Hello. this is about the turn signals of my Honda CB250. it's got regular incandescent bulbs and recently they stopped working. the problem was the signal relay itself. i fixed it with a new one.

1) upon checking my Honda wiring diagram i've got a small question. i made this simplified diagram of my turn signals setup.


For instance let's say we turned on the right side turn signal switch. so basically +12V goes through the 10A fuse and reaches the signal relay. then it goes to the turn signal switch. after that it goes to the two right side 21W bulbs and finally grounds. so those two right side bulbs begin to flash.
at the same time power goes to that 1.7W indicator bulb. but the only ground path for that indicator bulb lies through a left side 21W bulb. so my question is, isn't that bulb going to flash too with the indicator bulb ? but as we know, practically only the indicator bulb is flashing. can some one explain the theoretical reason for not flashing that left side 21W bulb.

2) i took out the circuit board of my old relay.


i found out that there are several types of signal relays like thermal, electromagnetic and solid state. as we see on the diagram one pin is a dedicated ground. so is this relay a solid state one ? just asked since we don't see some kind of coil on the circuit board too.

thank you so much.

Moderators note: uploaded images , external sites may dissapear.

 

Hi Auto
We cannot open your image links??

Moderation.

 

I already uploaded the images and modified the post.
@ericgibbs

 

The signal bulb is designed for a current of 1.75 A (21 W/12 V) whereas the indicator bulb is designed for only about 140 mA, so you have a resistance of the hot filament of about 85 ohm. This means the 140 mA won't heat up the filament of the signal bulb very much since it has a hot resistance of 6.8 ohm. But when it is cold, the resistance is less.

Yes, the current flows through the signal bulb, but it is not enough to heat the filament. It was a very common way to use a low resistance device as the return path to GND for a higher resistance device.

 

Hello. I also noticed that the two left signal lamps are in parallel, as are the right side. Since the left side is providing a pathway to ground, having them in parallel will divide the current between them and further reduce the resistance to ground. Remember, there are two signal bulbs to each side. If one lamp is 6.8Ω when hot then two bulbs in parallel will present 3.4Ω.

 

@ericgibbs sorry. now they are working.
@bertus thx for fixing the links.
@Rf300 thank you so much for your detailed explanation.
@B-JoJo-S thx a lot for your clarification. yea you are right.

 

RF300 provided an accurate explanation that is easy to understand AND totally correct!