Pulsed DC to Steady DC in Cars

Hello once again, so after my other plan for turn indicators as DRLs has failed, I have come up with another way to solve it. Here it goes:

The goal is to pull a fused 12V line from the battery and wire it to the normally closed terminal of an SPDT relay. The normally open terminal will be connected to the factory wire coming from the flasher relay unit. The common terminal will go to the bulbs. I will be adding a simple RC circuit that will convert the pulsing 12V DC coming from the flasher relay into a steady 12V DC to control the relay. Which means, when turn indicators are being used, the SPDT switches from the resting state (powering DRLs) to energized state (powering the bulbs from factory lines). Circuit diagram is below (I hope its clear as I just mocked it up in Gravit Designer).

Assumed Values:

• Power source: 12V
  
• Capacitor's capacitance: 0.01F
  
• Resistor's resistance: 20ohms
  

So a couple of questions starting from the simple one:

1. What kind of diode should I use to prevent the capacitor back feeding the flasher relay?
   
2. If you noticed it already, at the SPDTs the factory lines (red: one for each side [left/right]) are basically open circuit. Which means when the indicators are activated there will be no load (or infinite load: not sure which one) here which will cause hyper-flashing. You that super quick flashes your indicators make when one of the bulbs is blown? That. When this happens, pulses of 12V are very brief (period also). Will it be enough to charge the capacitor and the SPDT eneregized?
   
3. Are the capacitance and resistance values I used feasible (too high? too low?)?
   

Thanks,

Tamrat

Ok, after searching a bit on the web I found an interesting write on how to build a "switchback circuit" from Instructables. The post was EXACTLY what I was looking for. So let me just answer my own questions for other interested people (follow the link to see his version).

1. What kind of diode should I use to prevent the capacitor back feeding the flasher relay?

Based on the Instructables post, a 1N4007 will do the job. The 1N400X is a family of diodes where the X can be from 1-7 depending on the current requirement. Personally I think even the 1N4001 will do as the relay draws very little current but using the 1N4007 is also fine.

2. [...] Will it be enough to charge the capacitor and the SPDT energized?

It should be enough. I put together the components roughly and tried feeding the relay with a brief +12V (like a quick tap on the battery positive). The brief 12V was enough to energize the solenoid and also charge the capacitors. The relay was left energized for about 2 seconds before it clicked back as the capacitor ran out of charge. In theory, if the relay is energized, the OEM turn indicator lines will be connected to the bulb and therefore no hyper-flashing.

3. Are the capacitance and resistance values I used feasible (too high? too low?)?

The capacitor's value (0.01F) is a ridiculous amount. Most capacitors are in the uF range. Based on the post from Instructables, a 6600uF will get the job done and I confirmed this. The resistor isn't required at all.

With that out of the way, here is the improved wiring diagram for the complete circuit. The pulsing-to-steady DC conversion is achieved by capacitors.



Tamrat,

Hi there,

A 6600mf capacitor?? Are you sure?? That is huge i would have thought.. [as in massive] [nay, gigantic!!]

john...

apprentice87 said:

Hi there,

A 6600mf capacitor?? Are you sure?? That is huge i would have thought.. [as in massive] [nay, gigantic!!]

john...

Click to expand...

Hello John, its not mF (milli farads), it's uF (micro farads). 6600uF is the total capacitance I used (as per the steps in the link). I couldn't find a single 6600uF unit therefore I also purchased 3 capacitors each having a capacitance of 2200uF (but they were marked with 16V on them [hmm....]) and wired them in parallel to get a total capacitance of 6600uF. I don't know much about capacitors so I don't know what their voltage rating play here. Maybe its the maximum potential difference that can be stored on the capacitor. When I tested it, the values were perfect, I have no other word for it. A brief tap on the positive terminal of the battery was enough to charge up the capacitors to 12V and they kept the relay energized for about 2 seconds. Which is more than enough considering power will come and go in about 0.5s (maybe) intervals when turn indicators are being used.

Tamrat,