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Newbie needs help understanding a tachometer's monitor 
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#1
Feb1512, 07:14 PM

P: 4

Ok let's start with what I have.
I have a 1960s era StewartWarner 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 orangeorangered 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 


#2
Feb1512, 07:35 PM

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The coil is an induction coil and its purpose is to create a voltage spike sufficient to jump the spark gap in the spark plug. So the connection from the coil is producing electrical spikes in time with the engine. The voltage is quite high. The capacitor keeps DC voltage from flowing through the circuit and the resistor is a ballast resistor. Its high resistance prevents much current from flowing through (even AC current from the pulse).
The resistor's voltage rating is not too big an issue given the pulsed input. The whole purpose of this circuit is to provide a moderate voltage low current signal to the tachometer. Inside the tach. should be a circuit converting frequency to (small) proportional current which deflects the meter in proportion to the engine speed. What is your ultimate issue/goal/problem? 


#3
Feb1512, 09:01 PM

P: 4

I think you covereed that thanks, but have a few more. Does the fact that the resistor is reading slightly more resistance (3.6 KOhms instead of 3.3 Kohms) mean that the tach will be reading slightly slower. Is the ignition coil more like a step up transformer from 12v to 25k+v rather than a capacitor? That would make sense. What does the value of .068 MFD really mean? What is a micro ferrand? So the capacitor ensures a D/C current going to the tach right? The black band denotes the negative side of the capacitor right? Remember I am a novice to this stuff so be gentle. Trying to get a better understanding. I'll need to read up more on resistors...I don't think I have the right understanding of them, nor of the difference between Voltage and Current. Thanks Chris 


#4
Feb1512, 09:57 PM

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Newbie needs help understanding a tachometer's monitor
Current is the charge flow rate, 1 Ampere is 1 coulomb per second. Take that as a velocity analogue. Voltage is the force analogue (Newtons * m/s = watts; volts * amps = volts * coulombs/sec = watts). In the analogy assume the "belts" have almost zero mass so any force if not resisted or mediated by inertia will create a great deal of motion. Resistors are the friction analogue. They obey Ohm's law V = IR. That's the mechanical analogue of a velocity dependent drag, Force = resistance * speed. Viewed another way resistors dissipate energy VI = Watts power, so I^2 R = heat generated by a resistor for a given current. Capacitors act like springs but the capacitance is comparable to the reciprocal of the spring constant. C = Coulombs per Volt, compares to 1/k = meters per newton. 1/C = Volts per coulomb corresponds to the force per distance constant of a spring. Inductors, as I mentioned, are an inertial analog. They can be thought of as flywheels in the pulley analogue. Their units are Henry's (L symbol) which gives the voltage for a rate of change of current V = L * dI/dt which corresponds to Newton's F = mass * acceleration in the mechanical analogue. Now AC you can think of as oscillatory motion of the mechanical analogue (like a belt driving the agitator of a washing machine ) and DC is steady one way motion (spin cycle!) So you can imagine the "monitor circuit" as running a belt from the pulsing of the coil through a pulley with a tight spring on it (low cap = tight spring) and then through a damper (high resistance means its highly viscous). The spring ensures only a "kick" is transmitted through the circuit and the damper ensures the "kick" doesn't move very far and doesn't "bounce" producing false signals. The coil is a transformer but you can think of it as a heavy flywheel. Steady belt force brings it up to speed and a sudden locking of the belt causes it to transmit a strong pulsed force along a secondary belt. A practical analogue is say an impact wrench. Unfortunately this mechanical analogue breaks down when you want to describe a transformer. For sinusoidal AC you can sort of get by thinking in terms of a gear system trading off speed and force but there are subtleties not represented by the analogue, most importantly the fact that the transformer output voltage is proportional to the change in input current. It can't work in DC the way a gear works for continuous motion. I'll try and think up some mechanism which mimics a transformer. But other than that you have my "quickndirty" electronics review. Hmmm.... 


#5
Feb1612, 09:12 AM

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Back in 1960s usually the input circuit was a low pass filter.
The points produce a 12 volt pulse train with huge inductive spikes at each opening and i'm guessing it's a RC lowpass to remove the spikes so the tach doesn't try to count the oscillations of coilcondenser ringing.. 


#6
Feb1612, 12:48 PM

P: 4

wow Jambaugh that was in depth explanation.
I thought I 'd attach a picture of what the pieces look like. Is that black band on the capacitor the negative side? Thaks for letting me know that the resistor's reading is fine....scared me abit. if the cap is producing a/c and the ignition terminal is producing 12v DC, is the resulting AC 12volts too? I need to go back to school :) 


#7
Feb1612, 06:42 PM

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Well, so much for my Low Pass idea!!!
a picture's worth a thousand words. ignore that post, that's not what the cap does , and i apologize. The black stripe  That type capacitor is insensitive to which lead is positive or negative. A capacitor is made from two strips of foil, separated by an insulating film and rolled into a tube. The two lead wires are each connected to one of the foils. The black stripe indicates which end connects to the foil that's on the outside of last wrap. It's not important in your application which way it goes.. Were it a filter, the outside foil ought to be the one grounded. old jim 


#8
Feb1712, 12:09 AM

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I did a bit of math... (I might give this as a problem for my ODE students.)
Note the unit of capacitance, the Farad equals one second per ohm. Thus the RC product gives a characteristic time for the circuit. With the values you posted [itex]3.6K\Omega \cdot 0.068\mu F = .2448 [/itex] milliseconds. Call this T. The behavior of a series RC circuit connected to a low resistive load will be to exponentially decay any change in voltage. [tex] V_{out}(t) = V_{jump} e^{t/T},t>0[/tex] (t = 0 at the jump.) Think of this as a decay with halflife ln(2) T (= 69.3% T = .17 milliseconds). For high resistive load the characteristic time will be be [itex](R_{load}+R)C[/itex] and the output voltage will be attenuated by a factor of [itex]\frac{R}{R_{load}+R}[/itex]. Take an ohmmeter to your tach to see what resistance it shows. I think the resistor wouldn't be necessary in the second case so I'll wager its there to limit current through the low resistance Tach input. I believe the purpose is to make sure the tachometer is only being input changes in voltage on the order of the characteristic time with moderated transient current. 


#9
Feb1912, 07:22 PM

P: 4

Thanks all for the replies, interesting stuff.
Jambaugh, too complicated for my knowledge, you lost me with the formulas...no worries. I like the anology to mechanical things, that helped alot. Chris 


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