Managing parallel loads in 12v circuit

[Paul Larned]

Hello,

I am trying to build a novelty turn signal unit for an ebike. When energized, I want a 12v-24v solenoid to push an aluminum lever from vertical to horizontal, and, at the same time, activate a flasher-led combination situated on the end of the lever. I have no trouble activating the solenoid, but I cannot seem to add the flasher-led combo into the circuit. Each circuit (solenoid and flasher-led) works fine when connected separately to the 12v dc supply, but connecting the two circuits in parallel across a 12v dc supply does not work. How can I isolate the two circuits, yet still connect them to the same 12v battery supply?

 

[Nanren888]

Probably want to look at it a little differently.
.
You are correct in that if the supply stays as 12 volts, it should make each or both in parallel go.
Most likely when you have both connected, the total load is greater than the supply allows; that is the total current is greater than the supply rating.
I'd expect an e-bike to be able to supply significant current. so this is a little surprising.
Is this a supply outlet of the e-bike? What is the supply intended for? Is it an accessory supply?. Does the documentation suggest a current limit?
.
If you can measure the supply voltage, say with a voltmeter, across the supply outlet that you are using, when you connect each load and when both are connected, you may find the supply is below 12 volts with both connected.
If this is the case, you'll have to arrange another supply in some way.
Not familiar with e-bikes, so sorry can't suggest how this might be done conveniently.
.
When you first connect a solenoid, the current drawn can be quite large, supplying the energy to move the core, then drop off. There is a chance that it is a transient effect, but then I'd expect it to work after a small delay, unless it triggers the supply to shut down from over-current. This brings us back to how you are getting the 12 volt supply to your circuit.
.
Also, solenoids often create a voltage spike when disconnected, so often some form of protection is included in circuits that switch solenoids. The voltage spike can damage other circuits, particularly semiconductor circuits such as LEDs. The protection is often in terms of a reverse polarity diode across the solenoid.

 

[Paul Larned]

Probably want to look at it a little differently.
.
You are correct in that if the supply stays as 12 volts, it should make each or both in parallel go.
Most likely when you have both connected, the total load is greater than the supply allows; that is the total current is greater than the supply rating.
I'd expect an e-bike to be able to supply significant current. so this is a little surprising.
Is this a supply outlet of the e-bike? What is the supply intended for? Is it an accessory supply?. Does the documentation suggest a current limit?
.
If you can measure the supply voltage, say with a voltmeter, across the supply outlet that you are using, when you connect each load and when both are connected, you may find the supply is below 12 volts with both connected.
If this is the case, you'll have to arrange another supply in some way.
Not familiar with e-bikes, so sorry can't suggest how this might be done conveniently.
.
When you first connect a solenoid, the current drawn can be quite large, supplying the energy to move the core, then drop off. There is a chance that it is a transient effect, but then I'd expect it to work after a small delay, unless it triggers the supply to shut down from over-current. This brings us back to how you are getting the 12 volt supply to your circuit.
.
Also, solenoids often create a voltage spike when disconnected, so often some form of protection is included in circuits that switch solenoids. The voltage spike can damage other circuits, particularly semiconductor circuits such as LEDs. The protection is often in terms of a reverse polarity diode across the solenoid.

Thanks for your reply!

Currently, I have the circuit set up on the bench with a DC power supply supplying the power. I am using a two-pole (B abd L) car flashing unit, and 6 2v leds connected in series for a 12v load. I find that I have to increase the dc supply to 14v in order to begin flashing, but this circuit then works well. If I connect the solenoid after the flasher circuit, it energizes off-on with the flasher circuit, which is not what I want. If I connect the solenoid first, the voltage drops to a variable 6-8 volts downstream, and the flasher-led circuit will not work. I don't know how to "isolate" the two circuits so that they will both function when energized.

The ebike battery has connectors for various 12 v systems on the bike, such as power control and lights. I will need to get a Y connector to tap into the battery 12v system. I don't think this will be a problem once I can figure out how to isolate the two circuits. The two circuits will be energized by an on-off-on switch, so I think I need to deal with a single tap from the battery.

Thanks for any help you can provide!

 

[Nanren888]

Ok, I will try again.
"isolate" is not a fruitful direction, for any interpretation that I think likely.
Load currents add. You must supply the sum of them. I try to explain, below.
.
Sounds as if your bench supply cannot supply the current required by the combination of both circuits.
If one circuit at 12 volts draws 1amp and the other draws say, 1.1 amps peak, then the supply must be able to supply 2.1 amps to run both loads, both circuits.
.
If the supply is unable to supply this, by exceeding its capacity, or a built-in safety current limit, it will drop the output voltage, and/or shut off,
.
With reduced voltage, both circuits, solenoid and LEDs would be expected to misbehave in a connected way as each changes the current that it draws and causes the voltage to change accordingly affecting the other circuit.
.
With an e-bike battery, as supply, likely a lot of current can be supplied; you are unlikely to get this problem.
This also means that you might want to include a fuse in your ciruit, so that an issue, for example a short, in your new circuit does not create a short directly across the battery. IF that happened, things might get hot and could be dangerous. Your ciruit would preferably be connected after a fuse if one already exists.
.
For experimants, do you have another (large-ish) battery you can use as a test supply, or a larger DC 12 volt supply?
.
Maybe also note the comment I posted about protecting things from the solenoid kick when switched off.

 

[Paul Larned]

Ok, I will try again.
"isolate" is not a fruitful direction, for any interpretation that I think likely.
Load currents add. You must supply the sum of them. I try to explain, below.
.
Sounds as if your bench supply cannot supply the current required by the combination of both circuits.
If one circuit at 12 volts draws 1amp and the other draws say, 1.1 amps peak, then the supply must be able to supply 2.1 amps to run both loads, both circuits.
.
If the supply is unable to supply this, by exceeding its capacity, or a built-in safety current limit, it will drop the output voltage, and/or shut off,
.
With reduced voltage, both circuits, solenoid and LEDs would be expected to misbehave in a connected way as each changes the current that it draws and causes the voltage to change accordingly affecting the other circuit.
.
With an e-bike battery, as supply, likely a lot of current can be supplied; you are unlikely to get this problem.
This also means that you might want to include a fuse in your ciruit, so that an issue, for example a short, in your new circuit does not create a short directly across the battery. IF that happened, things might get hot and could be dangerous. Your ciruit would preferably be connected after a fuse if one already exists.
.
For experimants, do you have another (large-ish) battery you can use as a test supply, or a larger DC 12 volt supply?

Maybe also note the comment I posted about protecting things from the solenoid kick when switched off.

Thanks again for your reply!

My bench power supply is capable of 10 amps, and, with both the solenoid and flasher-led circuits in parallel, appears to draw just over 1 amp. But given what you explained, maybe that is not an accurate reading of the current flow. I also wondered whether the flasher, designed for car incandescent bulbs, might be the wrong one for the job.

The fuse is a great idea, as is the diode, and until I can get both circuits working properly, I'm not going anywhere near the bike battery!
.

 

[Kiwi]

Car flasher units are usually built to operate two 21watt bulbs in parallel, that's about 3.5amps. Most won't work with the very small current that LED's draw.
You can get special flasher units for LED's.
I would suggest that you could try the six LED's in parallel with a series resistor, instead of directly in series to the power supply.
The trafficator system you are building was common in the mid 1900's on cars such as the Morris 1000.

 

[Bluejets]

The trafficator system you are building was common in the mid 1900's on cars such as the Morris 1000.

Yes, I remember Dad's "Rolls Royce", a six eighty Wolsley had them.
I think early VW might have as well.

 

[Paul Larned]

Thanks! I ordered a flasher designed for leds. I wanted to do this project because I used to have a '55 import VW with the little chicken-wing turn signals, and I wish I had never sold it! Next big problem is mechanical--trying to find a 12v solenoid strong enough to push a 3/8" x 4" piece of aluminum from vertical (hanging) to horizontal.

Thanks, all, for your input!

 

[Bluejets]

Shouldn't be too difficult to do even with trial and error.
You could use for example an rc servo in place of the solenoid, or in fact a small wheel retract unit.

If you do a search for trafficator you will find more info.
Just don't go trying to buy a second hand one, the price will astound you..........

Examples........

http://www.angliaobsolete.com/trafficator-text.html