Newbie needs help! Relay diagram question

mik3

Joined Feb 4, 2008
4,843
Disconnect the 20K from the drain and connect only pin C. Maybe pin C does not require a pull up to the positive supply but only to be switched to ground.
 

mik3

Joined Feb 4, 2008
4,843
Well, I don't know what is going on but anyway. Use the MOS to drive a relay and use the relay contacts to pull pin C down to zero volts.
 

mik3

Joined Feb 4, 2008
4,843
Maybe they will work don't build one from discrete components. Buy a chip which contains NOT gates inside and use it. Take care to buy a CMOS one as to be able to connect it to 12V.
 

Thread Starter

merzatt

Joined Jan 15, 2009
43
mik,

What I discovered today may give a clue why strange things happened to this circuit:

1) voltage on pin C of the radio with nothing else connected to that pin is 11.6 volts and 0.3mA. This is regardless of the parking lights ON or OFF.

2) while the multimeter leads connected to PinC and ground, I switched the multimeter from voltmeter to ohmmeter or ammeter. And guess what, the LCD went to DARK mode!
I switched back to voltmeter or disconnect the lead from the multimeter, it goes back to DAY mode. Regardless of what the parking lights ON or OFF!
 

mik3

Joined Feb 4, 2008
4,843
mik,

What I discovered today may give a clue why strange things happened to this circuit:

1) voltage on pin C of the radio with nothing else connected to that pin is 11.6 volts and 0.3mA. This is regardless of the parking lights ON or OFF.
I thought of that and that is why i told you to remove the resistor between drain and 12V.

2) while the multimeter leads connected to PinC and ground, I switched the multimeter from voltmeter to ohmmeter or ammeter. And guess what, the LCD went to DARK mode!
I switched back to voltmeter or disconnect the lead from the multimeter, it goes back to DAY mode. Regardless of what the parking lights ON or OFF!
This is happening because in the ammeter function the multimeter has a very low internal impedance and pulls pin C to ground.

Is it working now with the circuit you posted on post #42 ?
 

mik3

Joined Feb 4, 2008
4,843
Yes you can use a BJT with a 2.2K resistor between its base and pin A. If pin A provides PWM then you will need to use a latch circuit like a RS latch to keep the output voltage constant.
 

Thread Starter

merzatt

Joined Jan 15, 2009
43
I wanted to start with the safe approach and used 2.2k. And it worked on the first try! That means pin A was PWM and that's why the previous circuits did not work.

Here is the latest diagram. Pin A comes from a dimmer and I actually jump wired it therefore also connected to the backlight LEDs and microprocessor too. Using 500ohm vs 2,2k effect the brightness of the LED by any chance? Let me know if I need to change anything else.

You are very patient and awesome. Thank you so much for staying with me to figure out this. :)
 

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mik3

Joined Feb 4, 2008
4,843
I wanted to start with the safe approach and used 2.2k. And it worked on the first try! That means pin A was PWM and that's why the previous circuits did not work.

Here is the latest diagram. Pin A comes from a dimmer and I actually jump wired it therefore also connected to the backlight LEDs and microprocessor too. Using 500ohm vs 2,2k effect the brightness of the LED by any chance? Let me know if I need to change anything else.

You are very patient and awesome. Thank you so much for staying with me to figure out this. :)
By replacing the 2.2K by a 500 ohm won't affect the brightness of the display and may destroy the processor, so leave as it is. Also, remove the 10K resistor which is connected between the base and the emitter.
 

Thread Starter

merzatt

Joined Jan 15, 2009
43
Also, remove the 10K resistor which is connected between the base and the emitter.
I thought the 10K first resistor limits the discharge rate into the base, so the transistor stays on between PWM pulses. The second 10K resistor ensures the transistor turns off when the input is off. Am I wrong?
 

mik3

Joined Feb 4, 2008
4,843
I thought the 10K first resistor limits the discharge rate into the base, so the transistor stays on between PWM pulses. The second 10K resistor ensures the transistor turns off when the input is off. Am I wrong?
You are right about the first resistor it should stay there. The second 10K resistor just discharges the capacitor more quickly and may cause problems if the PWM duty cycle is reduced much. If you remove it the transistor will turn off but in a bit more time than if the second 10K is there which is not a big problem.
 

Thread Starter

merzatt

Joined Jan 15, 2009
43
You are right about the first resistor it should stay there. The second 10K resistor just discharges the capacitor more quickly and may cause problems if the PWM duty cycle is reduced much. If you remove it the transistor will turn off but in a bit more time than if the second 10K is there which is not a big problem.
If I turn the dimmer all the way down, transistor doesn't turn on. But just a touch of dimmer increase is sufficient. Is this the problem you are describing because of that second 10k?
 

mik3

Joined Feb 4, 2008
4,843
If you remove the second 10K you will be able to decrease the dimmer more. If the transistor has enough gain, then you can increase the first 10K to a higher value and be able to decrease the dimmer more.
 

Thread Starter

merzatt

Joined Jan 15, 2009
43
If you remove the second 10K you will be able to decrease the dimmer more. If the transistor has enough gain, then you can increase the first 10K to a higher value and be able to decrease the dimmer more.
I removed the 10k. Just like you said, I am can move the dimmer lower now.

I also increased the 10k to 15k. But if dimmer is at the lowest end, transistor still does not turn on. Should I increase it more than 15k or just leave it there and don't turn the dimmer to the lowest end?
 

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