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WOW. 7A. That is 84W for a solenoid. I do not expect that. That is still very small wattage compare to the total energy a battery can hold.
Additional info: It was my mistake, I worked with switching relay before and normally the current is small. The solenoid needs to push the pinion gear in position, may be that is why it needs more energy. When I read the Wikipedia, it says "A small electric current flowed through the starter relay coil, closing the contacts and sending a large current to the starter motor assembly." May be it is not that small after all. Thanks for the info. |
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Here's one http://www.ietlabs.com/IET/LOM-510.html Note the range: 1 micro-ohm to 200 ohms 1 micro ohm is 1/1,000.000 of 1 ohm. |
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I dont know where they would be, if there are any, for the nss and the ignition switch. Quote:
If you were going to do a voltage drop test for the nss and ignition switch, what connection point would you use? |
You were kinda doing it when you figured out that you dropped 0.7 volts across the solenoid wire. You can measure all your voltages from a reference point like the negative battery post, or you can measure from one end of a wire to the other end. Whenever you have current flowing through a resistance you'll see a voltage drop which can be measured.
If you want to measure across the NSS and the start switch look in the schematic you should be able to do that at that junction box on the fender. You may have a few connection points thrown in the path, but you'll get a good idea of what's going on. With the solenoid connected, put one lead from your meter on the right hand pair of connections and put the other lead on the left hand connection where the solenoid connects. When you crank the engine, the voltage drop will tell you if you need to look at those switches. |
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To get the voltage drop across the NSS, you connect your meter leads on the terminals of the NSS. You don't isolate anything, you simply measure voltage drop across any individual element in the circuit. |
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Okay, the solenoid activation circuit can be tested in the same manner as the high current portion of the circuit. Just put your voltmeter leads across the item under test and see if it is dropping voltage. If it is open, then virtually all of the battery voltage will be dropped across that point. It's DIFFICULT to describe things electronic without a schematic or black board, but let's say that you have B+ going to an ignition switch, then a neutral safety switch and then to the coil of a solenoid. If everything is correct, when the ignition switch is turned to start you will have virtually all (99+%) of B+ ACROSS the solenoid. That is with one lead on ground and the other on the solenoid power wire you will have B+, say 12 Volts. NOW, let's say the resistance across the coil (yes I know it is an inductive load, but I'm trying to keep it simple) is 2 Ohms, then 12 Volts across 2 Ohms gives you 6 Amps although exactly how much current is in the circuit doesn't matter in what I'm trying to explain. So NOW, let's say the NSS is open. Put your voltmeter leads across the NSS and turn on the ignition switch to start and you will see 12Volts across the switch. WHY? Because the resistance across the switch is infinite, while the resistance at the solenoid remains 2 Ohms. So you have BEAUCOUP resistance across the switch and relatively nothing across the solenoid. Forget the exact calculations, just remember that 99% or more of the circuit resistance is at the NSS, SO.. the Voltage is dropped across the NSS in your test. Now take that same thinking to a situation where there is a dirty connection in the circuit causing say 2 Ohms resistance when carrying full current. That would mean that this dirty connection is dropping the same amount of voltage as the solenoid. So, the dirty connection would be dropping 6 Volts leaving only 6 Volts across the solenoid, which probably wouldn't pick it. Does this help? |
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I just want to make it clear: I understand the theory (V=IR, In a series circuit the summation of Voltage drop at individual elements is equal to voltage applied to the entire circuit) I understand the general methodology (apply dmm leads in parralel i.e. across element you want to determine Voltage drop on, looking for voltage draw at element while cranking, this is done sequentially on each element ). I don't know the specifics of application for an '83 300sd, primary ignition circuit (e.g. nss, and ignition switch). I don't know where connect my leads to execute the test. Thats all I dont know about this test. Thats what would be helpful. |
Okay, great. You're more than half way there. Understanding what you're testing and how to test it is the hard part.
NOW, what you're looking for is the location of the switches. I wish I could be more help with that. It seems like someone earlier in this thread or another thread recently described where to find the NSS. My 123 cars have been manual transmissions except one that my wife had and I never had to deal with this particular problem. I would expect that if you get underneath and look at the transmission shift linkage you can find the switch. Maybe someone on here has an FSM for your car. I am assuming that you have already checked for a voltage drop at the starter solenoid. That one is relatively easy to get at. |
Probing the test points can be a PITA if they are hard to access.
You said jumping the termimal block (small screw to big screw) nothing haopens. Did you try jumping small screw to battery + terminal? put a 20 A ammeter between the small screw and battery + terminal. The should pull in the startier solenoid and draw 7A.. |
One comment - high current DC ammeter is very expensive as ammeter is inherently intrusive to the circuit. It needs a very small value and precise resister to do the measurement. Most DIYers do not have one. Need plan B.
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So where did using an ammeter come into the discussion?
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