Diagram, converting rear turn signals to run lights.(2 relays, 2 diodes)
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I did take up your challenge but stopped when I realised that the ground had to be switched.
Usually the bulb holder is grounded by virtue of the indicator housing being metal, or in some cases plastic that has chromed film, yes it does conduct.
This is my version. (L is a switch, normally open, closed by a relay when left indicator is powered )
When running, both lamps at half power.
Left indicator on, left lamp full power, right lamp off.
And same for the right indicator.
Usually the bulb holder is grounded by virtue of the indicator housing being metal, or in some cases plastic that has chromed film, yes it does conduct.
This is my version. (L is a switch, normally open, closed by a relay when left indicator is powered )
When running, both lamps at half power.
Left indicator on, left lamp full power, right lamp off.
And same for the right indicator.
I redrew your diagram to understand it better. If my "your" diagram is done right, it will not work because right bulb will blink with left turn.
1.Run is fine.
2.right turn is fine
3.Left turn will make right bulb blink. (off/half-bright)
Sorry, try again!!!!
When relay is closed, everything is fine. (left is full on, right no current)
When relay opens (it works like flasher), right bulb and left bulb go in mode "run".
So, in your "left turn " mode left bulb will blink full/half, right one will blink half/off
It can't be done. Your circuit will change the brightness of the lamp on the other side.
1. When running both lamps are in series hence half brightness.
2. When one side is indicating, that bulb is at full power so it is impossible for the bulb on the other side to be at half power.
1. When running both lamps are in series hence half brightness.
2. When one side is indicating, that bulb is at full power so it is impossible for the bulb on the other side to be at half power.
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Run is 12v from running lights, connected to a resistor which is connected to the lamp.
Value of resistor chosen so lamps are at half power.
L and R are power to indicator (the circuit is duplicated, one side has L, other R)
When either L or R are powered, bulb is at full brightness, else at half brightness.
Note the resistor could be the second filament of a 2 filament bulb so basically the circuit someone posted a earlier.
I'm working on a circuit that when one side is indicating, the bulb goes off completely.... watch this space. (it may take a while as I'm trying to use the least number of components)
Value of resistor chosen so lamps are at half power.
L and R are power to indicator (the circuit is duplicated, one side has L, other R)
When either L or R are powered, bulb is at full brightness, else at half brightness.
Note the resistor could be the second filament of a 2 filament bulb so basically the circuit someone posted a earlier.
I'm working on a circuit that when one side is indicating, the bulb goes off completely.... watch this space. (it may take a while as I'm trying to use the least number of components)
One possible circuit is shown below (one for each side)
TR1 and TR2 (PNP Darlington pair) operate as a switch that connect the indicator to running power via resistor R2. R2 is chosen so that lamp is at half power so may not be 12 ohm.
When indicator power is applied, R2 is bypassed by diode D2 so that lamp is more or less at full power. At the same time C1 gets fully charged. The high voltage on base of TR2 turns it off, turning off TR1.
When indicator power is turned off, capacitor C1 discharges via resistor R1. This causes the voltage on the base of TR2 to drop until it is low enough to turn TR2/TR1 on so the lamp, which was off, is again at half power.
The value of R1 and C1 are chosen to be a little bit more than the indicator 'off' time.
When indicator is off, the lamp runs at 1/2 power.
When indicator is on, the sequence is full power, off, full power, off ......
When indicator is turned off, after the last lamp off sequence, the indicator returns to run at 1/2 power after the time determined by C1/R1. You will need to play with value of C1 and R1.
An approximate analogy using water to represent electricity is below.
The outlet pipe (R1) empties the bucket (C1). Increasing the diameter of the outlet pipe is equivalent to lowering the value of R1. Increasing the size of the bucket is equivalent to increasing the value of C1. Both of these factors determine how quickly the bucket is emptied. Inside the bucket is a float on a pivot that operates a switch (TR1/TR2) when the float drops to a specific level.
The bucket is constantly being emptied but when indicator power is applied, it gets filled instantly. Once indicator power is removed, the water level (voltage on C1) starts to drop and eventually the switch is operated so the lamp, which was off, runs at half power.
TR1 and TR2 (PNP Darlington pair) operate as a switch that connect the indicator to running power via resistor R2. R2 is chosen so that lamp is at half power so may not be 12 ohm.
When indicator power is applied, R2 is bypassed by diode D2 so that lamp is more or less at full power. At the same time C1 gets fully charged. The high voltage on base of TR2 turns it off, turning off TR1.
When indicator power is turned off, capacitor C1 discharges via resistor R1. This causes the voltage on the base of TR2 to drop until it is low enough to turn TR2/TR1 on so the lamp, which was off, is again at half power.
The value of R1 and C1 are chosen to be a little bit more than the indicator 'off' time.
When indicator is off, the lamp runs at 1/2 power.
When indicator is on, the sequence is full power, off, full power, off ......
When indicator is turned off, after the last lamp off sequence, the indicator returns to run at 1/2 power after the time determined by C1/R1. You will need to play with value of C1 and R1.
An approximate analogy using water to represent electricity is below.
The outlet pipe (R1) empties the bucket (C1). Increasing the diameter of the outlet pipe is equivalent to lowering the value of R1. Increasing the size of the bucket is equivalent to increasing the value of C1. Both of these factors determine how quickly the bucket is emptied. Inside the bucket is a float on a pivot that operates a switch (TR1/TR2) when the float drops to a specific level.
The bucket is constantly being emptied but when indicator power is applied, it gets filled instantly. Once indicator power is removed, the water level (voltage on C1) starts to drop and eventually the switch is operated so the lamp, which was off, runs at half power.
Your diagram should show:
2 bulbs
3 inputs: run, left turn, right turn.
My diagram has 2 diodes and 2 relays, so if yours has more components, you cannot claim title "Best turn signal converter in the world"
Transistors are definitely better than relays, please specify if they are normally open or closed and if they are engaged by voltage or ground.
2 bulbs
3 inputs: run, left turn, right turn.
My diagram has 2 diodes and 2 relays, so if yours has more components, you cannot claim title "Best turn signal converter in the world"
Transistors are definitely better than relays, please specify if they are normally open or closed and if they are engaged by voltage or ground.
No. Your diagram does not work. I explained this in my post on Saturday at 5:58 PM
When Left indicator is fully on, the right indicator goes off.
When right indicator is on, left indicator is off.
So in both cases when the indicators are flashing, one side goes from on to half power; the side that is not meant to be indicating goes from on to off.
In the latest circuit I posted has the side that is indicating go from fully on to off with the other side unchanged at half power. (I know it works as I mocked it up on a breadboard).
re: my circuit above, 1 component less. Remove R2 and select R1 so that it limits current through the bulb to be half power. R1 will depend on the Hfe of TR1 and TR2, the value of C1 will need to be reduced.
When Left indicator is fully on, the right indicator goes off.
When right indicator is on, left indicator is off.
So in both cases when the indicators are flashing, one side goes from on to half power; the side that is not meant to be indicating goes from on to off.
In the latest circuit I posted has the side that is indicating go from fully on to off with the other side unchanged at half power. (I know it works as I mocked it up on a breadboard).
re: my circuit above, 1 component less. Remove R2 and select R1 so that it limits current through the bulb to be half power. R1 will depend on the Hfe of TR1 and TR2, the value of C1 will need to be reduced.
