Using a diode to lower the heat output of a food dehydrator

[paternoster2012]

I recently purchased a Nesco food dehydrator.
When plugged into the wall, it powers a fan and a heating element. Together they blow warm arm over food to dry it out.

The top of the machine has a 'thermostat' with temperature settings from 32°C (90°F) to 68°C (155°F).
However, when the machine is actually in use, the temperature as read with a thermometer is 75°C (167°F).
Turning the 'thermostat' has little to no effect on the temperature.

This temperature is much to hot to dehydrate food, and actually cooks it. So I needed to lower the heat.

The first thing I did was to install a power switch on the heating element. This gave me the choice between too hot or too cold.

So I pulled the guts out of it and this is what I came up with in its current state. (I added the power switch)
**broken link removed**

The fan is working just fine and I don't intend to mess it.
The heating element is what I want to modify.

The 'thermostat' appears to be a potentiometer, but when I try to get a resistance reading I get a reading of .8 to 1.0 ohms no matter what position the dial is in.

The element is 112 Ω, which at 230 VAC is 478 watts, plus whatever the DC motor pulls. The dehydrator is rated at 230 VAC and 500 watts, so I assume everything is as it should be aside from the potentiometer.

Now here is my question I would like to pose.

My country is 220 VAC at Hz, but my wall socket reads 230 VAC.

I was wondering if I could use a toggle switch with a diode to halve the voltage flowing to the element. 115 V at 112 Ω is 118 watts.
I'd like to believe this would give me a nice warm air without being too hot.

Here is my proposed diagram
**broken link removed**

I was thinking of bypassing the potentiometer all together seeing as it doesn't seem to be working very well (cheap production quality).
I can't really replace it, as the dehydrator is kind of built around it.

118 watts isn't much, but I don't exactly need much either to warm some air a bit above ambient temperature.


Am I on the right track here? Is there perhaps another option I am over looking?

Thanks for your help

 

[MikeMl]

If the thing you call a "potentiometer" is really a just a variable resistor, it would have to be capable of dissipating about 110W, and would look like this:

More likely, the thing you call a "potentiometer" is really an SCR or TRIAC dimmer that has failed shorted.

How about just getting a $10 incandescent lamp dimmer at the home store, and wiring it to control the heater. That way, you can "tune" the power needed to dry the food...

Putting in a suitable rectifier would drop the power in the heater from ~460W to ~230W, which might still be too hot...

 

[paternoster2012]

Yes, it is probably a triac or a rheostat. I got lost looking for the correct item in the circuit diagram program. The device itself is surrounded by molded plastic so it's hard to see exactly what it is, I just know it doesn't work very well.

The problem with the dimmer switch is it would be hard to install it in the body of the dehydrator. I could replace the power cord with a lamp dimmer cord, but I'm not sure how that would affect the 12v transformer driving the fan motor.

I have a spare dimmer cable lying around I could use, but I'd kind of like to maintain the appearance of the factory dehydrator.

 

[MaxHeadRoom78]

The 1n4007 is rated at 1amp so this would be a little underated.
Best to go with a 5a version.
Max.

 

[JimB]

The "potentiometer" may be something called a "SimmerStat".
I have not seen one of those things for years. It has a heating element (low resistance) connected in series with the thing which it is controlling. The heating element heats a bi-metal strip which bends and opens a pair of contacts, breaking the circuit through its own heater and the thing it is controlling.

As the bi-metal strip cools, the contacts close and the heating cycle starts again.

Simple (crude ?) on/off control.

JimB

 

[AnalogKid]

That's what I think, it's an adjustable mechanical thermostat whose contacts have welded themselves together. Turning the knob pushes the two strips apart, so the bendable one has to bend farther to make contact, requiring higher heat. Only a few companies make them, but they pump them out by the millions.

A 6A6 diode is a common part in home appliances.

ak

 

[chemelec]

Why Not run Two Wires Off to a External Dimmer?
Or Put it in Place of your Power Switch?

 

[paternoster2012]

Here is how the the thing looks, and the appearance I would like to try to maintain
**broken link removed**

and with the cover off, you can see the molded in 'thermostat' or whatever it is.
**broken link removed**

The dimmer cable could work, but honestly I don't feel I need much temperature control. Just a lower temp than 75 C.


So If I put a 5A diode in series with the element it would drop the element to 118 watts? (115 V x 112 Ω) I'm not sure what sort of temperatures it would provide me with, but it must be better than what I'm working with now. I would probably bypass the 'thermostat' altogether going this route as it seems to be malfunctioning.

*note* I don't play with electricity on that stainless steel table, I just dissembled it there

 

[MikeMl]

... Just a lower temp than 75 C.
So If I put a 5A diode in series with the element it would drop the element to 118 watts? (115 V x 112 Ω) ...

Not quite:

 

[AnalogKid]

That looks like a standard shaded-pole AC motor running straight off the power line. What makes you think that that is a 12 V fan motor? Where is the transformer?

ak

 

[chemelec]

Yes it is a 220 VAC Shaded pole Motor.

So your DIODE or a Light Dimmer Must NOT Connect to the Motor.

 

[paternoster2012]

Thanks for the replies so far.
That motor is indeed a shaded pole motor (I'm not well versed in electronic motors and didn't pay much attention to it as I don't want to alter the fan in anyway)

My mains is 230 VAC, so I assume the motor is 230 V as it is connected directly to the power line.

I am just looking for the best way to lower the heat output of the element so the machine is somewhere between room temperature and 75 C

 

[chemelec]

Are you SURE your Dehydrator is Rated for 220 VAC? (Not 110 VAC)

If so, than In MY OPINION, The BEST Way would be to Connect a Dimmer Control in either the Black or White Wire going to the Heater Element.

It would than Give you FULL Temperature Control.
The Dimmer can be Mounted in a Separate Box, OUTSIDE of the Unit, so as to Not really Changing it.
Just a Small Hole to bring the Wires through.

 

[AnalogKid]

The "right" way is to resurrect the thermostat you already have. Separate from that, a series diode will decrease the heater power by approx. 75%. No idea how that will translate into chamber temperature, other than it will be lower.

ak

 

[paternoster2012]

Yes, I am positive it is rated for 230 VAC. I purchased it here, the sticker says it is rated for 230 VAC, the box as well. I also had it plugged in for hours and it didn't self destruct.
It is just too hot at the moment. I don't need temperature control. The thing cost 30 USD, so I'm not looking to invest a lot of time effort or money on it.
If I can add a diode to it and have it function safely, at a lower temperature, I would be happy.

I'm heading to the electronics market tomorrow to get some other stuff, and would like to pick up something to have this thing function with warm.

 

[paternoster2012]

The "right" way is to resurrect the thermostat you already have. Separate from that, a series diode will decrease the heater power by approx. 75%. No idea how that will translate into chamber temperature, other than it will be lower.

ak

The thermostat is probably cheap junk, which is why it is broken in the first place. I'm almost positive its a chinese knockoff of a nesco machine, but it's all that is available over here in SE asia.

75% less than too hot is still probably warmer than room temperature. So I'm leaning toward the series diode. 6A6-T would be the part number?

 

[AnalogKid]

Only you have see the inside of this thing and know the space you have to work with. IIR the 6A6 comes in a fat axial package and a 2-lead TO-220. The 220 can move more heat if it is bolted down to a metal surface, but the tab might not be isolated from both leads.

ak

 

[MikeMl]

I have a (USA) NESCO FD-50 Food Dehydrator, rated at 120Vac 500W. It has a built-in thermostat, settable in degrees F. The nominal setting appears to be ~130deg F; that is where we set it for drying food.

I plugged it into my Kill-aWatt. During initial warm up, the FD-50 draws 485W. After about 5 min, the thermostat began cycling, and after 40 min, it settled to a heater on/off duty cycle of about 9sec on, 19 sec off. The power input to run just the fan motor is 24W. The power input to run both the motor and the heater is 485W.

The calculated average power required to keep the exhaust air at 130degF (measured with an accurate lab thermometer) is (9s*485W + 19s*24W)/(9s+19s) = 172W. The inflow air was 65degF.

I did not have any food loaded in the FD-50. I'm guessing that if I did, the initial warm-up time would be longer (to heat up the mass of the food). However, since loading the trays would reduce the net air flow through the box, the power required to maintain the outflow air at 130degF would be less than I measured, say ~150W.

So, my conclusion is that simply adding the diode will not sufficiently reduce the power input. If you go back and look at my post #9, adding the diode reduces the initial heater power by only 40%, leaving a power input of over 280W with the diode in place.

Looks like you either fix the thermostat or add an outboard dimmer, like the 230Vac version of this one (the linked one is for 120Vac).


ps: The Kill-a-Watt reports 370Wh in 2h 5min, or 370Wh/2.083h = 177W average power input. This is the steady-state measurement, beyond the initial warm-up.

 

[jjw]

Not quite:

View attachment 98176

View attachment 98177

Should'nt the averaging interval be multiple of the 50Hz cycles, 20ms, 40ms, 60ms ... , not 50ms.

 

[MikeMl]

Should'nt the averaging interval be multiple of the 50Hz cycles, 20ms, 40ms, 60ms ... , not 50ms.

Good catch! I repeated the LTSpice runs with an integration interval of 60ms.

Without the diode, at a line frequency of 50Hz, the power pulses in the heater occur at twice the line frequency = 100Hz, so an integration interval of (1/100Hz) = 0.01s = 10ms or integer multiples of 10ms is correct. The average power in the heater of 472.18W is the same if the integration interval is 50ms or 60ms.

For the run with the diode, the power pulses in the heater are occurring at the line frequency of 50Hz, so the integration interval must be (1/50Hz) = 0.02s = 20ms, or integer multiples. Using 60ms as the interval, the power in the heater drops to 234.46W, which is slightly less than half of the of the power without the diode because of the dissipation of the diode itself...

Back in post #2, I said:

Putting in a suitable rectifier would drop the power in the heater from ~460W to ~230W, which might still be too hot...

so I expected putting the diode in would halve the power. In posting the LTSpice runs, I was trying to show the TS that his statement:

... Just a lower temp than 75 C.
So If I put a 5A diode in series with the element it would drop the element to 118 watts? (115 V x 112 Ω) ...

was not correct. I should have followed my intuition that the power would be halved....

Based on the measurements of post #18, just adding a diode would still make the food dryer too hot.