120v to 100v 800+ watt step down converter

[on1]

Hey everyone,

I have a Japanese table-top heater-based appliance I want to run (100v 50/60 hz ~600W) and I live in the USA. I need a 120v -> 100v step down converter, but I am a bit concerned with the available brands I have found online.

Is anyone familiar with VCT? I have found those on Amazon, but a few of the reviews spoke about bursting into flames. Given the obvious dangers of a poorly designed converter, I am not sure I can trust this brand.

I was wondering if anyone had any recommendations? I was thinking about just rolling my own, but thats a bit more work I want to put in before Christmas time.

Thanks.

[JacquesBBB]

What is your appliance ?

It may well not matter wether you use it on 100V or 120 V. Specifications are not that tight depending on
the device.

[Richard Crowley]

You could use a 12V @ 10A transformer connected to "buck" the mains voltage down to ~100V
For example:
http://www.digikey.com/product-detail/en/VPS24-5400/237-1277-ND/666163

[Delta]

If it's purely a resistive heater with a mechanical thermostat, it might well be fine.  The heater will simple spend less time "on" to maintain the same temperature, so the average power consumed should remain the same.

[Tomorokoshi]

I've used several methods when running Japanese equipment:

1. If it's a wall plug-in that doesn't auto range, replace with an appropriate one.

2. If it has feedback of some kind, like a thermostat for heaters or uses AC motors, I might run it as is. For instance, the mochi maker. If it seems to have an auto-ranging switcher then I will also use it as is. For instance, the rice cooker. I also use a takoyaki maker directly plugged in. It may have a thermostat somewhere in it; not sure. Need to watch it to make sure it doesn't burn things too much.

3. If it seems like it will be somewhat sensitive about it, I've used one of these. Make sure the output is rated for 100V and not 110V like this one:
http://www.ebay.com/itm/AC-VOLTAGE-REGULATOR-DACTRON-AUTO-STAC-SERVO-MOTOR-ST1000W-1000VA-110V-220V-/131363725310

4. Transformers are fairly available. Not sure about the eBay ones. When running cookers and heaters these can get to be quite big.

5. If it seems like it will be quite sensitive then I use a variable voltage / frequency power converter.

6. Finally, one might have to resort to replacing elements and components to match the power.

Any of the converter methods are generally fairly large and unwieldy. For occasional use they work out, but it can be trouble if it always has to be in the kitchen or something.

If using direct 120V watch the power cord. They can get hotter than expected if they went cheap on the wire gauge.

[Ian.M]

7. If it doesn't have any electronic controls, motors or solenoids you can simply reduce the power with a dimmer rated for at least 30% more watts than the appliance.  Replace the potentiometer with a multi-turn preset, (or remove the knob and lock the shaft with glue after setting), and 'dial-in' the correct power with a true-RMS voltmeter.

[NiHaoMike]

If it's only a resistive load with plenty of headroom, you can try a power diode. You won't get the full power but it might be enough for your application.

[ConKbot]

You could use a 12V @ 10A transformer connected to "buck" the mains voltage down to ~100V
For example:
http://www.digikey.com/product-detail/en/VPS24-5400/237-1277-ND/666163

Cheap and easy solution is right here.  No emi, no electronics to burn out, No distorted waveform,  will work with any 100v load.

[on1]

Thanks for all the responses.

I am fairly certain it is a purely resistive load, basically this: http://www.amazon.com/gp/product/B00YX3NWD2

I think I will just go ahead and make a little box with a transformer

Thanks.

[Delta]

Thanks for all the responses.

I am fairly certain it is a purely resistive load, basically this: http://www.amazon.com/gp/product/B00YX3NWD2

I think I will just go ahead and make a little box with a transformer

Thanks.

Just whack that onto the 120 supply, it'll be fine.

[on1]

I read that some people tried that and the heating element burned out on theirs.

[on1]

You could use a 12V @ 10A transformer connected to "buck" the mains voltage down to ~100V
For example:
http://www.digikey.com/product-detail/en/VPS24-5400/237-1277-ND/666163

Sorry if this is a dumb question, but how would you wire that transformer up to get 100V out?

Never mind. I didn't see the tap arrangement in the datasheet.

[Richard Crowley]

You could use a 12V @ 10A transformer connected to "buck" the mains voltage down to ~100V
For example:
http://www.digikey.com/product-detail/en/VPS24-5400/237-1277-ND/666163

Sorry if this is a dumb question, but how would you wire that transformer up to get 100V out?

Never mind. I didn't see the tap arrangement in the datasheet.

You wire the primary (input) for your 115V mains voltage).
Then wire the secondary (output) windings in parallel for 12V@ 10A

THEN you take that secondary voltage and wire it IN SERIES with the mains power so that it SUBTRACTS 12V from the mains voltage.

Here is an example with slightly different voltages, but the principle is the same...



That is a rather interesting page on transformers....
https://www.ibiblio.org/kuphaldt/electricCircuits/AC/AC_9.html

[Delta]

http://sound.westhost.com/articles/buck-xfmr.htm#s30

[on1]

Thanks.

Do you have any recommended chassis/box makers for putting this thing in? I was thinking a metal hobby box and installing a 3-prong AC connector for the input with appropriate fuse and tying the chassis to earth ground. Not sure where to buy the 2-prong japanese connector though

[Richard Crowley]

I would use a box something like this....

http://www.homedepot.com/p/Carlon-6-in-x-4-in-Junction-Box-E987RR/100404096

[on1]

Thanks for all your help.

I have decided to go ahead and try the buck configuration. I want to go ahead and fuse the input to prevent the transformer from catching fire in the case the resitive load decides to become not so resistive.

In doing the calculations for the maximum current through the primary coil, would this be 800W/120V =  7A ? The maximum current rate on the secondary side of the transformer I am using is 8.8A @20V, I just rounded down to 8A max for the heck of it.

I am a bit scared that the inrush current from plugging in this simple transformer circuit would lead to a not so happy fuse, but I am not familar enough with in-rush current calculations.

Thanks again.

[Ian.M]

The max primary current for a 600W load will be about 0.85A but that subtracts from the load current so the supply current should be about 4.2A.  If you were to load it to its full 8.8A rating, it would draw about 6.2A with a primary current of about 1.3A so I would suggest a 6A fuse. You'll need a mains rated ceramic time-delay fuse similar to the ones used in Microwave ovens. 

[on1]

Could you enlighten me on that calculation?

Since the load would be about 600W (6A  at 100V) this means the secondary would need to supply 6A of current... the power drop across the secondary could would be 20V*6A  which is 120VA/120V =1A on the primary side?

When I did my original calculation I realized I only focused on the power through the load... not the secondary coil..

I am not sure I understand " load it to its full 8.8A rating, it would draw about 6.2A with a primary current of about 1.3A so I would suggest a 6A fuse.". Where is it drawing the 6.2 A?

Thank you

[Ian.M]

It draws less from the supply than it supplies to the load.   You can consider it as a funny sort of autotransformer, where insead of a tap on the winding for the load, there is another wire wound turn for turn beside the top part of the primary.  The fact its actually separate from the primary with insulation in between doesn't significantly affect the first order analysis.

The odd and non-intuitive thing about autotransformers is: when wired to step-down the current in the two parts of the winding is in opposite directions!  (If you think about it, its the only way that (neglecting losses) VA out can be equal to VA in.) 

[on1]

Right. The easiest way to calculate the primary current load is conservation of energy tho, right? I am just confused if we would want to use total power on the secondary side vs. just power though the coil. To me it makes sense that there would be greater primary-side current if there is a greater power dissipated on the secondary side, no?