Schematic Review - Flyback Diode to protect Reed switch from Relay Coil

[randallamoore]

I'm new to designing circuits, and have some very basic questions.

I've designed a pcb to light a sign when a door is closed. I've used an alarm sensor reed switch in the door, which can handle up to 300 milliamps at 30 volts.

When the door is open, the NC leg of the relay sends current to a Green led. So you know the system is on, and the door is open.

When the reed switch closes it energizes the coil in the relay, and the Red led lights up.
It worked fine for awhile, then stuck on Red - the reed switch was stuck closed.

A little googling has led me to believe I need a flyback diode across the relay coil to prevent it from emitting a burst of excess voltage when the coil is switched off. The diagram below shows my circuit.

1. Will the Schottky Diode in my design protect the Reed switch? It looks to me like it will prevent excess current from traveling down the ground path and damaging the LEDs.

2. Does it matter that the reed switch in on the Positive path of the circuit? Yeah, I'm that ignorant. I assume it makes no difference. But I saw a circuit while researching that had the reed switch on the negative leg.

3. Would it make sense to add a diode between the reed switch and the relay? It seems to me that would definitively stop any current from passing back to the reed switch and damaging it.

Thanks in advance. I sure appreciate having a place to post this question.

 

[ronsimpson]

Hope you add a resistor to the LEDs.
It is very common to have a diode across a relay coil. A silicon diode is also fine.

Schottky Diode in my design protect the Reed switch? It looks to me like it will prevent excess current

When the reed switch opens the voltage on the coil will snap down to a large negative voltage. The diode will prevent excess voltage. The large voltage is only there for a very short time.

Positive path ……. on the negative leg.

Dose not matter. Just interrupt the current.

Would it make sense to add a diode between the reed switch and the relay?

no

 

[gophert]

I'm new to designing circuits, and have some very basic questions.

I've designed a pcb to light a sign when a door is closed. I've used an alarm sensor reed switch in the door, which can handle up to 300 milliamps at 30 volts.

When the door is open, the NC leg of the relay sends current to a Green led. So you know the system is on, and the door is open.

When the reed switch closes it energizes the coil in the relay, and the Red led lights up.
It worked fine for awhile, then stuck on Red - the reed switch was stuck closed.

A little googling has led me to believe I need a flyback diode across the relay coil to prevent it from emitting a burst of excess voltage when the coil is switched off. The diagram below shows my circuit.

1. Will the Schottky Diode in my design protect the Reed switch? It looks to me like it will prevent excess current from traveling down the ground path and damaging the LEDs.

2. Does it matter that the reed switch in on the Positive path of the circuit? Yeah, I'm that ignorant. I assume it makes no difference. But I saw a circuit while researching that had the reed switch on the negative leg.

3. Would it make sense to add a diode between the reed switch and the relay? It seems to me that would definitively stop any current from passing back to the reed switch and damaging it.

Thanks in advance. I sure appreciate having a place to post this question.

Add resistor as shown.

 

[randallamoore]

ronsimpson I do have resistors for the LEDs, I left them out since they weren't part of my question. Sorry to have introduced the uncertainty. The circuit actually feeds a single Red/Green led with appropriate resistors, AND sends current on the Red leg on to a strip of Red LEDs inside the sign.

gophert Is that 100 ohm resistor for the LEDs? That's clever to put in on the input. I've got 2 on the output. One for each color on the LED.

Thanks for the feedback. I really appreciate it.

 

[gophert]

Yes a 100 ohm resistor.

if your Green LED has a 3v forward voltage and the red LED is 2V, then you can simplify your circuit with this trick.

 

[Visitor]

The current to operate the relay & the LEDs is probably more than the poor lite reed switch can handle.

A transistor driver can solve the problem or get one of these relay modules from ebay.

 

[randallamoore]

Visitor Your comment interests me. The whole point of the relay was so the current for my real load, a string of LEDs, is NOT pulled through the reed switch. Here's a more accurate schematic. Note that my drawing tool doesn't allow me to properly indicate that the 5V input is tied to the input on the Relay, and the NO terminal on the Relay feeds both the red status led, and the string of LEDs which light up the sign (the real load)

I was using the relay specifically because the Reed switch can't handle the 500 milliamps the LED string requires.

 

[Visitor]

The pertinent question is how much current does the relay require? Sorry I misspoke about the LEDs.

 

[randallamoore]

If I'm reading the datasheet correctly, the coil uses 30 mA at 5V. Well below the 300 mA at 30V the reed switch is rated for.

I'm assuming that the reed switch closes and current flows through the coil. The relay activates, and the the LED string is able to pull the 500 mAmps it needs from the load side of the relay. The Reed switch should be unaffected by that load.

Is that correct?

 

[gophert]

Is that correct?

Yes, that is correct.

 

[rjenkinsgb]

In addition to the flywheel diode && other mods people have already suggested:

Add a low value resistor in series with the reed switch, at the switch end of the connecting cable.
Reed switches are very sensitive to load capacitance and simply shorting the capacitance of a length of cable can cause rapid contact degradation or failure.

Fitting a low value resistor, eg. 10 to 33 Ohms, at the switch end limits the current to a safe level and prevents the switch being damaged.

Many commercial "alarm" type reed switches fitted in enclosures already have a limiting resistor included, or are very high rated; this is only important if you are using a bare glass switch.

 

[Visitor]

My mistake on the current rating of reed switches. A quick look shows many rated at 250mA or more.

 

[randallamoore]

Thank you to all of you for your helpful feedback. This is the absolute best part of the internet. Kind people willing to help out a stranger with their expert knowledge. Thanks for taking the time to read and respond to my post.

Randall

 

[Pommie]

If your LED strings realy take 0.5A then your 100R resistors are too high. Do you have a link to the LED strings?

Mike.

 

[randallamoore]

Pommie . The 100 ohm resistors aren't in the circuit with the LED string. They are between the power and a single status LED.
The LED string has about 20 LEDs, each with its own resistor. It's meant to be powered by USB, and I'm sending 5v from a usb charger, through the relay, straight to the LED string.

For context. I've included a picture of the sign. We just remodeled a bathroom that backs up against the garage. We added a door into the garage, through the bathroom (I KNOW, too weird). The sensor is in the OTHER door. So when someone walks into the bathroom from the house side hallway, and closes that door, this door in the garage displays the Occupied sign. This is a convenience for someone entering from the garage - so they don't walk in on someone using restroom.

Below is the door (we just finished, I haven't painted yet). The second picture is the pcb I had made by OshPark. $12 for 3 board. You can't beat it. you can see where I re-routed a resistor, and added a jumper. I'll get a new board printed with the info I've learned here. It just occurred to me that I can use a single 100 ohm resistor on the status LED's cathode, rather than putting a resistor on the red and green anodes!

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