add light to 1930s fuel pump

[duke37]

You need to understand the difference between a parallel circuit and a series circuit.
Parallel
The two circuits (components) are connected to a common power input. They have a common voltage but the currents can differ enormously.
Series
When connected in series the current has to pass through one device, then the other. The current is common but the voltage can vary a lot. A led will need 20mA, a pump 4A, i.e. 200 times so a series connection is just not possible.

A monostable will give an output pulse when it is triggered. For an indication of whether the traficator is energised, you could use a constantly lit led or an led driven by a 555 BISTABLE to make it flash. You can also buy a flashing led.I suggest that you do not use a flashing led for the headlights since it can be distracting.

The trafficators and headlights will be powed with one connection the chassis so only one wire will need to be fed from the switch.

The trafficators on my Austin van were reluctant to operate and needed a thump to make them extend. I do not know whether there is an internal switch to reduce power when the arm is extended.

Edit
The ignition leads have a very high voltage pulse which is picked up by the wound wire acting as a capacitance. This can be used to drive a 555 or any other suitable circuit. You cannot get any significant signal out of a low voltage circuit without a direct connection.

 

[morris8]

I think my examples were ill thought out, I've adjusted the drawing to rephrase the question - essentially can the circuit be placed anywhere after the switch/points? (first picture)
My thinking is no because the 2 connecttions (A & B) are essentially the same point... in which case, if I were to put something inbetween A & B, say a bulb, so that it was in parallel to the LED circuit - then would that work? If it were not a bulb, but an inductor - would that work? The bit I'm struggling with is identifying what, if anything is needed inbetween A & B and why - resistance, voltage drop, interruption of electron flow... if there's a term to describe it I shall be happy to google.

 

[duke37]

The battery + will be at +6V relative to chassis.
When you connect to a device, coil (pump or trafficator) or bulb there will be a voltage across the device. This voltage can be used to light a led.

The first diagram shows A and B connected together so there will be no voltage to drive the led.
The second diagram shows a coil and bulb in series with the led connected across the bulb. The 6V will be shared by the coil and bulb so there may not be enough voltage to light the led and there may not be enough voltage or current to activate either device.

The circuit as first given has a capacitor to hold some charge when the pump solenoid is energised for a very short time. This means that the led will light for a longer time and so will be seen better. I think you said that it would last almost a second.
If you are wanting to monitor a light, then the capacitor is not necessary as the led will light whenever the light is on. In theory the diode is also not necessary when monitoring a bulb but it does protect against reverse polarity which coils can generate.

 

[morris8]

Ah, thanks - it's coming together now... the LED or whatever is supplied by the voltage difference of the device(s) it is in parallel with. If voltage drop is the relevant term, then it's the voltage drop across devices I need to plan for rather than the battery voltage - both of which happen to be the same in this one coil circuit. I kind of knew all these little bits but they hadn't gelled, hopefully they have now, although I'll need to think of another project to prove it...

thanks a lot duke, you've been very helpful...

 

[duke37]

You are welcome. It is nice to help someone who is willing and able to learn.