Ok. Let me see by removing the diode. Dont get angry. I am new to this field. I always welcome all the replies and suggestion all you made. Will take care of this. Thank you

No one including me said to remove a diode. Look at your own drawing in post #13. Take out C1 and the connection(Not the diode) from the bottom of D2, the connection to ground (negative) at the bottom of D2. The connection is shorting pin#5 directly to negative, making D2 useless.

 

No one including me said to remove a diode. Look at your own drawing in post #13. Take out C1 and the connection(Not the diode) from the bottom of D2, the connection to ground (negative) at the bottom of D2. The connection is shorting pin#5 directly to negative, making D2 useless.

@shortbus
I do not follow you. D2 is properly connected as a flyback diode across the coil of the relay. When the "turn indicator" signal from U2:1 is turned off (i.e. open or ground), then the voltage at pin2 of the relay will fly back negatively (relative to ground) and will be clamped at ground by D2. At the same time, D1 will be forward biased...except that its anode is either open or grounded. If D1's anode is grounded, then D1 performs in parallel with D2; if D1's anode is open (ignoring D4), then no current will flow through D1.

NOTE: I am not saying the circuit performs the intended function; I am only saying that D2 is properly connected to the relay coil.

 

I am only saying that D2 is properly connected to the relay coil.

Look again at the schematic in post 13, unless he has edited it. The diode is correctly connected as a flyback, EXCEPT, for the connection between the anode of D2 and the negative wire, that shorts out the path through D2. Either that or I(and a couple of others) are missing what you and the TS are saying.

 

Look again at the schematic in post 13, unless he has edited it. The diode is correctly connected as a flyback, EXCEPT, for the connection between the anode of D2 and the negative wire, that shorts out the path through D2. Either that or I(and a couple of others) are missing what you and the TS are saying.

U mean i will remove the negative wire from D2 anode. If so the how the relay will run? If possible u make a circuit comple and post it pls

 

Look at the circled area

 

Look at the circled area
View attachment 185517

Yes . Saw. Will i remove that wire?

 

Yes . Saw. Will i remove that wire?

Maybe you need to draw a schematic or even just note what the connections are in what your calling U2 and U3?
"U" is commonly and usually signifying "integrated circuit or IC" on a schematic, so then tell what U is.

 

Maybe you need to draw a schematic or even just note what the connections are in what your calling U2 and U3?
"U" is commonly and usually signifying "integrated circuit or IC" on a schematic, so then tell what U is.

@shortbus
Come on! What the TS has labeled as "U2" is obviously (to me) a connector that brings the turn signal voltage to its pin1. I personally would have labeled the connector "J2" but different people have different conventions.

I stand by my previous statement. If pin1 of the "assumed" connector does bring a voltage signal (e.g. +12V and/or ground), then the diodes are okay as shown and the ground is okay as shown.

 

I stand by my previous statement. If pin1 of the "assumed" connector does bring a voltage signal (e.g. +12V and/or ground), then the diodes are okay as shown and the ground is okay as shown.

I would too if it wasn't for that pesky pin #3 on U3. But then I'm old.