proper resistor to use for rear window defogger

Uh,OH, Houston, we have a problem...

Ohm's law: Voltage across resistor =Amperage thru it x Resistance in Ohms
It can be rewritten the usual ways: V=A x R or A=V/R or R=V/A

Power in Watts = V x A or A^2 x R or V^2/R

12V/220 Ohms = 0.054 A pretty close to your 0.047 A.

Old grid:
12V/3.5Ohms = 0.8A
12V x 0.8A = 9.6 W ~= 10W

Your new grid:
12V/0.8 Ohms = 15A
15A x 12V = 180 Watts, Yup, that's gonna get very hot!

To get about 10W out of that 0.8 Ohms you'll need:
A = SQRT(W/R) = SQRT(10/0.8) ~= 3.5A

Total resistance = 12V/3.5A = 3.4 Ohms

Resistor = 3.4 Ohms total - 0.8 Ohms grid = 2.6 Ohms

However, while the grid will now be dissipating 10 W, the resistor will also be dissipating 3.5^2 x 2.6 = 31.5 Watts!

You'll need a metal cased 100W power resistor to keep it's temperature down, and it'll need to be bolted to something that can absorb & dissipate the heat!

You can make it work.

IMHO, That grid line kit is a very poor match to this application. Whoever sold it to you did you a real disservice.

 

You might want to be careful assuming that the steady-state "on" resistance (at high current/high temperature) would be equal to the resistance you measure at room temperature with a multimeter (at very low current/low temperature). Maybe not as large a change as a filament bulb sees, but I'd bet that the resistance increase at high temp is noticeable (10~15%).

It's actually even worse than Verell's calculations indicated (the 12V/3.5 Ohm in the "Old grid" should be = 3.4 A and the resulting power is ~42W) so any added external resistance solution is going to be wasting an unacceptable boatload of power (136W) in the external resistor.

Any possiblity/option to rewire a few of the resistive elements in a more series configuration rather than parallel?

 

"If I left it in parellel would 2.7ohm 25 watt resistor work" = No. Per Verell's calculation method (which I think is the right way to look at it -- you want near-equal power dissipation in the heating element), you would need at least a 150W-rated 2.7 Ohm resistor to put in series with your 0.8 Ohm grid. (And this would be crazy so don't look for one!)

"I could rewire so that from the 12v terminal to the ground terminal would be one long wire" -- how many individual "lines" do you have and what is the resistance of an individual "line"? Do you have a jpeg (or sketch) of the "grid" in the configuration where you measure the 0.8 Ohms?

 

I would "rewire" the pattern so that each line goes into the next as a series circuit in a "zig zag" pattern. Start at the top and send the current through the first line, then down to the next line and back across the window, down to the next line and across, etc. This places all 4 of the .8 ohm lines in series, giving you 3.2 ohms total. You'll have a cool "zig zag" defroster pattern, no wasteful resistor getting hot, no broken glass and no excess wear on your alternator.



- +
|-----------| <---What you have now
|-----------|
|-----------|
|-----------|



+
----------------,
,---------------' <-----series connection
'---------------,
----------------'
-

If that's not hot enough, run two pairs of series connected lines in parallel. If the original wires had 3.4 ohms each and they were in parallel, then you had about .85 ohms total load for the whole circuit. Putting a pair of these new .8 ohm lines in series gives you 1.6 ohms, and putting two of these series pairs in parallel gives you .8 ohms for the entire window, which is the same as stock.

Birdman

 

OOPs! Good eye 91tr,
I really blew my arithmetic! That throws all subsequent calculations off!
Here's the corrected version:

Old grid:
12V/3.5 Ohms = 3.4A
12V x 3.4 A = 40.8W

Your new grid:
12V/0.8 Ohms = 15A
15A x 12V = 180 Watts, Yup, that's gonna get very hot!

To get about 40W out of that 0.8 Ohms you'll need:
A = SQRT(W/R) = SQRT(40/0.8) ~= 7A

Total resistance = 12V/7A = 1.7 Ohms

Resistor = 1.7 Ohms total - 0.8 Ohms grid = 0.9 Ohms ~= 1 Ohm

However, while the grid line will now be dissipating ~40 W, the resistor will also be dissipating 7^2 x 1 = 49 Watts!

You'd need a metal cased 100W power resistor to keep it's temperature down, and it'll need to be bolted to something that can absorb & dissipate the heat!

BETTER SOLUTION:
However, now that you've fully described what you've done, a much better solution may be possible: You have 4 lines, each at about 0.8 Ohms, and you want them to dissipate about 40W each.

THIS SHOULD WORK:
If you just put two of the lines in series, the resistances add which puts you at 1.6 Ohms. And 12^2/1.6 = 90 Watts for the pair. That's only ~10% higher power/line, and doesn't require an external resistor.

This could then be repeated for the 2nd pair of lines, with the same effect for that pair.

BTW, the formula for resistors in parallel is: Rtotal = 1(1/Ra + 1/Rb +1/Rc...)

Thus the total resistance of this combination of 2 pairs of 2 lines in series would be 0.8 Ohms.

THIS CIRCUIT ALSO HAS 45W/LINE BUT MIGHT BE EASIER TO CREATE:
since you've already got all 4 lines bussed together, I suspect it would be simpler to do the following:

On one side, break the connection joining the upper pair of lines to the lower pair of lines. Then connect power and ground to the upper and lower sections on that end. You end up with something like this:

+12V<=================|
GND <=================|

The wo 0.8 lines in parallel will have a combined resistance of 0.4 ohms., then two more in series with bring the resistance up to 0.8 Ohms again.

POWER SEEMS TOO HIGH:
The above circuits produce 180 Watts (4 lines at 45W/line)!!! That's a lot of power for one small section of glass. I'm comfortable with 40 or 50 W for the total circuit, that's enough so that the glass is going to be fairly warm, but this is ~4x that!

What size is the fuse for the window heater? The above circuits are going to draw 14Amps. I would expect a 30A fuse, or at least a 20A fuse. Since this is a 308, the largest fuse is going to be 16A, implying more like a 7A load.

IMHO: 4 LINES S/B IN SERIES:
If all 4 lines were in series you'd have 3.2 Ohms, and 45W total power.

 

I see that Birdman & I must have been typing at the same time. We also reached the same conclusion.

 

getting a bit late for school-boy physics here so maybe this is rubbish but, -

old window is 4 strips at 3.4 ohm gives 0.85 ohms for the window (1/R=1/r+1/r....) and a power of 169 watts (V*V/R)

new window was 4 strips at .8 ohm gives 0.2 ohm for the window and a power of 720 watts (mmmmm hot)

don't know how they expect you to wire that stuff up, but you need the old resistance of 0.8 ohms. Strangely enough if you wire it up with two lots of two wires in parallel combined in series if you see what i mean (told it was late) then you'll get 0.8 ohms and bingo no problems.

+
¦----------¦
¦----------¦

join the right hand ends together, can't get this to look right

¦----------¦
¦----------¦
-

so resistance would be 1/(1/.8+1/.8)+1/(1/.8+1/.8) =.8


might be an idea to check the sums though, you might blow the window out the back of the car

best of luck...........

 

(Edit: Obviously, the other's previously posted the same thing, but thought I'd post the figures anyway)

I think Verell and I both misunderstood your resistance values of 0.8 and 3.4 as being for the whole network with all lines in parallel rather than for an individual line. It doesn't change the conclusion that adding an external resistance (in order to keep the four 0.8 lines in parallel) is not going to work well, but it does mean that the rear defroster must be drawing a lot more current (unless the resistance changes more when hot) so if a 328 Owner with a working rear defroster could please:

1. Unplug one side and measure the DC resistance of the combined 4-wire rear defroster grid (should be fairly low like 0.85 Ohms if I'm understanding the previous description correctly), and, the bonus question,

2. Put an ammeter in series with the rear defroster and measure the total current when it's "on" (is it really 10~15A?).

If you can do kind of a combination series + parallel wiring arrangement for the four 0.8 Ohm lines as shown below ("proposed") I believe it would have a very similar total current draw and power dissipation (does require moving both connections to one side).

 

Think we've all been typing at the same time having reached the same solution

I'm off to bed..

 

180W (45W/line) ALTERNATIVE:
The 45W/line alternative is to cut 1 buss bar in the middle creating 2 buss bars with 2 lines each on that side. All 4 lines remain bussed together on the other side of the glass. 12V & ground end up on one side of the window: 12V is connected to one of the newly created 2-line buss bars, and ground to the other 2-line buss bar.

45W ALTERNATIVE:
All 4 lines are connected end to end into one long serpentine line. as shown by Jonathan's 'series connection'. It too can be created by only cutting bus bars. Again,12V & ground end up on one side of the window.

This is the alternative I'm most comfortable with. If it isn't warm enough, the 180W alternative can be created by re-connecting some buss bar sections & splitting others.

 

All you need to do is ask the people on FerrariChat.com and you get some great answers.

Good luck David

Doug Axt

 

Wow! Not only are you guys REALLY good, you ALL help at the same time.

Thanks Fchat!

 

No, that would put all 4 lines in series -- please wait, and I will post a sketch...

 

All 4 in series is what I recommended. 4 lines in series will provide about 45 Watts of heat. That should be more than enough to demist/defrost the window. With 45W, the entire rear glass is likely to be warm to the touch after a few minuites.

Still wish we had current meas'ts from an actual 3x8 defroster circuit.

 

Would anybody consider making a powersupply (switchmode?) reducing the 12V to a level where the desired wattage (10W?) is dissipated?

 

(Even if we're overestimating the true power dissipation in each case) I can't say that I follow your logic that if the stock unit puts out "X" total watts then a "0.25X" total watts configuration is therefore OK too? I actually like best what you said in your first post: "That grid line kit is a very poor match to this application", and wonder if cutting/rewiring is worth it...

Anyway, here are some possible configurations for the 180W total configuration: