FIXED: Power Designs 2005A repair

[motocoder]

Would not trust that tape as insulation, it tears too easily. The kapton tape will do for that.

Ok, I found some wrap that is used for insulating exhaust pipes. I have that and the two types of tape ordered from Amazon.com. Should be here on Wednesday.

BTW - if it turns out that the SCR is bad, do you have any suggestions on a replacement? The local electronics store has an NTE5404, which looks like it would fit the bill.
(Link to NTE 5404 datasheet: http://www.nteinc.com/specs/5400to5499/pdf/nte5400_06.pdf)

Thanks again for your help.

[SeanB]

Yes it will work, though I would use a C106D there as it will survive transients better. Just remember this heater is at mains voltage, so needs to have good insulation. Your heater resistance will have to be around 1kOhm or higher or it will heat up too fast and burn out the reference diode. Your length of resistance wire has to be long enough and thin enough so that the length wound on the inner is both covering it fully yet is insulated from shorting out either turn to turn or to the casing, and that the insulation is capable of handling 120v in operation while hot.

Myself I would modify the oven to use the 26VAC secondary and this would reduce the requirements that the heater must withstand AC mains standoff of 400V for a minute ( standard AC insulation test voltage), and then replace the pilot lamp with a LED, and replace the SCR with a C106D to handle the increased current.  Resistance will then only need to be around 60R, easy to wind with bare wire.

[robrenz]

Some info on a 2005 heater repair starting here in this thread

[motocoder]

Yes it will work, though I would use a C106D there as it will survive transients better. Just remember this heater is at mains voltage, so needs to have good insulation. Your heater resistance will have to be around 1kOhm or higher or it will heat up too fast and burn out the reference diode. Your length of resistance wire has to be long enough and thin enough so that the length wound on the inner is both covering it fully yet is insulated from shorting out either turn to turn or to the casing, and that the insulation is capable of handling 120v in operation while hot.

Myself I would modify the oven to use the 26VAC secondary and this would reduce the requirements that the heater must withstand AC mains standoff of 400V for a minute ( standard AC insulation test voltage), and then replace the pilot lamp with a LED, and replace the SCR with a C106D to handle the increased current.  Resistance will then only need to be around 60R, easy to wind with bare wire.

Ok, now this sounds like a lot more funthan winding 30 feet of fine gauge wire without having shorts between coils. I think I will look into this.

[motocoder]

Yes it will work, though I would use a C106D there as it will survive transients better. Just remember this heater is at mains voltage, so needs to have good insulation. Your heater resistance will have to be around 1kOhm or higher or it will heat up too fast and burn out the reference diode. Your length of resistance wire has to be long enough and thin enough so that the length wound on the inner is both covering it fully yet is insulated from shorting out either turn to turn or to the casing, and that the insulation is capable of handling 120v in operation while hot.

Myself I would modify the oven to use the 26VAC secondary and this would reduce the requirements that the heater must withstand AC mains standoff of 400V for a minute ( standard AC insulation test voltage), and then replace the pilot lamp with a LED, and replace the SCR with a C106D to handle the increased current.  Resistance will then only need to be around 60R, easy to wind with bare wire.

SeanB - I assume you are referring to using transformer secondary taps 3 & 4 (or 4 & 5), which give about 26VAC (RMS).

I took some resistance measurements on the old heater wire, and it matches almost exactly what robrenz posted in the other thread - 674 ohms.

Assuming my LTSpice model is vaguely close to reality, here are my calculations on heater power with the original design, and what would be required for an equivalent design using the secondary 26VAC output:

With the original design, peak heater current is 249mA, with corresponding peak instantaneous power of around 42W. Now this is a half wave rectified sine wave, and due to turn-on time of the SCR, not quite that, so if you check average power in LTSpice it is about 10 watts.

The simulation shows that the heater control circuit should still work just fine with the lower input voltage. To achieve a similar average power value with the 26VAC off transformer taps 3 & 4, we need to scale the current by 120/26, or about 1.1A, and a heater resistance of 30 ohms. Do you think this transformer can source that much current in addition to the current it is alreay sourcing for the load? Is there some way to test that? Also, would you then add a fuse to this circuit, or is the fuse on the primary sufficient?

Some info on a 2005 heater repair starting here in this thread

robrenz - Are you referring to the guy that added a microcontroller PID loop to control the temperature? Thanks, but no thanks!

[motocoder]

I think I will fuse this circuit as the consequences of the heater coils shorting could be a burned out transformer secondary or even a fire.

For the LED mod, it seems sufficient to just replace the lamp with an LED and change the series resistor to 2k2. The LED will be flickering at 60 Hz, but I don't think that will be visible.

[robrenz]

robrenz - Are you referring to the guy that added a microcontroller PID loop to control the temperature? Thanks, but no thanks!

I only meant to see pictures of the insides and the discussion on what to use for replacement insulation, not the pid mod.

[motocoder]

robrenz - Are you referring to the guy that added a microcontroller PID loop to control the temperature? Thanks, but no thanks!

I only meant to see pictures of the insides and the discussion on what to use for replacement insulation, not the pid mod.

I see. The pictures are useful. However, I didn't see where he ever posts back with resolution on insulation.

[SeanB]

The secondary winding should handle the 20W load with little issue, as it only is going to be on until hot, then it will only be on to make up losses in insulation. I would make the heater a 20W unit, so cold resistance would be around 25R, so it will take a little longer to heat up, but will get to the same set point.

You lose the standby heating capacity, but then you just leave it on when needed. Neon replaced with red ultra bright LED means you do not need to change the resistor, it will be bright enough. To reduce the loss in the winding simply connect it so the diode CR26 ( replace with a 1N4004) connects to pin 4, and the other end to pin 3 or 5. That way it will only draw current through the winding half when it is not supplying current to the main rectifier Cr13 or CR14.

[motocoder]

The secondary winding should handle the 20W load with little issue, as it only is going to be on until hot, then it will only be on to make up losses in insulation. I would make the heater a 20W unit, so cold resistance would be around 25R, so it will take a little longer to heat up, but will get to the same set point. You lose the standby heating capacity, but then you just leave it on when needed. Neon replaced with red ultra bright LED means you do not need to change the resistor, it will be bright enough.

I am not sure where you are getting 20W. Of course, my simulation results could be wrong, but based on the measurements I took on the old heating wire, and both my simulation and a back-of-the-envelope calculation, the average power of the original heater circuit is around 10W. Now peak value is 40W, so maybe that is what you are referring to? If that's the case, you are saying run the new circuit at half power? That would require a resistance of 60 ohms, not 25. 25 ohms would actually draw more power than the original.

To reduce the loss in the winding simply connect it so the diode CR26 ( replace with a 1N4004) connects to pin 4, and the other end to pin 3 or 5. That way it will only draw current through the winding half when it is not supplying current to the main rectifier Cr13 or CR14.

That is a very clever idea!

[SeanB]

20W load is around half the original heater, so will take longer to heat up but will stress transformer less from added load. Thermostat will regulate it fine. 25R cold will increase somewhat when hot, so it will start out doing a little over 30W but as it warms up it will drop likely to around 20W just as the thermostat kicks out. As the heater is only going to have half wave AC applied it will actually draw less power, probably around 15W dropping to 10W near cut out, and will hover around that with the thermostat regulating. You will need to insulate it a little better to drop the power further, but just using glass wool and an outer cover of glass wool or even just a card shield it will run with lower loss.

[motocoder]

20W load is around half the original heater, so will take longer to heat up but will stress transformer less from added load. Thermostat will regulate it fine. 25R cold will increase somewhat when hot, so it will start out doing a little over 30W but as it warms up it will drop likely to around 20W just as the thermostat kicks out. As the heater is only going to have half wave AC applied it will actually draw less power, probably around 15W dropping to 10W near cut out, and will hover around that with the thermostat regulating. You will need to insulate it a little better to drop the power further, but just using glass wool and an outer cover of glass wool or even just a card shield it will run with lower loss.

Actually, the original heater is 40W peak. If you do the math, because it is half-wave rectified average power is 0.318 of peak, or 12.7W. And in reality, because the SCR does not turn on right away, the actual average power of the original heater is somewhat less. I estimate 10-11W average power. All that is based on the cold resistance values;I assume it goes down as the heating element heats up.

So, in any case, I will stick with my numbers, which are 30R to be equivalent to original circuit, higher than that if I want to drop the peak current and power a bit.

[motocoder]

Any idea what the connector inside the heater tube is? Further inspection shows that the inside is actually badly damaged. Looks like the heater must have shorted out and generated a lot of heat when it failed. The insides are burnt and the plastic is cracking. Several connections are no longer working.

To get access, I had to destructively remove the inner heater tube. So I will need to rebuild that part. The board itself, and it's connector is fine, so if I can find the female version of this connector, I can leave the board as-is.

[motocoder]

I found an oven for another model on eBay, and got the seller to accept a reasonable offer. I'll savage the mechanical parts from that.

[SeanB]

If the donor still works sans heater will you post the remains to me, I would like to try to fix that.

[motocoder]

If the donor still works sans heater will you post the remains to me, I would like to try to fix that.

Gladly. It is the least I can do after all the help you have provided me on this thread.

[motocoder]

If the donor still works sans heater will you post the remains to me, I would like to try to fix that.

Ok, the replacement heater unit came today. I removed the board (which is very different), and verified that the connector inside is the same, and that the internal connector connects through to external pins labeled 1-8. Pin 9, 10, and 11 have a connection to the heater wires. I suspect it is the same as the 2005A unit with pin 9 and 10 going to the heater wire, and pin 11 to one side of the thermostat. The insulation on this unit was much more restrained, and I did not see anything resembling asbestos, nor any burn marks.

Question #1 : Where is the thermostat located? I am guessing it must be down inside the area with the connector. Opening that bit is how I broke the last heater assembly, so as long as its working I'm content to leave it there - just curious.
Question #2 : How would you route the connection to the NEW heater coil from the inside of the tube to the outside of the tube? The existing connections are not accessible, unless I want to try and solder to the little stub of wires from the old heater wire. I am almost certain these will eventually break, so I'd rather not try that (they also have significant resistance probably > 20 ohms combined).

[motocoder]

Disregard question #2 above; I cut away some of the plastic insulating ring at the bottom, and drilled a small hole in the PCB on the underside. With those two things I can get access to both places I need to solder the heater wires. I'll fill them both with some high temperature epoxy potting compound once it's all confirmed working.

[motocoder]

Well, heck. I put the heater wire on there (33.4 ohms). Hooked it up to a DC power supply but with no board inside. I ran it all the way up to 70C, and then measured the resistance between pins 11 and 9, but it does not change from the cold value.

So it seems the thermostat is not working in this new oven assembly. 

[motocoder]

I have a thermistor that I salvaged from a coffee maker. I think I will just make my own thermostat board out of that, a comparator/op-amp, and a 2N7000 MOSFET. Also will need a little voltage regulator and a few other components to generate a stable 5VDC for the op-amp.

Only open question is how to get the thermistor inside the tube.

[SeanB]

Look for a PTC thermal switch, you get a 70C one which is small and which will fit inside there. Otherwise you also get really small thermal switches that will fit.

[motocoder]

I'm not aware of a device like that which can be used for thermal control. I am aware of protective devices, reset-able thermal fuses, but those have a wide temperature swing and are not suitable for this purpose.

If you know of such a device, please post a link.

[SeanB]

http://uk.farnell.com/texas-instruments/lm26cim5-sha/thermostat-trip-70c-smd-sot23-5/dp/1312675

Small, but needs a 5V supply, which can easily be derived from the heater circuitry. Will need a small amount of extra componentry to drive the heater control thyristor.

http://www.farnell.com/datasheets/1771291.pdf

Page 10 of the pdf shows using it to control a heater, and you could use it to drive the thyristor in place of the transistor using a C106D, powering it off the 26VAC circuit via a single diode, a capacitor and a 5v1 0.4W zener diode and a resistor. Capacitor would be a 220uF 63v unit, zener a 5V1 400mW one and the resistor a 1k8 1W resistor.

[motocoder]

Thanks for the link. That's basically a single-chip implementation  of the circuit I was planning on building myself.

At this point I've got the thermistor all nicely wired up inside the tube. I have all the parts I need to build this circuit already here, so I'll probably just go ahead and do that rather than waiting another week for another part.

Another option I am considering is just driving a power MOSFET from the comparator directly, which means I would no longer need the SCR or transistor in the current heater circuit.

[SeanB]

That will work, and so long as the power device is rated for 50V or more it will do.