Project AAR resto-mod, Rebuilding the AAR electrical prototype Options

[pgollender]

I am trying to expand off Dave Darling’s old posts on the Pelican and from info archived from Brad Anders pages. Many thanks to Paul Wiles who supplied me with a few dead AAR’s to experiment with.
The mechanical rebuilding process has been explained in detail including a nice piece by Ian Karr.
The problem with many of our AAR (Auxillary air valve regulators) is that the 12volt lead has broken off or the NiChrome heating element has broken off or burnt out. Depending on whether your valve is stuck open or frozen shut will affect either the warmup process or performance of your 914 after it has reached operating temperature. This is assuming that your cold start valve is also functioning properly. The insulated 12volt lead passes through the bottom of the AAR then through a thin pressed fire-retardant material and is riveted to a copper pin on the inside of the ceramic heat-sink. From the pin connection, a 13 ohm wound niChrome heating element (resistor) exits counter-clockwise to a grounding copper rivet that grounds to the bottom of the AAR. The ceramic heat sink is also riveted centrally to the bottom of the AAR. In the original design the wound NiChrome wire is epoxied to a small round kerf in two places. It is 99% impossible to remove the epoxy/wire without either destroying the wire or the round kerf mounting surface. I have included a few pictures to help look at the structures. Per Dave Darlings Pelican notes I ordered some ceramic “5 watt bathtub resistors” and will connect them in series to obtain 13 ohms. It’s been along time since I used Ohm’s law so I’m not 100% sure if I am doing this right. Anyway you have to remove the ceramic tub to have enough clearance for the metallic spring to fit.
I haven’t completed the project yet but wanted to post the project so far. I still need to secure the resistors in place and find a suitable 12volt wire insulator to pass through the narrow channel and then be able to ground it to the AAR chassis properly.

[Superhawk996]

I ordered some ceramic “5 watt bathtub resistors” and will connect them in series to obtain 13 ohms. It’s been along time since I used Ohm’s law so I’m not 100% sure if I am doing this right.

You are forgetting the power side of ohms law.

I’m going to use 13v as nominal voltage to keep the math simple

Current = voltage / resistance = 13/13 =1 amp

Power = current ^2 x resistance = 1x1x13 =13 watts total. You only have 10 watts (max) with your two 5 watt resistors.

The bigger problem is that power formula also applies for each resistor individually.

So resistor #1 - 10 ohms

Power = 1x1x10 =10 watts. Problem is your resistor is only 5 watts and that assumes it’s in free air and able to cool via air. Your 5 watts total resistor is eventually going to burn out trying to dissipate 10watts inside an enclosed canister.

The AAR is continuously powered so that only makes matters worse with a 100% duty cycle.

Resistor #2 - 3 ohms

Power = 1x1x3 =3 watts
This resistor would be ok in theory but again packing it along side the 10W coming off resistor #1 in an enclosed space is going to overheat it too.

You can redo the math at 12v nominal but you’ll see that resistor #1 is still at 8.5 watts.

If you could squeeze three 5 ohm 5 watt resistors in there the math works in theory but I suspect that it will still burn out pretty quickly due to poor heat transfer from resistors to the AAR housing and then to air. Hard to say for sure how long it might live but that would be the way to try.

[pgollender]

I ordered some ceramic “5 watt bathtub resistors” and will connect them in series to obtain 13 ohms. It’s been along time since I used Ohm’s law so I’m not 100% sure if I am doing this right.

You are forgetting the power side of ohms law.

I’m going to use 13v as nominal voltage to keep the math simple

Current = voltage / resistance = 13/13 =1 amp

Power = current ^2 x resistance = 1x1x13 =13 watts total. You only have 10 watts (max) with your two 5 watt resistors.

The bigger problem is that power formula also applies for each resistor individually.

So resistor #1 - 10 ohms

Power = 1x1x10 =10 watts. Problem is your resistor is only 5 watts and that assumes it’s in free air and able to cool via air. Your 5 watts total resistor is eventually going to burn out trying to dissipate 10watts inside an enclosed canister.

The AAR is continuously powered so that only makes matters worse with a 100% duty cycle.

Resistor #2 - 3 ohms

Power = 1x1x3 =3 watts
This resistor would be ok in theory but again packing it along side the 10W coming off resistor #1 in an enclosed space is going to overheat it too.

You can redo the math at 12v nominal but you’ll see that resistor #1 is still at 8.5 watts.

If you could squeeze three 5 ohm 5 watt resistors in there the math works in theory but I suspect that it will still burn out pretty quickly due to poor heat transfer from resistors to the AAR housing and then to air. Hard to say for sure how long it might live but that would be the way to try.

I think there is room for 3 x 5 ohm resistors in there. Also I was thinking of just casting them in high temp ceramic with a small space between the outer housing like the original design. I never understood why Bosch left this as an always on circuit. How is it that the original thin gaged ni-chrome filament can last so long in many of these 50 year old cars with the circuit always on?