- #1
cfauvel
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newbie needs help understanding a tachometer's "monitor"
Ok let's start with what I have.
I have a 1960s era Stewart-Warner 970 series tachometer in my 1969 Mach 1.
It uses an "ignition monitor" model number 990B.
The monitor has a terminal for a lead that goes to the negative terminal of the ignition coil.
The monitor has a terminal for a lead that goes to the tachometer.
The ignintion coil, I think acts like a big ol' capacitor. It has a positive terminal,
a negative terminal and a big wire to the distributor cap. The coil typically puts out 25,000 to 42,000 volts
(not sure if it is DC or not, but I imagine it is )
The positive terminal of the coil is powered by ignition switch, I would imagine 12votls since that is the car's normal voltage.
The negative terminal of the coil is connected to the distributor's points. When the points close I guess the
circuit closes and let's loose the high voltage from the coil.
I opened up the monitor and was surprised as to how little was inside. Actually there are TWO componets
to make the "monitor"
1 capacitor with the markings of .068 MFD 200V.D.C. It is cylindrical with a wire lead on each end. One end has a black band,
which I assume denotes the negative side.
1 resistor with the markings orange-orange-red and silver, Through research I found that it is a
3.3KOhm resistor. Currently it meters out to 3.6 KOhms
The circuit goes like this
Neg Coil → neg end Capacitor → resistor → tachometer
My questions are these
What exactly are the componets doing?
How bad is that the resistor rated for 3.3kohm is actually reading 3.6 kohm?
Through my research I think the capacitor is changing the voltage from A/C to D/C
and supplying a constant voltage. Not sure why the rating of 200V.D.C. seems a high to me.
No clue what .068 MFD really is, I know it is Micro Ferrands, but that is still greek to me.
I am guessing the resistor is steppiing down the voltage, not sure if there is a formula that
given a voltage on one side of a resistor will produce x amount of voltage. Hence the questions here.
Anxiously waiting for your expert teachings.
Chris
Ok let's start with what I have.
I have a 1960s era Stewart-Warner 970 series tachometer in my 1969 Mach 1.
It uses an "ignition monitor" model number 990B.
The monitor has a terminal for a lead that goes to the negative terminal of the ignition coil.
The monitor has a terminal for a lead that goes to the tachometer.
The ignintion coil, I think acts like a big ol' capacitor. It has a positive terminal,
a negative terminal and a big wire to the distributor cap. The coil typically puts out 25,000 to 42,000 volts
(not sure if it is DC or not, but I imagine it is )
The positive terminal of the coil is powered by ignition switch, I would imagine 12votls since that is the car's normal voltage.
The negative terminal of the coil is connected to the distributor's points. When the points close I guess the
circuit closes and let's loose the high voltage from the coil.
I opened up the monitor and was surprised as to how little was inside. Actually there are TWO componets
to make the "monitor"
1 capacitor with the markings of .068 MFD 200V.D.C. It is cylindrical with a wire lead on each end. One end has a black band,
which I assume denotes the negative side.
1 resistor with the markings orange-orange-red and silver, Through research I found that it is a
3.3KOhm resistor. Currently it meters out to 3.6 KOhms
The circuit goes like this
Neg Coil → neg end Capacitor → resistor → tachometer
My questions are these
What exactly are the componets doing?
How bad is that the resistor rated for 3.3kohm is actually reading 3.6 kohm?
Through my research I think the capacitor is changing the voltage from A/C to D/C
and supplying a constant voltage. Not sure why the rating of 200V.D.C. seems a high to me.
No clue what .068 MFD really is, I know it is Micro Ferrands, but that is still greek to me.
I am guessing the resistor is steppiing down the voltage, not sure if there is a formula that
given a voltage on one side of a resistor will produce x amount of voltage. Hence the questions here.
Anxiously waiting for your expert teachings.
Chris
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