> 12v transformer to give me isolated variable low voltage supply.

Load = nominal 1K At 5v with no diode current = 5mA (not surprising) At 10v with diode is series current = 4.7mA

As the forward diode drop (0.7) is significant relative to the low voltages I threw caution to the wind and used 115v and 230v straight off the Variac:

Load = nominal 100K At 115v with no diode current = 1mA At 230v with diode in series current = 1mA

Diode used was a 1N4007 current measured with AVO8 voltage measured with Fluke

CONCLUSION:

The simple series diode does indeed work to run 115v resistive loads on 230v

Interesting debate with loads of people contributing for which I thank you - just shows, I should have done the experiment first

AWEM

It seems you are confusing mean with RMS The MEAN voltage of a half wave rectified sine wave is indeed half that of a full wave rectified sine wave.

However, the RMS value is 1/SQRT(2).

this might help .....

The "proposal" applies 2 times the voltage for half the period. Now doubling the voltage also doubles the current, therefore the power is quadrupled. So we have 4 times the power for half the period, which gives an average of

2 times the heater's power dissipation.

Ian

But the voltage 230 and 115 is already Vrms, and is not Vpp.

Nick

Now you need a big diode (and if you attach a heatsink to the tank you can use the waste heat too ;)

Do you still want the txfmr to do a 'proper job'?

Dave

driving

straight

measured

Dave,

If transport can be arranged then yes please as it really is the proper solution.

AWEM

Did you check how hot the resistor got in each case ?

(But don't touch the resistor until you switch off the supply)

suppose 115kOhm resistor.....

115v (RMS) no diode gives 1mA (RMS) current and 115mW power dissiation 230v (RMS) with diode in series gives 2mA half cycles and therefore 230mW power dissipation. (Double the current, 4 times the power, for half the time)

You will still measure approximately one milliamp because the AVO (if I remember rightly it's a moving coil meter ?) is measuring the mean DC current in this case. double the current for half the time) . The DC current would in fact be 2*SQRT(2)/PI milliamps if you neglect the voltage drop due to the diode.

Unless you're happy that your heater can survive working at double its rated power I seriously suggest you don't continue with this project (Notwithstanding the "issue" of DC current in the mains supply )

Ian

PS I think an experiment with a 47 ohms half watt resistor in conjunction with your variac and 240/12V transformer might be more illuminating.......

5 volts no diode should give you about the resistor's rated power dissipation.

7.7 volts (approx. allowing for the diode volt drop ) with the diode should also get teh resistor to the same temperature

10.7 volts and your resistor, depending on it's type may begin to show signs of darkening whilst dissipating twice its rated power

It certainly is ! (however interesting the other proposals might have been)

Ian

I stand corrected - I had allowed for quadrature but was careless with the simple arithmetic! 200uf should do nicely.

Jim

>

NO - the current is the same, but the voltage is doubled, and as power equals voltage times current, the power is also doubled.

Note, a diode halves the power to an ac resisitive load - but running at twice the voltage _quadruples_ the power, so overall doubling the voltage and using a diode doubles the power.

Two 115V 1.5kW heaters wired in parallel and run from 230V would each have

115 V across them, and would run at 1.5kW each, so overall you would have 3kW total - you could then half the power by adding a diode, giving 1.5 kW total.

~~~~~~~~ Series?

Nick

Pay attention to what Ian is saying! Avo's do not show true RMS on the AC scales, they show average* 1.4. If you can find a moving iron meter (that reads RMS) or a true RMS reading electronic meter, then you will find that your 1mA average current is 1.4mA RMS.

If the fluke voltmeter reads true RMS (not guaranteed, depends on model) then use it to measure the voltage across the resistor as an indication of the RMS current.

As an alternative experiment that is visibly more satisfying... try running a

5W 12V tail lamp bulb from half wave rectified 24V.

Have fun

Mark Rand RTFM

Problem with this is that it isn't the same mA !

The AVO8 doesn't really measure RMS current. The Fluke might, it depends on the model. This is because RMS requires that you integrate the area under the curve of the waveform. Traditional RMS meters heated a thermistor bead to measure the heating effect of the waveform. Modern meters digitise the waveform and make an approximation. Their ability to get a good approximation depends on the shape of the curve and the rate of oversampling.

The AVO is even cruder - it will rectify the AC signal, take the mean value of the result, and apply a fiddle factor such that a sine input will give the right number. For non-sinusoidal waveforms this will give a predictable result but it won't be RMS.

As some others have said, to do the test properly you need to measure the temperature rise of the heater element.

-adrian