Gypsy
Well-Known Member
To measure the current you put your meter in series with the line/wire. In this case check the ranges of your multimeter, for "A"mp measurement there will be some low ranges such as 10mA, 50mA, 500mA and possibly one for 10A which requires you to put the meter +ve lead into a separate socket on the meter. Start off with the highest range and take the yellow lead off the B+ point then attach the +ve of your meter to the B+ terminal and the -ve of your meter to the yellow wire. Hook it all up before you start the motor. If the reading is very low, as I suspect it will be, change to a lower range on your meter but turn off the engine if you have to unplug the probe - it shouldn't cause a problem to lift the sense lead off when the motor is running but why tempt fate.
Just in case your batteries are fully charged, make sure there is a decent load whilst you are making the measurement - put a few lights on.
As regards the 4 diodes in my fridge fan circuit, yes that is just what happens, 4x0.6=2.4V (give/take). Actually there is another in the cct so I get a 3V drop which allows the fan to run slower and circulates air around the fridge to even out the temperature. Without this I found that stuff next to the evaporator would keep cool but in the back corners the temperature rose quicker. But that is a whole new subject that you can get onto after you sort out your batteries.
Now before we get too carried away, remind me, what is the maximum voltage you have seen at the alt B+ terminal - 13.6V or 14.4V? Remember if this trick works it will boost the volts by 0.6 or 0.7 and that may not be advisable if you can actually get to 14.4V already.
Just to give you some real info on the process of the diode. Jim raised the question of the voltage drop if the current is low. I made some measurements this morning. We always talk about 0.6v across a silicon junction (diode) but it does vary a bit with voltage and temperature. Spec sheets typically show 0.6V to 1V. Here are my readings for a 1N4002 with 13V supply
I = Current thru diode
PD = Votage drop across diode (potential difference)
R = Series resistance from diode cathode to -ve
I PD R
1.0mA 0.584V 12kohm
2.8 0.627 4.7k
5.75 0.658 2.2k
67 0.766 180ohm
300 0.826 40ohm
So you can see that as the current increases the voltage drop does too, but not linearly. I expect the I through the yellow wire to be quite low but we won't know until you test it. The sense line on my Balmar regulator is fused at 1/8A (125mA) but that is only an indication of what to expect.
If the sense wire current is lower than 200mA your 1N4148 could be used. The trouble is we don't know the range of current drawn under all circumstances. Measure first, decide action next.
Good luck
Just in case your batteries are fully charged, make sure there is a decent load whilst you are making the measurement - put a few lights on.
As regards the 4 diodes in my fridge fan circuit, yes that is just what happens, 4x0.6=2.4V (give/take). Actually there is another in the cct so I get a 3V drop which allows the fan to run slower and circulates air around the fridge to even out the temperature. Without this I found that stuff next to the evaporator would keep cool but in the back corners the temperature rose quicker. But that is a whole new subject that you can get onto after you sort out your batteries.
Now before we get too carried away, remind me, what is the maximum voltage you have seen at the alt B+ terminal - 13.6V or 14.4V? Remember if this trick works it will boost the volts by 0.6 or 0.7 and that may not be advisable if you can actually get to 14.4V already.
Just to give you some real info on the process of the diode. Jim raised the question of the voltage drop if the current is low. I made some measurements this morning. We always talk about 0.6v across a silicon junction (diode) but it does vary a bit with voltage and temperature. Spec sheets typically show 0.6V to 1V. Here are my readings for a 1N4002 with 13V supply
I = Current thru diode
PD = Votage drop across diode (potential difference)
R = Series resistance from diode cathode to -ve
I PD R
1.0mA 0.584V 12kohm
2.8 0.627 4.7k
5.75 0.658 2.2k
67 0.766 180ohm
300 0.826 40ohm
So you can see that as the current increases the voltage drop does too, but not linearly. I expect the I through the yellow wire to be quite low but we won't know until you test it. The sense line on my Balmar regulator is fused at 1/8A (125mA) but that is only an indication of what to expect.
If the sense wire current is lower than 200mA your 1N4148 could be used. The trouble is we don't know the range of current drawn under all circumstances. Measure first, decide action next.
Good luck