There’s one in the LH footwell as well.

 

Uh Left Hand ? LOL

 

New Thought, Let me run this by yall. free to shoot it down.

EPB Elimination

Having run a test lab for 30 years Im pretty much up on electrical, not quite so much on electronics.

After seeing the new DTIC code today and the resulting out of spec measurement on the EPB motor / actuator. Im wanting to give this a try.

Reviewing the schematic for the EPBM to Actuator , there are 5 wires going to it. 2 for motor power and 3 for position (hall effect.)

The EPBM expects to see between .1 and .2 ohm in the motor circuit.

I propose to cut the RW and the GW wire between the connector and the motor. I would then solder in a 0.15 ohm, 1 watt resistor.

Circuit is fused at 30 amps for a locked rotor condition. Wiring appears to be either 16 or 14 ga. 16 handles up to 19 amp 14 up to 27 amp.
This would tell the EPBM there is a load there.
Then, verifying that the parking brakes are indeed unlocked , the hall sensor would allow a pass on the DTIC and the motor would not run but would fool the module

What say ye ?

 

The hall effect sensor's failed on mine and has failed on other members as well. Since removing the module or actuator renders the car undrivable, if what you suggest works, I reckon many others would seriously consider this idea.

 

Im going to give it a shot. take a few days , had to order the resistor

 

If you subscribe to Topix for an hour/day/week, etc, you can get access to the pinout diagrams. e.g. the attached is CV007.

https://topix.jaguar.jlrext.com/topi...cle/lookupForm

I do have them for most connectors on the 4.2's if you need any more.

 

Big Thx

 

Interesting web site. Bet its a Money maker

Do need one other Pinout Check resistance between the following points: • Pin CR50-3 and pin CV7-1 (pass value: less than 5 ohms). • Pin CR50-2 and pin CV7-6 (pass value: less than 5 ohms).

CR50 I cant find it on the Connectors file

 

Did you mean CA050?

 

Reconsider your analysis:

E = I R 0r i = E/R

Assume V = 13.5V

Given 0.15 ohms

Therefore: I = 13.5 / 0.15 = 90 A (fuse too small)

P = I E or P = E(squared) / R

Therefore P = (13.5) squared / 0.15 = 1215 watts

1 watt resistor will get very hot very quickly.

Recommendation: apply 12V directly to the actuator (very briefly) to test the actuator. Reverse polarity and try again.
If the actuator functions, use it (vs resistor) to check the Hall sensor and EPBM.

Note a 12 V portable drill battery works well for this purpose.

 

Understand your thoughts , but realize that in this circuit the Acutator is the LOAD. Since Im taking it out of the circuit there is no current to get that resistor hot. It just tells the EPBM that the circuit is live.

 

Its a quote straight from the manual.. I cant find CR50 either. I do know its one of the 2 connectors for the EPBM .

 

I can only guess its a misprint. They tell you to check continuity between those pins and the corresponding ones on CV7. Looking at the pinout you posted its gotta be that connector and the destinations that are 0.5 ohm, since there are no -2 and 3 pins, but pins 10 and 11 are correct matches.

Thx this now makes sense.

 

You state your intention to replace the actuator/load with a resistor. It appears that you intend to emulate the test/static resistance of the actuator for the resistor selection.
Since the circuit is protected by a 30A fuse, we might assume 24A design. But, for the sake of discussion, let's say 7.5A for the actuator. If 10V is delivered to the actuator under load, power dissipated (P = I E) would be 75 watts.
Your 1-watt resistor is still history.
You intend to replace a powerful actuator with a resistor that has the same initial resistance as the actuator.
Same resistance, same voltage, same current/power.
Assuming that the voltage to the actuator/resistor is 10V, then (P = I E = E square/R) your one-watt resistor must be at least (E squared / P) or 10 squared / 1 = 100 ohms

However, the resistor, at 100 ohms will see a voltage closer to full system voltage, due to the reduced current through the wire and EPBM.
So, ignore the other voltage drops and size your resistor based on 13.5 volts. Say 185 Ohms for your one-watt resistor.

Alternatively, use a very large watt resistor, consider an incandescent light bulb, a ni-chrome heater of some sort, etc.
Or you can just test the actuator separately and determine that it works and go from there.


PS: I am concerned that your post #43 will lead some unsuspecting forum member down a primrose path.
Although protected by a 30A fuse, creating a near short with a 0.15 Ohm resistor could destroy a perfectly good EPBM or other problems.

 

Wouldn’t be the first one. I had one that had me stumped for awhile.

 

Dont think so. That 30 amp fuse is there to protect from a locked rotor condition.

My rationale is , the EPBM SWITCH is right now on my car , commanding the EPB to be unlocked. Motor Resistance static is 12 OHM indicating an issue with brush or armature drag. The Only way current of any magnitude passes thru my proposed circuit change is if the PBS gets actuated to command the actuator to lock the brakes. At which time the hall sensor does not move and sets a DTIC.

Theres no possibility of a short. The .15 ohm resistor at 1 or 2 watts pulls about 2 amps. Had the resistor been in line with the motor and the motor ran then current could go way up. So I see no concern. You dont push amps through a circuit , you pull them. No demand no pull.

"let's say 7.5A for the actuator. If 10V is delivered to the actuator under load, power dissipated (P = I E) would be 75 watts." Delivered to the actuator is the key here . No actuator , no power dissipated..

Tie the RW and GW wires together for a direct short. EPBM IMMEDIATELY DETECTS Motor shorted to ground and sets a DTIC. However the EPBM has to see a specific R value to say OK. Thats all.

When this works , Im pouring concrete around the PBS so it never moves.

My $.02

 

MudDog,

I'm beginning to wonder if you are pulling my leg.

You seem to think that the wattage rating of the resistor somehow controls the current. It does not.
As you alluded, resistance and current are the primary drivers. E(lectromotive force) and R(esistance) establish the current, thus power.
The wattage rating of the resistor denotes its ability to dissipate energy in the form of heat. It does not limit power except by burning out.

 

Maybe a little

Like I said , Im fine with electrical, not as much with electronics, Gimme a reasonable solution to eliminate the actuator and not throw a code.

 

says
Im in agreement with this "Therefore: I = 13.5 / 0.15 = 90 A (fuse too small)"
.
So how is the manufacturer stopping the circuit from melt down ? 30 amp supply limit via fuse, wiring is good for 27 continuous. Motor probably pulls 20A. And manufacturer spec says 0.1 to 0.2 ohm
Go figure.

 

says
Im in agreement with this "Therefore: I = 13.5 / 0.15 = 90 A (fuse too small)"
.
So how is the manufacturer stopping the circuit from melt down ? 30 amp supply limit via fuse, wiring is good for 27 continuous. Motor probably pulls 20A. And manufacturer spec says 0.1 to 0.2 ohm
Go figure.