I wanted to post this in the electrical tech info section but I guess I don't have permission to do that.
My any load (within reason) turn signal flasher. If you don't want to read the story or how it works just look at the circuit diagram (don't laugh at my MSPaint skills) and my parts list.
This is a circuit that I built for my 87 Delta 88. I purchased some LED tail light turn signals and found that they no longer flashed. The reason is this: the stock flasher is a two pin mechanical type. This has a bi-metallic strip inside with one end fixed and connected to one of the pins on the outside of the flasher unit. The other end rests against a contact that is attached to the other pin on the outside. One pin is connected to twelve volts when the ignition is on and the other to the turn indicator switch. When the switch is put in either the left or right positions the flasher is connected to either the left or right indicator bulbs through the switch. The other side of the bulbs are connected to ground. At this point the bulbs turn on. The bulbs are a fairly significant load and this causes the bimetallic strip to heat up. Because this strip is made of two different metals that expand at different rates the strip bends away from the contact. This makes a “click” sound. Now that the circuit is broken the lights turn off. With no current flowing through the flasher the bimetallic strip cools down and snaps back into place. This makes the other “click” sound. If the indicator switch is still in either the left or right position the process starts again. The first time the bulbs come on they usually stay on a little bit longer than the following times as the strip is usually cooler.
Plugging the LED bulbs in just the rear reduces this load to the point that the bimetallic strip does not heat up enough to bend and break contact. So the bulb just stays on until the indicator switch is returned to the middle position.
There are a number of solutions to this problem. The easiest is to place load resistors in parallel with the LEDs. This will get the stock flasher to work, but it involves cutting the wiring harness to the bulbs. And if the LEDs were installed in the hopes of reducing the load on the electrical system, this would defeat that entirely. Places that sell LED bulbs will usually sell LED flashers as well. You have to be careful with these though and make sure you know what load they can handle. If something happens and you need to return the LED bulbs to standard incandescent lights you have to make sure that the flasher unit can handle that load. I also find these units to be a little overpriced.
As I have several boxes of spare electronic parts I decided to make my own. I did some hunting on the internet and found several circuit diagrams that had flasher units, but none of them would really work for me.
The only one I found that was designed for a two pin flasher is this one:
How to Build a Heavy Duty 12 Volt Flasher Unit? Detailed Description Using Circuit Schematic
I put this together on my breadboard (not using the exact same components, but equivalent ones). After some testing I found that it was very sensitive to changes in voltage. When testing at twelve volts it worked, but at fourteen it changed its flash rate and didn’t turn the LEDs off completely. I played around with it a bit but decided to go in a different direction. Also, this unit has no audible indication that it is working. Someone has modified the diagram and posted it with a buzzer that goes in time with the lights to fix this though.
The simplest one was to use a 555 timer to produce a square wave to drive a switching transistor. This is one of the most popular types I found while searching. Yes this will flash the lights, but it would be “on” all the time with only the output of the transistor going to the indicator switch. I started to play with this one a bit too. I used the transistor to drive a standard Bosch style relay though, which gave me the clicking sound. The trouble I had was getting this to turn on only when the indicator switch was in the on position (without re-wiring the cars turning indicator circuit, I want to be able to return this to stock if need be).
It was in trying to figure out this solution that I came up with my solution to the flasher unit. Here is my final circuit diagram:
Here is the circuit broken down:
+V is the hot in run from the fuse. It is usually twelve to fourteen volts. GND is chassis ground.
The relay is a standard Bosch style five pin automotive relay. The coil contacts are 85 and 86. When one of these are connected to ground and the other to a voltage source this activates the relay. Pin 30 is the common connection. Pin 87 is the normally open connection. When the relay is not active it is not connected to anything internally. When the relay is active Pin 30 and Pin 87 are connected. Pin 87a is the normally closed pin. When the relay is not active this is connected to Pin 30. When the relay is active this isn’t connected to anything internally. In this circuit pin 30 is connected to +V through D1 and D2, pin 87a is connected to the turn signal indicator switch, Pin 87 is left open, pin 85 is connected to +V and pin 86 is connected to the collector of Q2.
The two diodes D1 and D2 are just there to prevent the base of Q1 from being connected to +V all the time.
Q1 is a PNP transistor switch. The emitter (the part with the arrow) is connected to +V. The base is connected to R1 and the collector is connected to R2. When the base “sees” ground it turns on the transistor switch and +V gets through to R2. When there is no ground at the base the transistor switch is open and R2 is disconnected from +V.
R2, R3, and C1 make up an RC charging/discharging circuit. When Q1 is on, C1 charges through R2. When Q1 is off and the voltage on C1 is greater than zero, it discharges through R3. Internet search RC Charging circuit if you want to know about this. D3 is just to insure that C1 doesn’t discharge through Q1.
R4 and R5 make a voltage divider circuit. The voltage between the two can be calculated as:
In this case Vout is the reference voltage for the next bit.
U1 is an op amp configured as a voltage comparator. The input voltage goes to the “+” pin. The reference voltage goes to the “-“ pin. There are two more inputs and they are Vcc+ and Vcc-. The way this works is that it compares the input voltage to the reference voltage and outputs ~Vcc+ (if the input voltage is greater than the reference) or ~Vcc- (if the input voltage is lower than the reference voltage). The output is generally about one volt less than Vcc+ and one volt more than Vcc-. In this case Vcc+ is +V and Vcc- is ground.
The diode D4 is in place only to ensure that the second RC circuit does not discharged into the output of the op amp.
C2 and R6 make an RC discharge circuit. This time there is no resistor to charge through so the capacitor charges very quickly.
R7 and Q2 form an NPN switch. The base of Q2 is connected to R7, the collector is connected to pin 86 of the relay, and the emitter is connected to ground. Unlike the PNP switch, this one goes active when the base “sees” +V. So when the base sees +V, the transistor is switch on connecting the collector to ground through the emitter and activates the relay.
Put it all together and this is what it does:
When the car is on +V is at twelve to fourteen volts. When the indicator switch is put in either the left or right position the base of Q1 “sees” ground through the load (bulbs) and turns on. This charges C1 through R2. Once the voltage on the positive pin of C1 is greater than the reference voltage on the comparator, the comparator’* output swings to +V. This charges C2 very quickly and turns on Q2. This turns on the relay and removes +V from the indicator switch turning off the lights. With the base of Q1 no longer connected to ground (through the load) it turns off and C1 discharges through R3. At the same time C2 discharges through R6. Once the voltage at the positive pin of C2 is low enough it turns off Q2 which in turn turns off the relay. This reconnects pins 30 and 87a on the relay, and if the indicator switch is still on it starts all over again.
-When the indicator is first turned on it takes longer to turn off for the first time because the positive pin of C1 is near zero. After the circuit has cycled once C1 will not have time to discharge completely so it won’t take as long for it to reach a voltage greater than the reference voltage.
-To increase the amount of time the light is off, increase the value of R6.
-To increase the amount of time the light is on, increase the value of R2.
-The load that this circuit can drive is limited only by the relay. In my case I’m using a 30Amp relay, which is more than enough for incandescent bulbs and/or LEDs.
-When power is first applied Q2 trips the relay for a second. I have yet to fix this. It isn't all that important as I am unlikely to need the indactor to work within the first second of turning the car on.
-Not pictured in the circuit but something that is a good habit to get into is to put a diode in parallel with the relay coil with the stripe pointing to +V. This would protect the transistor Q2 from voltage spikes when the coil discharges.
Here are the parts that I used (bear in mind that I didn’t actually do the proper calculations for the base resistors on either transistor, I just threw in some parts I had on the bench):
D1, D2, D3, D4: 4N1001
R1, R5: 3.3 KOhm
R2: 100 KOhm
R3, R6: 180 KOhm
R4: 10 KOhm
R7: 15 KOhm
C1, C2: 100 uF (both rated to 25V)
Q1: General purpose low power PNP transistor (can’t read part number)
Q2: NTE36 (way overkill)
Again, these are all parts that I had laying about, so equivalents can be used.
I apologize for the cell phone pics and vids.
Here is a picture of the proof of concept on a bread board:
Here is a video with an LED bulb (note the power switch of the power supply represents turning the car on and the top switch on the red switch block represents turning the indicator on):
20130606_090824_zpsbe53e098.mp4 Video by Berg9987 | Photobucket
Here is a video of the same circuit with a standard bulb (note the current needle on the right jumps much higher than with the LED bulb):
20130606_090917_zps0f118f9e.mp4 Video by Berg9987 | Photobucket
(I couldn't get the vids to embed)
One last note: I the D4, R6, C2 part of this circuit was an afterthought. Originally I was trying to use a feedback resistor from the output of U1 to the input to set up timing through hysteresis. However I found that this was very twitchy and opted instead to use the second RC circuit to hot Q2 on.