How to trigger a relay when the signal voltage slowly rises from 0v to 12v?

ElectricSpidey

Joined Dec 2, 2017
2,758
Assuming the PWM is low side and the new lamp is PWM dimmable.

AAC_Add_on_Lamp.jpg

The source is connected to positive.
The drain is connected to the lamp.
The gate is connected to the resistors.
 

Tonyr1084

Joined Sep 24, 2015
7,853
Still too much dust on the brain cells. I'm trying to figure out the MOSFET situation but I'm seeing things backwards. Must be my aixelsyd.
 

Tonyr1084

Joined Sep 24, 2015
7,853
OK, yeah, @ElectricSpidey has the right solution. I was missing the low ohm resistor in my circuit. Use his resistor values or use 10 times higher resistors (both) for less wasted current and unnecessary heat.
1640019309270.png
 

MisterBill2

Joined Jan 23, 2018
18,176
I suggested the diode, capacitor, and relay arrangement because it is both simple and forgiving and uses only two parts that might be unfamiliar to the TS. Also, it could be asembled with crimp connections.
 

Tonyr1084

Joined Sep 24, 2015
7,853
I suggested the diode, capacitor, and relay arrangement because it is both simple and forgiving and uses only two parts that might be unfamiliar to the TS. Also, it could be asembled with crimp connections.
As I showed, I already suggested the capacitor in post #2. Didn't suggest the diode because I see the cap as catching any back EMF. If I'm wrong - I'll gladly accept the critique.
 

MisterBill2

Joined Jan 23, 2018
18,176
As I showed, I already suggested the capacitor in post #2. Didn't suggest the diode because I see the cap as catching any back EMF. If I'm wrong - I'll gladly accept the critique.
The diode is mandatory so that the other loads do not discharge the capacitor. The capacitor holds the charge so that the relay will stay pulled in. The diode is absolutely needed. During the pulse on time the capacitor charges, during the pulse off time the diode keeps the rest of the loads from discharging the capacitor, so that the capacitor can keep the relay activated. So the diode allows the capacitor to stretch the pulses.
 

vu2nan

Joined Sep 11, 2014
345
Thank you for sharing this circuit. However as said above, the cabin light has constant positive supply and the dimmer sends PWM signals through ground when the door is opened. So it would be great if you could help me with the circuit for this setup. Thanks in advance
Hi Jayendra,

Here's the circuit, taking the 'ground-end dimmer' into consideration.

1.png

Nandu.
 
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Thread Starter

jayendra

Joined Dec 20, 2021
8
Assuming the PWM is low side and the new lamp is PWM dimmable.

View attachment 255634

The source is connected to positive.
The drain is connected to the lamp.
The gate is connected to the resistors.
The new lamp is a RGB Strip and uses a dedicated controller, which requires 12v - 24v. So it is not PWM dimmable. Using the above circuit, can I add a relay after the MOSFET? Will that work?


The diode is mandatory so that the other loads do not discharge the capacitor. The capacitor holds the charge so that the relay will stay pulled in. The diode is absolutely needed. During the pulse on time the capacitor charges, during the pulse off time the diode keeps the rest of the loads from discharging the capacitor, so that the capacitor can keep the relay activated. So the diode allows the capacitor to stretch the pulses.
A rough sketch of the circuit would be greatly appreciated



Here's the circuit, taking the 'ground-end dimmer' into consideration.

1.png
Thank you. Will work on this and test it out. I assume this wouldn't cause any effect on the original dome lights and the PWM controller.
 

MisterBill2

Joined Jan 23, 2018
18,176
OK, here is a "rough sketch, for those devoid of any visualizing ability: From the non-ground, positive side of the 12 volt PWM power circuit, the anode of a diode. The cathode of the diode, the end with the stripe, connects to both one side of the relay coil and the positive terminal of a capacitor, at least 47mFD. The other side of the capacitor, and the other side of the relay coil, connect to the "ground"(negative) common to the PWM lighting circuit.

At every positive transition of the PWM signal charge flows through the diode into the capacitor, and also flows through the relay coil, continuing to flow when the PWM signal goes back to zero. At some point enough charge flows into the capacitor/relay coil segment so that the relay energizes and the contacts close.

When the duty cycle reaches 100% the relay stays energized and the capacitor remains fully charged, When the PWM slowly reduces, the charge on the capacitor decays through the relay coil until the point where the relay releases and the other circuit is switched off.

I hope this explanation is adequate for those who are unable to visualize very well.
 

Tonyr1084

Joined Sep 24, 2015
7,853
I hope this explanation is adequate for those who are unable to visualize very well.
Nope. It doesn't. Either it's me or it's the painter of the portrait. A "PICTURE" paints a thousand words. Your description only used 171 words. Lacking color and definition. The TS asked if you could draw a picture, not describe the Mona Lisa.

Why are you being so negative? Santa got you on the naughty list?
 

MisterBill2

Joined Jan 23, 2018
18,176
Nope. It doesn't. Either it's me or it's the painter of the portrait. A "PICTURE" paints a thousand words. Your description only used 171 words. Lacking color and definition. The TS asked if you could draw a picture, not describe the Mona Lisa.

Why are you being so negative? Santa got you on the naughty list?
I do not have drawing capability on this computer. And lightning got the other computer that did have drawing programs. And what was left out of that explanation that is so hard to understand??
 

Tonyr1084

Joined Sep 24, 2015
7,853
A pencil, paper and a cell phone with a camera will do. Draw it; snap a picture of it, upload it. No software needed.

For me - at least - your description seems to jump about a bit. Hard to follow and even harder to make sense of it. Sorry, it must be just me but your description doesn't exactly paint a picture for me.
 

MisterBill2

Joined Jan 23, 2018
18,176
Connects to both one side ? ? ? Got me confused.
"both one side of the relay coil and the positive terminal of a capacitor," . This is one sentence, with the word "BOTH" implying that two items follow.
And taking a picture and loading that onto my computer is a very big deal because of two things: First, my phone does not have a camera because some of my clients do not allow cameras in their buildings, and second, because my phone does not talk with that computer, nor does it access the internet.
Other folks who frequent this forum have been quite able to understand my descriptions and even produce correct drawings from them.
I offer no more comment about your complaints.
 

ElectricSpidey

Joined Dec 2, 2017
2,758
The new lamp is a RGB Strip and uses a dedicated controller, which requires 12v - 24v. So it is not PWM dimmable. Using the above circuit, can I add a relay after the MOSFET? Will that work?
No

But it will if you combine it with what MisterBilll2 is trying to explain.

This will place the load of charging the capacitor on the MOSFET instead of the PWM source.
 
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