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Multiplexing two voltages

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joe_1

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I got a solenoid which needs something like 16V to activate, but then it needs only 9V to hold it. It uses 1A.
So, I must activate the 16V initially for short time, then shut it off, and supply the solenoid with the 9V.
The solenoid has flywheel diode which must connect to one of them...which one ? No idea !
I cannot connect the two voltages together, or use diode for each (to separate them), because diodes will block the flywheel.
If anyone has any idea how to mux the voltages, thanks in advance.

You could use this circuit.

I don't know what you are using as switches, so I've drawn them as physical switches.

You could manage without D1. D2 will stop the 16 V feeding back to the 9 V. D3 is the freewheel diode.

Add a series resistor that can reduce the voltage to 9V at whatever current the solenoid takes at that voltage.

Connect a fairly large capacitor across the resistor.

That will initially provide full voltage as the cap starts at 0V, dropping to the 9V sustain level as the cap reaches full charge.

When the power is disconnected , the resistor will discharge the cap ready for the next pulse in a reasonably short time.

The flywheel diode across the solenoid stays in place, no change there.

That is a simple, common solution & should be fine unless you need a rather fast repetition rate.

View attachment 139560
You could use this circuit.

I don't know what you are using as switches, so I've drawn them as physical switches.

You could manage without D1. D2 will stop the 16 V feeding back to the 9 V. D3 is the freewheel diode.

The problem with this circuit is that when the solenoid is turned off, the voltage coming from D3 has no place to go because D1, D2 will block it.
And without D1, D2, the two power supplies are shorted together which is not good.
This problem seems easy at first look, but it is a bit tricky to resolve.

The problem with this circuit is that when the solenoid is turned off, the voltage coming from D3 has no place to go because D1, D2 will block it.
There is no problem.

There is no voltage from D3 to go anywhere.
D3 creates a circulating current in the relay coil, this circulating current dissipated the energy from the collapsing magnetic field in the relay coil.
The voltage across D3 during this discharge period (a few milli-seconds) is about 0.6v (the forward volt drop of the diode).

The circuit proposed by Diver300 will work well, as will the proposal by rjenkinsgb.

JimB

Do you actually need the two different supplies or can you just add a resistor in series with the solenoid to reduce its voltage from 16V to 9V?

If that consumes too much power, you could use a PWM circuit to start with the max 16V and then reduce that average solenoid voltage to 9V from the 16V supply, with high efficiency.

Below is the LTspice simulation of such a circuit using one quad CD4093 dual Schmitt-trigger NAND-gate package:

When the control input V2 goes to 16V (green trace), the circuit initially applies the full 16V to the solenoid so it's current (yellow trace) rises to 1A.

Then after a delay determined by the R3C2 value, the PWM is turned on, which reduces the solenoid current to about 0.56A, as determined by the PWM duty-cycle setting of pot U5.
Due to the efficiency of the PWM signal, the average current drawn from the V1 supply at that point is only about 320mA.

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The problem with this circuit is that when the solenoid is turned off, the voltage coming from D3 has no place to go because D1, D2 will block it.
And without D1, D2, the two power supplies are shorted together which is not good.
This problem seems easy at first look, but it is a bit tricky to resolve.
You are confused - the solution in post #2 is perfectly fine, you are making entirely incorrect assumptions.

It's not 'tricky' at all, it's dead simple.

Ok, thank you folks.

1. Do you have two separate power sources (16 V and 9 V)?

OR

2. Do you have one 16 V power source, and you want the voltage across the solenoid to decrease to 9 V (thus reducing its current) after some period of time?

If #2, then this can be accomplished with a large capacitor in series with the solenoid coil, and a 7 ohm resistor in parallel with the cap. At turn-on, the voltage across the cap is 0 V, and the full 16 V appears across the coil. As the cap charges up, a voltage appears across the resistor, decreasing the voltage across the coil. After three time constants, there is approx 7 V across the resistor and 9 V across the coil.

Is 1 A the holding current at 9 V or the operating current at 16 V. This has a critical impact on the size of the capacitor.

If you use an R-C timer with a power transistor, you will have a much smaller capacitor for the same time delay.

ak

look at LM1949...

If running form a single 16V supply, you can reduce the power use while holding at 9V by means of a series resistor.
You measure the current at the relay coil at 9V, that is the holding current. From that, you can get the coil resistance. Then, calculate a series resistor for 16V so the current will be the same. You put in a capacitor after the resistor but before the switch, to allow charging to 16V. When switch is closed, you get a 16V pulse to the coil, then current is held at a lower value. This circuit depends a lot on unknown factors, and the capacitor may have to be large. Also, this is not for repeated on/off sequences, but a one-of.
The main advantage here is that the current load on the +16V supply is reduced while holding the relay/coil.
Example circuit:

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