# Briefly switching a coil

We are going to build a telegraph which can communicate to an other telegraph. One of our communication systems exists out of a bell. The idea is that the sender can press a button and the receiver will hear one single ping. This bell will be operated by an electromagnet. We also want to send a few pings after each other.

12 V and ground are fed by a flat 6 pole RJ12 cable. There is one pole for 12 V and one pole for ground.

There are two features I would like to implement. I want to add a capacitor discharge unit to power the coil with as purpose not drawing too much current from the 12 V line. (we would see LEDs flicker)

The CDU on it's own can power the coil. But if one is too send three pings, he might deplete the CDU too fast if hold the button too long. And the hammer might be touching the bell too long

For the second feature I would like to add a capacitor in the receiving line to let the button switch cause a pulse which can trigger an opto coupler.

The operation in this circuit in order: One presses the button at bellTx, this causes a pulse on the receiver which triggers the optocoupler for a short time.

When the optocoupler is active, the MOSFET Q3 will be conducting which should sink the coil to ground.

When that happens the CDU should discharge the coil for as long as the pulse is active or the CDU still has energy.

What I want to know: Did I make any obvious mistakes in this circuit? I was not sure if I have to tie bellRX to ground via a resistor as well? How can I calculate the on time of the opto coupler when I press the the switch near bellTX

I do not yet know how long I should switch and how much energy I am going to need. I do can experiment with the opto coupler pulse, but I'd like to know how to calculate the resistor and capacitor values.

I figured it would also be a good idea to add a potentiometer somewhere in the pulse line. This would give me more control over the pulse time.

• What is the meaning of CDU? What do you mean by pole as in: 6 pole rj12 cable - do you mean wire or conductor? Why is the base of your transistor connected to the emitter? Circuit simulators are basically free so why not simulate? Sep 29, 2021 at 11:08
• O sorry. CDU stands for capacitor discharge unit. In off-mode the capacitors are charged through the transistor. When the load (coil) is connected to the ground all current comes directly out of the capacitor. This limits the maxium current draw on the 12V line and it prevents the 12V from dropping. Sep 29, 2021 at 11:23
• What Andy said, your question begs to be simulated. Sep 29, 2021 at 11:24
• I understand your comment. And I'll try but I also want to learn how to calculate such a pulse time. I can try various R/C values until it is 'okay' but I doubt it will make me much wiser Sep 29, 2021 at 11:26
• the 'old way' to discharge capacitors on a load is with SCRs… rugged and somewhat brutal but efficient:D however if your capacitors are not slowly charged so it doesn't work. The idea of a CDU would be to quickly dump the capacitor on the load without powering the load from the supply rail What's the purpose of Q2? a current limiter? Sep 29, 2021 at 13:08

Did I make any obvious mistakes in this circuit? I was not sure if I have to tie bellRX to ground via a resistor as well? How can I calculate the on time of the opto coupler when I press the the switch near bellTX?

Sharp-eyed Andy has pointed out a flaw in charging the large reservoir capacitors. Q2 does nothing currently...the capacitors are charged via R7. Charging via these resistors is a simple, likely sufficient way to limit current from the +12V supply.

Does only one +12 V DC supply exist, feeding both transmit end and receive end? If so, the 5 m path from transmitter to receiver consists of three wires:

• +12V
• GND
• Key signal

If two 12V supplies (one for transmit-side and the other for receive-side) are used, then only two wires are needed:

• Key signal
• GND

Calculating pulse width when using C5 involves a time constant calculation. Assuming C5 is discharged initially before the key is activated, charge current flows through C5, R8, and infra-red diode inside the opto-coupler, and back to GND.
Initial current flow will be about $$\ {{12V - 1.5V}\over R8}\$$. This current will activate the IRled inside the opto-coupler.
With key pressed, current will decay to zero with time constant mostly due to R8 and C5. $$\ R_8 \times C_5\$$, with units of ohms and farads yields time in units of seconds. The leading edge is quite fast, and initiates the bell's hammer. The trailing edge is slower...somewhat longer than one time-constant, and occurs when the phototransistor inside the opto-coupler decays its current so that the voltage across R9 falls through the MOSfet's threshold voltage (a few volts).

Recovery-time when the key is released involves a different current path through R6, and R5, discharging C5 to zero with time constant $$\ (R_6+R_5)\times C_5\$$.

This circuit barely meets requirements of the opto-coupler, which specifies absolute maximum reverse voltage across the IRdiode of 6 V. At the moment when the key is released, C5 is charged to 12V, and this is distributed evenly between R5 and R6, so the IR-diode momentarily sees a reverse-voltage of 6V (too close to maximum allowed).

Note that the "key" switch is assumed to be bounce-free. In practice, mechanical switches never open or close cleanly, with one edge, but bounce for a few milliseconds between the "open" and "closed" states.

As far as functionality is concerned, this circuit is overkill. While the desire to limit current drawn from the +12V source might be valid, it can be accomplished with simpler methods. An optocoupler isn't needed, nor is a MOSfet switch. Adding them might aid you in exploring their use in other circuits.

• Yes there is only 1 12V supply. The 6 pole telephone cable has in use 1x 12V, 1x ground, 2x switch lines and the remaining 2 are used to conrol leds with. Sep 29, 2021 at 19:19
• My Idea about the optocoupler was that I would need a small current to send signals. I was hoping it would prevent any EM related problems. The mosfet is there because the coil may draw like 2.5A. I do not yet know how long this should be, but I figured that 100ms ~ 200ms would be plausible, that is why I chose for a mosfet. And the CDU can supply this 2.5A where the 12V line cannot Sep 29, 2021 at 19:27