Hi all,
I have a 5V power regulator that provides power to my entire circuit. The regulator has an enable pin with a pull-up resistor. When this net is pulled down, the entire system shuts off as expected.

I am using the following toggle switch on my enclosure. In its ON position, it shorts to GND, and in its OFF position, it opens. Therefore, when I toggle it to ON, my system turns off, and when I toggle it to OFF, my system turns on.

I am working on a new PCB that will require the same switching option. What can I do in my circuit to reverse this logic? I am thinking a simple transistor circuit should be sufficient.

HUAREW Rocker Switch ON/Off KCD1 Mini Round Toggle Switch SPST 2Pin Snap-in Latching with Wires Pre-Wired AC 6A/250V 10A/125V,for car Boat(10Pcs) : Amazon.co.uk: Automotive

 

Connect the pullup resistor to ground, and the switch between the enable pin and +5V.

When the switch is open, the enable pin is pulled low. When the switch is closed, the enable pin is high.

 

Connect the pullup resistor to ground, and the switch between the enable pin and +5V.

When the switch is open, the enable pin is pulled low. When the switch is closed, the enable pin is high.

The 5V is the output of DC-DC. The input will be 12V.

Ok so as a default the DC-DC needs to off by the pull down. And once 12v is supplied to ENB pin it switches on.

So the switch pass 12V

 

The 5v regulator may not like 12v on the enable pin. Check the datasheet to be sure.

 

The input to the dcdc is 12v

 

I suggest that IF the toggle switch can be reversed, THAT would be the way to go.

Really, I do not understand the reason for the reversal. So can the TS explain in more detail the reason for the requested change??

 

This is my circuit below, which includes a module.

Unfortunately, I have purchased a large quantity of toggle switches and would prefer not to return them as they are already wired for a custom setup.The switches short the circuit when set to the ON position and remain open when set to the OFF position. Therefore, I need to reverse the operation to match the power supply's behavior according to the switch symbol; otherwise, it would function opposite to expectations.

I am considering removing R2 and replacing it with a 47kΩ pull-down resistor. Then, I could use the switch to provide 12V to the ENB port, possibly through a 10kΩ resistor in series.

 

Jon has given you the solution in post #2. Disconnect the top of R2 from Vcc and connect it instead to Gnd. Connect the switch between Vcc and En.

 

Ok, should I also keep a series resistor between Vcc and Enb? I am trying to limit current to the Enb pin.

 

Like this.
The enable input has an internal "weak" pulldown resistor no need for series resistor.
There will be more current through R2 which can possibly be set much higher or may not be needed.

 

I advise reading and understanding the data sheet for the regulator IC, and learn what voltage and current are required to enable and disable the regulator. And now I understand that the switch is a PCB mounted device, and not one that can simply be inverted. So the scheme suggested in post #2 should work. BUT, to verify that it is OK, you need to read the data sheet.

 

Why not ground the enable pin and put the switch on the power? You can't get power out if you're not getting power in.

 

Why not ground the enable pin and put the switch on the power? You can't get power out if you're not getting power in.

Would need to connect the Enable pin to the Vcc pin for that to work.
I suppose the switch can't handle the load.

 

Would need to connect the Enable pin to the Vcc pin for that to work.
I suppose the switch can't handle the load.

OK, yeah, I did get that backwards. Connect enable to positive and use the switch to turn the whole thing on and off. IF the switch can handle 10A @ 120VAC it can handle 3A @ 12VDC. I can't see that chip requiring or handling that much amperage. But one would have to investigate the circuit amperage. And yes, I know the switch doesn't say it can handle X amps at DC volts. 12 volts doesn't seem like it's going to arc across the contacts with less than 50 amps.

 

Didn't notice this before:

The selling label said 6A @ 250V, 10A @ 125V. Did not specify AC or DC. I took it to mean VAC.

 

Exactly. I think that's good to go.

 

Connect the pullup resistor to ground, and the switch between the enable pin and +5V.

When the switch is open, the enable pin is pulled low. When the switch is closed, the enable pin is high.

When I made this reply, the regulator type had not been disclosed. I see it has been now.

My answer in post #2 is correct... but it's even easier than I stated. The chip has a weak pull down. If no connection is made to the EN pin, the output is off. If the EN pin is connected to Vcc (the input voltage), the output is on. The maximum voltage on EN is the same as the maximum Vin voltage.

The EN pin is a logic-level input. There's no worry connecting it directly to Vin – it will draw insignificant current.

There's no need to complicate this. Connect the switch between EN and Vin to achieve the desired result. Connecting the power switch to control Vin is undesirable for two reasons:

● The switch has to carry the full load current (sure, it's rated for it, but why stress it?)

● Switching Vin may cause transients on the output voltage as the regulator starts up.

The simple solution is to use the chip as designed, which makes the switch markings correct.

 

J.C. Is totally correct!!
IN ADDITION, the switch is evidently a PCB solder-in mount type, and the PCB it is on does not carry the mains power. So the very simple circuit revision is the way to go!!

If the TS is not able to interpret such a straight forward explanation, Oh Well..
There is no reason to make such a simple change into a complex "KLUDGE".

 

IN ADDITION, the switch is evidently a PCB solder-in mount type

No it's not.
Snap in panel mount.
The link is in post#1

 

Snap in panel mount.
The link is in post#1

That doesn't not alter my correct, simple suggestion in post #2 nor my slight simplification of post #17 when the chip was identified.

But please, feel free to argue amongst yourselves