Resistor sizing - what am I doing wrong ?

superheat6k

superheat6k

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I have decided to make a replacement for the VP Alarm module from scratch. The repair job was simply awful to look at.

Using Mapilns superbright LEDs 3.6v fwd 100mA with a series connected 150 ohm 0.6w metal film resistor, as indicators only these work just fine (nominal 12v max 14.6v).

But for the Alternator excitation circuit I have used the same pairing plus a parallel connected 68 ohm 2W resistor. The circuit (mostly through the larger resistor) is pulling the correct current of 0.15 amps, but the parallel connectd LED / Resistor doesn't light up. Could this simply be too much power being drawn thorugh the larger resistor, if so should i simply fit a lower value resistor for the alt function.

I don't want to blow the LED, but I could connect it with say 68ohm.

The +VE feed passes through a fairly hefty feeder diode, perhaps I should simply measure the PD across it when the 2W 68ohm resistor is on and adjust the LED paired resistor for the reduced voltage. I guess the PD across the feeder diode would be minimal with indicators only drawing minimal current, but when the 2W load is applied the voltage drop becomes more substantial.

Have I just answered my own question ?
 
A bit tricky to diagnose without a schematic diagram, but I think you mean the 68Ω resistor is in parallel with the LED and 150Ω resistor series circuit. If so, this should not affect the LED at all, providing the source voltage is coming from a good connection. The voltage across the input diode will vary slightly, but not enough to cause the problem.

The series resistor of 150Ω will give you about 70mA, which is nicely under the maximum of 100mA. I wouldn't change it.

Things to check: polarity and functioning of LED, measured value of 150Ω resistor, input voltage with and without 68Ω resistor in circuit.
 
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Seems like the maths checks out on the other stuff in ohms law terms. So yeah, check the PD across the diode.

That said... was the 0.15 amps total a typo? That wasn't the figure I got*. Is this connecting to the alternators exciting coil? What's the resistance of that?

*LED resistance at 36ohms, plus 150 ohm = 186, vs 68, 1/((1/186) + (1/68)) = 50 ohms, 12/50 = 0.24 amps.
 
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If you are only pulling 0.15A through the circuit you describe then you don't have as many volts as you expect.

At least 2/3rds of that has to be going through the 68 ohm resistor so there is no more than 50mA in the LED, probably less

You could try dropping the series resistor to 110ohm but I am not sure that would fix it.

Also 0.6W is a little underspecified - at 100mA you will be dissipating 1.5W in that resistor it may even be the affect of the extra heat is raising the resistance
 
The LED / resistor pair work fine on their own, and realistically they are only drawing around 70ma, so on the limit of the resistor but not much over. However, I will directly meausre this. Generally the alarm lamps are only on for brief periods.

There is also a further small voltage drop across the feeder diode, so I don't think the 150 ohm 0.6W is too small.

The affected LED pair lights if I disconnect the 68 ohm 2W, but does not even flicker if the larger resistor is fitted when it draws the 0.15a

I will measure the PD across the feeder diode and larger resistor tomorrow and re-calculate. I may need a larger feeder diode.

Thanks for all the feedback.
 
As a simple test you could measure the voltage across the resistor / LED combo both with and without the 68ohm in parallel. It will give you a quick indication of the "internal resistance" of the supply (not quite the right phrase I know).
 
bedouin said:
At least 2/3rds of that has to be going through the 68 ohm resistor so there is no more than 50mA in the LED, probably less...
Click to expand...
I understand that they are in parallel, so what you say doesn't apply.

There has to be a source resistance for the volts to be dropping, such as a bad contact. Like the "internal" as you say above.
 
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nigelmercier said:
I understand that they are in parallel, so what you say doesn't apply.
Click to expand...
If two elements of a circuit are wired in parallel then the total current through the combination is the sum of the currents through the individual elements of the circuit.

Ohms law tells you that the current is divided in proportion to the reciprocal of the resistance of each leg. The effective resistance of the 150ohm in series with the LED is going to be roughly 200 ohms so if you are passing 150mA through that in parallel with 68 ohms the maths suggest that the LED will only be passing 35-40mA

Of course as LED are not linear devices the figures are very approximate but even if the LED is effectively zero resistance the maximum current would be 50mA. That may not be enough for the LED to light.
 
Those superbrights need 3.6v forward current not 0.6 or 1.2

Your dropping resistor is seeing 8.4v dropped across it ( 12 - 3.6 across the LED ) at your stated 70ma for 0.588 W so you are at the limit of that resistor. Put a 1W resistor in.

The charge light should not be on at 14.4 v if it was you'd be dissipating more then 750mW in it. ( 11 * .07 )

The feeder diode will drop around 0.6v
 
bedouin said:
... The effective resistance of the 150ohm in series with the LED is going to be roughly 200 ohms so if you are passing 150mA through that in parallel with 68 ohms the maths suggest that the LED will only be passing 35-40mA ...
Click to expand...

Total nonsense I'm afraid.

The current through the LED is fixed by the resistor and the input voltage, putting another load across this voltage does not affect this current in any way, unless the voltage also drops.
 
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I commented in the original thread about this.
Use a bulb for the alternator.
Or put one in parallel with the LED/resistor circuit if you must have the LED for cosmetic purposes.

The bulb is non linear, it has low resistance at low current which helps the alternator start.
The alternator, when not charging, may have several volts on the WL pin, when the bulb is putting current into it.

The bulb does not simply see 12V.
 
superheat6k said:
I have decided to make a replacement for the VP Alarm module from scratch. The repair job was simply awful to look at.

Using Mapilns superbright LEDs 3.6v fwd 100mA with a series connected 150 ohm 0.6w metal film resistor, as indicators only these work just fine (nominal 12v max 14.6v).
Click to expand...

Have you checked what the light output is like ? I use hyper white for boat mimics, fitting a 108 K resistor to stop it blinding the user.

Remember these are the LED's in torches and day time lights on cars.

Brian
 
nigelmercier said:
Total nonsense I'm afraid.

The current through the LED is fixed by the resistor and the input voltage, putting another load across this voltage does not affect this current in any way,.
Click to expand...
Only if you assume a constant voltage source with zero internal resistance - but it seems that is not the case. If you look at the thread you will see I have suggested measuring the voltage across the circuit with and without the 68 ohm resistor - given the symptoms described I would not be surprised to see a lower voltage with resistor in parallel.

But the maths of the first case is pretty straight forward - if you don't agree you tell me how much current is flowing through the LED given the draw for the entire combination is 150mA as given.
 
lw395 said:
I commented in the original thread about this.
Use a bulb for the alternator.
Or put one in parallel with the LED/resistor circuit if you must have the LED for cosmetic purposes.

The bulb is non linear, it has low resistance at low current which helps the alternator start.
The alternator, when not charging, may have several volts on the WL pin, when the bulb is putting current into it.

The bulb does not simply see 12V.
Click to expand...

That's certainly my understanding. It's a rather quaint way of doing things but it works! The resistance of the bulb is very low when not lit. Which is I suspect essential. Also alternators tend to be somewhat spiky so I'd be worried about the reliability of a high power LED.
 
I mentioned this before but it wasn't picked up - perhaps it's wrong and that was why!

The overall resistance of the parallel LED/resistor and resistor combo in the circuit is 50 ohms... a quick google suggests that the resistance of the exciter winding in the alternator could easily be as much as 20 ohms.... are you sure you haven't accidentally created a potential divider and ended up with only a (50/70)*12 = 8v voltage drop across the LED and resistor circuit as well as a drop in current caused by the higher than calculated resistance?
 
http://oljeep.com/gw/alt/edge_Alternator_Theory.html#Section_2

If you scroll down a bit there is a circuit for a typical alternator.
The site has some theory too, it's all I can find before I go to a meeting.

But the bulb current does quite a lot, it needs to bias up the reg circuit, and start the field (rotor), so that the stator starts producing volts, which then takes over driving the reg via the 3 smaller diodes....

Without the bulb current, you need a lot more rpm to kick off the process from residual magnetism in the rotor.
 
bedouin said:
But the maths of the first case is pretty straight forward - if you don't agree you tell me how much current is flowing through the LED given the draw for the entire combination is 150mA as given.
Click to expand...

I think you are confused, current doesn't work like you are suggesting. If the voltage is 12V, then the LED current is about 70mA. The current in the other leg is totally irrelevant, as is the combined current. You could hang any number of extra paths in parallel with the LED path, and providing the voltage is the same, the current will be also. Think of it this way, do your car brake lights glow any less brightly if the headlights are on?

Having read later replies, it appears that this LED is not fed from a 12V to ground circuit, so the voltage may be lower. In that case the dropper resistor will need to be recalculated. However, I would expect it to still light the LED even at about 5mA.
 
nigelmercier said:
I think you are confused, current doesn't work like you are suggesting. If the voltage is 12V, then the LED current is about 70mA. The current in the other leg is totally irrelevant, as is the combined current. You could hang any number of extra paths in parallel with the LED path, and providing the voltage is the same, the current will be also. Think of it this way, do your car brake lights glow any less brightly if the headlights are on?

Having read later replies, it appears that this LED is not fed from a 12V to ground circuit, so the voltage may be lower. In that case the dropper resistor will need to be recalculated. However, I would expect it to still light the LED even at about 5mA.
Click to expand...
But we have been told that the current through the pair is 150mA. You are saying the current through the LED is 70mA implying that the current in the "shunt" resistor is 80mA. But if you know anything about parallel circuits you will know that is nonsense.

Given you are so sure I am wrong what is your calculation of the current in the two legs of the parallel circuit?
 
bedouin said:
But we have been told that the current through the pair is 150mA. You are saying the current through the LED is 70mA implying that the current in the "shunt" resistor is 80mA. But if you know anything about parallel circuits you will know that is nonsense...
Click to expand...

I am saying that the current through the LED will be about 70mA if the voltage is 12V. As I said above, perhaps it is not. However, if the LED current is 70mA, and the total is 150mA, then it is not nonsense to say that the shunt is drawing 80mA. Why this may be so is another matter.
 
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