Creating a proper control signal

I'm making a circuit that involves a rotary latch mechanism. I'm going to use a 12V power supply for the latch and a 3.3V supply for the microcontroller.

The part I'm hung up on is the current draw on the rotary latch's control signal line. It says it will sink up to 25mA on the signal line when activated (granted that's at 24VDC, but it's the only datapoint they give). My original thought was to put a pull-up resistor from the signal line to 12V. Then an n-channel FET between the signal line and ground would control the latch. But if the signal line sinks even just a few mA, the pull-up resistor might drop too much voltage.

Next I thought of a push-pull circuit, but then there's the complexity of shoot-through prevention. Am I overthinking this, or is this actually a complex problem that a FET and a few passives can't solve by themselves?

Thanks!

They don't quite give you everything you need in the spec do they?

If you already have one test to see if it is a positive or a ground that activates it. My bet is a ground level picks the latch in which case a logic level fet or a transistor would provide the level shift you need for your micro.

Oh you're just saying let it float when it's inactive and just put a single transistor to high or low (whichever it is) to activate it. No need to drive it when it's inactive. That's a good thought.

I think I would short the control signal to ground to see if it runs. If it doesn't then shrot it to +12. Then we would know what polarity it wants to see to make it turn.

You could use a CMOS buffer such as the CD4050 which will drive both high and low on the output. Connect two or more of the buffer inputs and outputs in parallel to increase the load drive capability.

I think I would short the control signal to ground to see if it runs. If it doesn't then shrot it to +12. Then we would know what polarity it wants to see to make it turn.

Click to expand...

Agreed in concept. Unfortunately, there's a 30 day lead time to get the part. I've left a message with their tech support people, so hopefully one of their engineers should get back to me soon.

They don't quite give you everything you need in the spec do they?

Click to expand...

Agree completely, by the way. I find this is especially true when the part is heavily mechanical in nature but has an electrical component to it. It's like they bar the sparkies from the room when they lay out the datasheet for a new part. "Electrical specifications?? Pssh, potential customers don't need any of that. I'm sure each and every one of them will know exactly all of the assumptions and industry standards that we may or may not have employed."

I'm not sure the 4050 would interface with your 3.3 volt micro when running on 12 volts, but 3 of these guys would.

https://www.electro-tech-online.com/custompdfs/2013/02/1924109.pdf

Or if you don't mind burning some power you could just use a NPN like a 2N2222 with 470 ohms in the collector to +12 and about a 10K in series with the base to your micro output. The problem with this is the 470 ohm needs to be 1 watt.

My impression is that the external control signal must source current from a positive supply which is at least 12V. The source current required is 25mA if the source supply is 24V. Can we assume that it would half that (12.5mA) if the source is 12V?

Since it takes 12 to 24V to power the lock, then I would use that supply as the source of the control signal current. Here is my approach:

I wholly agree with MikeMI's approach. The statement "It says it will sink up to 25mA on the signal line when activated (granted that's at 24VDC)" gives sufficient data.

Hmm, I guess it does say that.
Input Signal Current Draw: 25mA
MAX at 24 VDC

It seems quite clear as to it's operated in the Wire Color Code / Connector Pin Assignment section. Coil requires 8-26V, and the control signal needs 8-26V to operate; there's a single ground. MikeMI's post (#8) shows an appropriate drive circuit.