Date: Thu, 21 Jun 2012 21:23:48 -0400
Reply-To: David Beierl <dbeierl@ATTGLOBAL.NET>
Sender: Vanagon Mailing List <vanagon@gerry.vanagon.com>
From: David Beierl <dbeierl@ATTGLOBAL.NET>
Subject: Re: CO Adjusting Screw. PO Set it All The Way In? Base Setting?
In-Reply-To: <CAB2RwfieUSzcXbpTu4F1n2RFGa5EJVXYVd6F-h-tSG3x5sHXnQ@mail.g
mail.com>
Content-Type: text/plain; charset="us-ascii"; format=flowed
At 08:27 PM 6/21/2012, neil n wrote:
>Ok. I see that. Thanks for hint on resistor. (but why a yellow LED?
>< kidding >) But....
Actually because I once used a small yellow LED with no resistor at
all for exactly this purpose. Yellow because of the low-voltage
(red/green/yellow) LEDs it has the highest forward voltage. It
pulled the line down to two volts and change, but the ECU still saw
that as an open switch. Adding a resistor in series limits the
current through the LED in case it's a bit too high, and also lets
the ECU line run higher; but my little yellow guy seemed happy and of
normal brightness, and the ECU was still interpreting the line state
correctly. Since that line is grounded when the switch is closed I
wasn't worried about overloading anything inside the ECU. I ran it
that way for several months.
>By "shunting", I meant connecting an LED in parallel at wires going
>to the TVS (Throttle Valve Switch) plug.
Yes. Except I'm still questioning where exactly the throttle valve
is. Is it like a spit valve? ;-) Unless this is Canadian-specific
terminology I think it's wrong. -- Oh, that's what Bentley calls the
throttle, throttle plate, throttle butterfly (US), choke
(British). Silly me. My eyes have been skipping over that word
since forever. And they call the throttle body the throttle valve
body. Well, well. Pleh.
> Maybe I'm still misunderstanding things. Seems to me the ECU
> supplies both the - and + to the throttle valve switch
> plug. (otherwise why can one measure voltage there, ignition on?)
It supplies +5 in series with an internal resistor, and its own
ground (which is closely related to and derives from chassis ground).
>Would shunting in an LED in parallel complete the circuit so ECU
>thinks the throttle valve switch was closed 100% of the time?
If you used a piece of wire, yes. The yellow (or blue, or white) LED
will pull the line down to whatever its forward voltage* is, but (by
experiment) the resulting 2+ volts is still considered an OPEN by the
ECU. IC logic chips have a voltage above which the answer is
definitely yes, another below which the answer is definitely no, and
an undefined area in between. This may be set up that way or it may
just be a comparator that trips at some defined voltage. Since it's
an essentially DC circuit and the expected values are +5 and 0 it
doesn't have to be very concerned about thresholds.
*If you reverse-bias a diode, i.e. connect it in reverse polarity
across a voltage source, essentially nothing happens until the source
gets strong enough to force the diode to break down, which depending
on the diode and circumstances may or may not destroy it. Some
diodes, called zener or reverse-avalanche diodes are designed to
operate under breakdown conditions. LEDs typically have a rather low
breakdown voltage so it's not a good idea to hook them to 12 V backwards.
If you forward-bias it, essentially nothing will happen until your
source reaches the "forward voltage" of the diode, at which point the
diode will suddenly start acting like a variable resistance that will
keep the voltage across its terminals at or very near the forward
voltage or die in the attempt. The Vf of a red LED is typically
about 1.8 V; a yellow one might be around 2.3 V at some rated current
like 10 mA. Since the ECU supply to the throttle switch will only
supply a limited current an LED doesn't commit instant hara-kiri if
you hook it up directly, but it will pull the line down to 2.3 V or
so, the rest of the 5 V being dropped across the series resistor
inside the ECU.
When you want to hook up an LED to a source that can deliver more
than ~20 mA for the little ones, you have to supply a
current-limiting resistor in series whose value is calculated to fit
the crime. For example, you've got a red LED with Vf of 1.8 V and a
max current of 20 mA. You want to supply it with 15 mA max from
automotive "twelve volts."
First take your maximum reasonable system voltage. Call it 16 V to be
safe. The diode will be using 1.8 V so subtract that:
16 V - 1.8 V = 14.2 V. Your limiting resistor will have to permit 15
mA with 14.2 volts across it. So...
E (volts) = I (amps) x R (ohms) -->
R = E/I --> R = 14.2 V / .015 A = 947R. Nearest larger value is
1000R so you use that.
Then you had better check the power that resistor is going to dissipate:
P (watts) = E x I = 14.2 V x .015 A = .213 W. You could squeak by
with a quarter-watt resistor but much better to use a half-watter or
use two 2,200R quarter-watt in parallel. Rule of thumb is don't
exceed half the component rating.
Yours,
d
