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Re: Turn signals flasher

To: Chris.S.Mottram@ecc.com, alpines@Autox.Team.Net, tigers@Autox.Team.Net
Subject: Re: Turn signals flasher
From: SDeane7000@aol.com
Date: Sat, 2 Aug 1997 12:19:07 -0400 (EDT)
Hello Chris,

Please bear with me, as you may imagine it's as difficult to describe
"electricity" as it is to understand it.  Warning:  this class in electicity
101 is not for the light of heart. It's pretty boring to those not

Think of volts as a potential difference from ground.  Start at the battery,
you have 12 (actually something greater like 13-14) volts to ground or the
negative post.  All 12 volts will make it to the load (light bulb) if there's
nothing between the battery and the bulb to impede the flow of electrons (the
wire and all connections to battery and light are perfect).  The current that
flows (the electrons are like water flowing through a hose) is a function of
what is connected across that potential difference (the device connected at
the end of the hose).  The current is much like measuring the pressure in a
hose.  With water flowing through the hose and one end of the nose open,
pressure in the hose is low - almost zero.  Put a restriction like a spray
nozzle or close the other end of the hose and pressure in the hose is high.
 The pressure (or current) is a function of the thing that impedes the flow,
not the pumps CFM or potential to develop pressure.

Think of a 25Watt light connected to a 100 volt household outlet.  Connect a
100 Watt bulb and the light output is much greater.  Why?  The voltage
(potential) is the same.  It's the current that is different.  It is
different because of the construction of the bulb: the fillament in the 25 W
bulb is a very thin type of wire, the 100 W bulb has a thicker fillament.
 The thicker fillament "draws" of allows more current to flow through it than
the thinner fillament.  The fillament is known as a "resistor".  The material
it is constructed of "impedes" the flow of electrons or resists the flow of
current, much like a knot tied in a water hose restricts the flow of water.
 The current flowing through the restriction (resistor) can be thought of as
causing friction which causes heat which causes the wire to get white hot -
which gives off the light.

If the potential (volts) were increased from 100 to 120 V, the bulbs would of
course get brighter - the current goes up as the voltage goes up.  In fact
there is a simple equation that describes the relationship between voltage,
current and resistance known as Ohm's law: current = volts divided by
resistance.  If volts go up, current goes up.  If resistance goes up (thinner
fillament) current goes down.  It's important to note that you can change the
voltage and resistance but not the current.  The current will be the result
of the potential (voltage) on the resistance (load).

I could go on forever on this subject (ask an engineer what time it is and
we'll tell you how the watch works).  Boring!!!!

How's this apply to your question you might ask (I know, good thing I finally
got back on-track).  Given a fixed potential (charged battery that doesn't
"droop" when loaded), there must be something less than perfect between the
battery and the flasher bulb:  wires from the plus battery terminal,
connections to the turn signal switch, the turn signal switch, switch
contacts in the flasher module, the connector/lamp socket to the bulb, THE
GROUND WIRES from the bulb back to ground (don't forget these!!!).  In other
words if the voltage isn't getting to the light, there is another resistor
between the battery and the light or between the light and ground.

You can find the resistor in one of two ways.  You can measure the resistance
of the circuit (or parts of the circuit) with an Ohm Meter - I won't go into
this now but if you attempt to measure Ohms, disconnect the battery  (the Ohm
meter has it's own battery to supply the potential).  The easier method would
be to use a volt meter.  Look at the equation above,
current=volts/resistance.  This also tells us what voltage will be developed
(or really dropped) across the resistor for a given current.  This is simple
for the single light example - 100 volts are across the bulb 'cause its the
only bulb in the circuit (known as a parallel circuit: the bulb is parallel
to the battery).  If two bulbs are connected to eachother, one end to the
other or in "series" with eachother, each light get very dim - half the light
from each.  Each bulb, if equal wattage (thus equal resistance) will drop or
have across it 1/2 the voltage - 50 volts for each.

There I go again!  Bad engineer!!   Anyway - hook the minus terminal of the
meter to the battery ground and leave it there.  With a long wire connected
to the plus terminal of the volt meter, start reading the voltage at the
battery terminal, then follow the wiring from one point down the line to the
next - WHILE THE flasher light is operating thus causing current to flow
through the circuit thus allowing us to observe the voltage drop across the
unwanted resistor (the fault).  At some point the voltage will drop.  This is
your extra resistor:  perhaps through a bad connection, partially broken wire
or dirty light socket (very difficult to measure as it's covered by the light
- clean it and try again).  

Putting the volt meter across the "bad spot" will also read a voltage drop:  

The problem can also be isolated by putting the volt meter across portions of
the light circuit.  Say from the plus terminal of the battery to the flasher
terminal (before the flasher contact).  No appreciable volts should be
dropped or the problem is in this "leg" of the circuit.  With the flasher
light still activated, connect the volt meter from the flasher light to the
flasher (after the flasher contact), there should be no voltage drop even
when the lamp lights or this is your bad leg.

Well, please don't give me any flack about this tirade because I did warn
you.  How many of you lived long enough to make it to this point?


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