possible 1N007 fault

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Moffy4

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Hi guys , Im new here . thanks to all for being here to ask.

Ive got a pair of simple pedicure warming boots not functioning properly. 240v 3pin plug. only a couple of years old.
They barely get warm on both settings anymore, although, do get slightly warm on both settings.
The boots are both functioning at the same heat level. and are wired to the controller.
They are currently drawing 40w on pos 1 and 60w on pos 2

The controller is much the same as an electric blanket... off , pos 1 and pos 2.
Ive opened it. it seems very simple. just an led with resistor, a glass fuse and what looks like a diode between pos 1 and 2 on the switching circuit.

Im just a newbie with limited basic knowledge.
Am i right in thinking this diode could be the culprit?

Now heres my real difficulty:
If i could get a 1N007 ( im reading that off the component ...its definately not saying 1N4007 ), I would cheerfully swap it over to try it.
However I can find no trace of this component anywhere online.
Anyone got any clues please?

Cheers
Moffy
 
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A 1N4007 would be perfectly fine, and is probably what it actually is, regardless of mis-labeling

Presumably the diode is simply switched IN circuit for low power (only using one half of the mains) and switched OUT (actually shorted out) to provide full power, using both halves of the mains.

So the only two possible failure modes are:

1) Diode S/C - full power regardless of switch position.

2) Diode O/C - no power at all, regardless of switch position.

So EXTREMELY unlikely to be a faulty diode.

As it takes 40W and 60W, that shows that the diode is working as well, as you get different powers accordingly.

I would imagine the most likely problem is the heating element, probably made of a number of elements in parallel, and some of those have failed.

You mention electric blankets- what do you do with faulty electric blankets? - you throw them away
 
If it were simply a diode being switched in/out of circuit you would expect one power to be half the other. 40W is not half of 60W so perhaps the 1N007 is something other than a simple diode?
 
Thanks all so much for responding.
Nigel, I think ill take your advice and ditch them!

I really hate giving up on fixing something tho !!
cheers again
moffy
 
alec_t said:
If it were simply a diode being switched in/out of circuit you would expect one power to be half the other. 40W is not half of 60W so perhaps the 1N007 is something other than a simple diode?
Click to expand...

That's not the way it works - adding a series diode reduces power, but not by half.

The older members may remember the use of a half wave rectifier with old soldering irons, along with a micro-switch. You arranged it so when you put the soldering iron down (in a stand or whatever) the switch was triggered and put the diode in circuit, thus reducing the power, and when you licked it up the switch shorted the diode out. This increased bit life on non-temperature controlled irons.
 
Moffy4 said:
... I'll take your advice and ditch them!

I really hate giving up on fixing something tho !!
cheers again
moffy
Click to expand...

better they are in the trash heap than turning your toes into little rosted sausages (in case your repair proved to be "overly successful"..
 
gophert said:
better they are in the trash heap than turning your toes into little rosted sausages (in case your repair proved to be "overly successful"..
Click to expand...

Going back a number of decades, people used to bring electric blankets in to work for testing and servicing (the instructions state this should be done at frequent intervals for serious safety reasons) - we used to have to send them off to the manufacturers, and as far as I can recall every single one had the same result. They simply sent a letter back saying it's not worth repairing, and offering a discount on a new blanket.

Personally I wouldn't entertain the idea of an electric blanket, it seems such a seriously dangerous idea.

At least battery powered boots sound a bit safer - no 240V mains round you while lying in bed.
 
Nigel Goodwin said:
Personally I wouldn't entertain the idea of an electric blanket, it seems such a seriously dangerous idea.

At least battery powered boots sound a bit safer - no 240V mains round you while lying in bed.
Click to expand...

the newer lithium batteries can dump tens of amps and cause serious burns. I would be very concerned about a high-amp-hour lithium battery in my pocket than an AC powered "boot" for a pedicure.

new generation heater wire inks with temp-dependent resistance are arriving on the market (Already or very soon). These inks allow an inherently safe control system to avoid burning the wearer Of a heated coat, boot or glove - the resistance goes up dramatically when the printed conductor reaches the designed temperature.
 
alec_t said:
If it were simply a diode being switched in/out of circuit you would expect one power to be half the other. 40W is not half of 60W so perhaps the 1N007 is something other than a simple diode?
Click to expand...

Does the part look like the device shown in the link below, where it actually is just a resistor with only a partial part number stamped on it? A photo would better identify the part.

RHXH1N007R5F4 Johanson Dielectrics | Mouser

RHXH1N007R5F4 Johanson Dielectrics Thick Film Resistors - Through Hole 7.5Ohms 35W 1% datasheet, inventory, & pricing.
www.mouser.com
 
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So maybe the 1N4007 rectifier diode has gone 'kaput', and changed into a leaky diode, and now is a resister of unknown value. But I agree with Nigel that most likely a set of the heating elements may have failed. Do they heat properly on the high setting?
 
Nigel Goodwin said:
That's not the way it works - adding a series diode reduces power, but not by half.
Click to expand...
I don't understand why this is so (assuming perfect diode). Is it because the reduced power increases the resistance or something else?

Mike.
 
Basically because a semiconductor diode is not a linear device, and the foreseen 'resistance value' changes per the frequency and the amount of current flowing thru the device; since we are dealing with 240v AC, and not DC. Besides the fact it is only passing 50% of the sine wave, producing pulsed DC.
 
Pommie said:
I don't understand why this is so (assuming perfect diode). Is it because the reduced power increases the resistance or something else?
Click to expand...

Because ohms law etc. tells you so

240V mains, 240 ohm load - power = 240W

Now stick a half wave rectifier in:

120V mains, 240 ohm load - power = 60W
 
But it isn't 120V mains, it's still 240V mains but only switched on half the time and the current in the (on) half cycle will be the same. Into a resistive load I can't see why this isn't half power.

MTJB, I stated "assuming perfect diode".

Mike.
 
Pommie said:
But it isn't 120V mains, it's still 240V mains but only switched on half the time and the current in the (on) half cycle will be the same. Into a resistive load I can't see why this isn't half power.
Click to expand...

240V is 240V RMS - adding the diode makes it 120V RMS (only using alternate half cycles), and halving the voltage quarters the power.

You're confusing volts and watts.
 
The half cycle is identical to the full cycle 240V (only a half cycle) so the instantaneous current will be the same - so won't the instantaneous power be the same for the half cycle it's on and zero for the other half cycle? Sorry to harp on but I can't see how this works.

Mike.
 
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