Special car relay diagram question

car relay diagram

This is the relay that operates the ECU of a Golf Mk3.
Besides that diagram and the number "30." (on top of the relay), it doesn't give anymore explanation.

I've got two questions:
1. How can I find out what Amps it's rated at (normal car relay = 30A)
2. What is the practical difference between port 86 and 86A? Why is there a diode on one, and not on the other port?


WHY? => I want to use this relay for wiring a small electric motor if possible...



**broken link removed**

Thx in advance,

kn.

what does a car relay look like?

looks like a standard 25-30amp contacts type

port 86 has a diode to stop reverse flow of current
as some measure of protection

you may find that port 85 floats then this dioded will 'switch'
to protect the pins input and also to stop the contact 'chattering'
as the voltage climbs and falls at ignition time

i think this relay is aok to use as a motor switcher

the diode will be connected to ground {port 86 to gnd}

diodes like this are normaly straped accross the coil

but in this case the vehicle is always neg ground
so the relay has been setup so its easier to test it

negative relay digram

Thx for the info so far,

VSMVDD said:

the diode will be connected to ground {port 86 to gnd}

Click to expand...

I don't quite understand the last part of your post.
86 = switched positive (control of the relay),
85 to ground.

That's what I thought, and what testing with a 12V supply confirmed.

typo of yours, or am I missing something?

golf mk3 relays diagram

king nero said:

typo of yours, or am I missing something?

Click to expand...

bttt please?

what is the k in a relay diagram

My vote's for

86A=+V
85=86=-V (ground)

where the diode is a kickback diode to protect the switching circuit from the coil's current stored.

(A coil resists change to current, so at the opening of the current path, it's going to do what it can to keep current flowing in the direction it was flowing before the loss, from 86A to 85. To do ths it will produce a negative voltage spike at 86A (as 85 is at ground). This spike could look nasty to your switching device as it opens, for the high R of the openng switch will develop the high voltage spike across it. The diode, if it is used, will conduct (as its anode is at 0 potential if grounded), shunting the current away from your opening switching device. Hence, the diode kills the kickback from the collapsing coil.)

As far as identifying the current rating, try Audi (the symbol looks like Audi) or drop the numbers into a search engine would be my best suggestions.

Added after 3 minutes:

Oops! You also have a Volkswagon symbol.

P.S. Let us know how it goes.

Added after 6 minutes:

Incidentally, to anyone out there, is that a load across the coil? Or is it what I've heard called "a snubber", like is put across a transformer?

If it is a snubber, could you briefly explain its purpose, how it works, the R and C?

I'd really like to know more about that, whatever that is that's across the coil.

Thanks!

why is there a diode across pins 85 and 86

When I apply +12V to 86 and ground to 85, the relay closes.
Resistance is to "eat" up the energy spikes when the coil changes potential according to me.

I still do not understand why I got two different replies to ground pin 86, as flow direction (see diode) isn't logic: from ground to coil? That won't happen?

and neither do I think that 86A = 85, as the coil is inbetween both.

As I see it, it's a relay with tho different triggers (86 and 86A), diode's purpose is to keep the second trigger (86A) to go back into the first line.

But I'm far from sure on this one...

how much current to trigger car relay

86 goes to ground


85 is ground to switch it

86A is always +12v

its as plain as day to me

how to on car relay gnd

Perhaps a diagram will clarify what we have said.

+V(86A) ----------- snubber Z ---------- -V (85)
+V(86A) ---------------- L ---------------- -V (85)
+V(86A) ---------------- K --------------- -V (86 = 85)

The three components above, the coil (L), the snubber Z, and the diode (K), are all in parallel across the source +V. (I used a K to show a diode, as it kind of looks like a diode with anode to the right, and I called the other thing a snubber, because that's my guess for what it is.) Notice that, based on the diagram on the relay, the diode's cathode is permanently tied to the coil at terminal 86A -- the above text schematic shows why. Also, 86 = 85 means connect 86 to 85, making them the same potential, ground (0V).

You aren't supposed to power the relay through the diode but, rather, across the coil. Yes, the relay will still switch runnng current through the diode, but that's not how it's supposed to be done, as now you have a diode drop in series, and you're still going to get the negative spike at turn off.

Maybe this will help:

At turn on, the coil sees

+12V ------- L -------- 0V (ground)

This sends current ----->

At the removal of Vsource (which you've said is Vsource=12V), the coil looks like

open ------- L ------- 0V (ground)

The coil then collapses and sends current --------->

To send current to the right, the collapsing coil must look like a battery that looks like

-V ------- L ------- 0V

and 0V, as it's tied to ground, must be 0V, even though it's now the positive terminal of the coil "battery."

Hence, you now have a negatve voltage spike of -V for a brief instant, relative to ground. If the diode is across the coil, that negative voltage spike is going to forward bias it, limiting the spike to only a diode's drop from 0V. If that diode is just a single diode, the spike will be limited to approximately -0.65V. But if the diode isn't there, your switching device is going to feel it!

Hopefully that makes more sense.

Capacitors become instant voltage sources. Coils become instant current sources. Either way you get instant batteries.

car relay diode

Thanks for the explanation. I didn't see the coil as a current source.

Wiring like this is what I ought to do, correct?
**broken link removed**



And if I were to pu the switch S1 between the car battery and port 86A (instead of putting the switch where it is now, between 85 and chassis), wouldn't it be better?
As that way, the relay wouldn't see a direct power feed when car is switched off, for fire hazard reasons...

car relay diode 85 86

Switch open and closed the +12v connection to 86A. Ground is permanently connected.

But I'm guessing you're wanting a low side switch, based on your drawing. If so, then connect the anode of the diode (86, if I remember) to the low side of the coil (85, right?), and then switch the common line of 85 and 86. Let me try to diagram this.

You want this for the high side switch:

+V (86A) ------- L ------- Gnd (85)
+V (86A) ------- K ------- Gnd (86 = 85)

+12v ----- switch ------ +V (86A)

or

You want this for the low side switch:

+V (86A) ------- L -------- common (85)
+V (86A) ------- K ------- common (86 = 85)

common (86 = 85) ------ switch ------ Gnd


In my explanation before, I was figuring you were intending a high side switch, but either will work to pass or open current through the coil.

Notice that in either case, when the switch opens, the coil's magnetic field collapses and creates a negative (with respect to your 12v supply battery) potential that appears across the open switch, that's if you don't have the diode in parallel with the coil when its field collapses.

In other words, your circuit will look like

nodeA ----- Vsource ------ open ------- -Vcoil ----- nodeA

for the high side switch, and

nodeA ----- Vsource ------ -Vcoil ------ open ------ nodeA

for the low side switch.

That circuit (both!) can also look like

nodeA ----- -Vcoil + Vsource ------ open ----- nodeA

It's interesting, yes?

It's all about Kirchhoff. The sum of the voltages around a closed loop always must add to zero. Hence, the open feels -Vcoil + Vsource

Current leaves the high potential terminal of the source, travels through the load, and then back into the source's low potential terminal.

Normally we'd think of the open switch as a broken wire. However, as the switch is opening, it's actually a resistance that increases towards infinity. When the coil's field begins to collapse, it (the coil) does every thing it can to keep that switch closed, keep the current going, even though you are trying to stop it. It likes the current, and it's going to produce as high a voltage as it needs to keep the current constant, relative to the initial current, even as that switch's resistance is getting higher and higher. It will not stop generating voltage (potential, which gets lower and lower) until its field is gone.

Sorry, but I just find it fascinating: V(coil) = L (di/dt)

What this means is, when your switch initially opens, the rate of change of the current, with respect to time (di/dt), is abrupt, so Vcoil is large. It's this Vcoil that can exceed the rating of your switching device, as it is opening, becoming a higher and higher resistance, be it a mechanical switch (ever seen burned contacts, evidence of the arc that developed?) or a transistor acting as a switch.

Technically, the collapsing coil is supposed to be viewed as a current source, but it's hard to see current on a scope. So, instead, I think of it as an instant voltage source oriented in the direction that keeps the current flowing as it was before the collapse. ...my pea brain can only handle so much and the scope is my friend...

P.S. You can actually read the voltage potential created by the collapsing coil if you use a differential input to your dual channel scope. Make ChA the "high" side, ChB the "low" side, and then invert ChB and add ChA and ChB, which creates a single trace of V(ChA) - V(ChB), assuming you made the traces inito one on a reference=0 line prior to making the measurement. Normally you use one channel referenced to ground, but that only shows you relative to ground, some common point to everything, but you want to see the potential at the "high" side of the coil relative to the "low" side.

Ok, I'm done now.

why would a diode be put across a 12v coil

Thank you very much! very detailed explanation...