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LED Flasher with light bulb break detection. Sat Jan 29, 2022 5:55 am
Nent2
New member
I was able to replace the flasher with an LED.
Failer light bulb of the flasher valve is detected by U243B of TELEFUNKEN Semiconductors.
The current flowing through the front and rear flasher bulbs flows through the 30mohm resistor
between pin6 and pin7 of the U243B, causing pin7 to generate a voltage according to the current.
If the voltage is less than 81 mV, disconnection is detected and the blinking speed becomes faster.
Replace that 30mΩ with a register suitable for the LED current.
The current consumption of the flasher relay, flasher indicator lamp and flasher relay unit was 0.21A.
(I am replacing the flasher indicator lamp with a 20mA BA9S type LED.)
The current consumption of the LED bulb made in China purchased from Amazon Japan was 0.2A/1pcs.
So, the current with 1pcs LED is 0.21A + 0.2A = 0.41A.
And the current with 2pcs LED is 0.21A + 0.2A + 0.2A = 0.61A.
Assuming that 0.45A, which is between 0.41A and 0.61A, is the disconnection detection current.
(Since the LED that I bought is driven by a constant current, the current changes from 0.2A to 0.16A
when the voltage rises from 12V to 14V, so the disconnection detection current was set to 0.45A.)
The resistance value at which 81mV is generated when a current of 0.45A flows is calculated.
R = 81mV / 0.45A = 0.18Ω ( I make 0.18Ω with 0.2Ω and 1.8Ω parallel connection.)
Note: LEDs are not standardized, so you need to measure the current consumption of your LED
to determine the parameters.
And inexpensive Chinese LEDs can get hot, so it is safer to add an overcurrent protection fuse
to flasher bulb circuit (or inside of LED bulb that you bought).
Disable the 30mΩ resistor.
Cut the 30mΩ resistor and remove the vertical metal rod.
Install a 0.18Ω register here.
best regards.
Failer light bulb of the flasher valve is detected by U243B of TELEFUNKEN Semiconductors.
The current flowing through the front and rear flasher bulbs flows through the 30mohm resistor
between pin6 and pin7 of the U243B, causing pin7 to generate a voltage according to the current.
If the voltage is less than 81 mV, disconnection is detected and the blinking speed becomes faster.
Replace that 30mΩ with a register suitable for the LED current.
The current consumption of the flasher relay, flasher indicator lamp and flasher relay unit was 0.21A.
(I am replacing the flasher indicator lamp with a 20mA BA9S type LED.)
The current consumption of the LED bulb made in China purchased from Amazon Japan was 0.2A/1pcs.
So, the current with 1pcs LED is 0.21A + 0.2A = 0.41A.
And the current with 2pcs LED is 0.21A + 0.2A + 0.2A = 0.61A.
Assuming that 0.45A, which is between 0.41A and 0.61A, is the disconnection detection current.
(Since the LED that I bought is driven by a constant current, the current changes from 0.2A to 0.16A
when the voltage rises from 12V to 14V, so the disconnection detection current was set to 0.45A.)
The resistance value at which 81mV is generated when a current of 0.45A flows is calculated.
R = 81mV / 0.45A = 0.18Ω ( I make 0.18Ω with 0.2Ω and 1.8Ω parallel connection.)
Note: LEDs are not standardized, so you need to measure the current consumption of your LED
to determine the parameters.
And inexpensive Chinese LEDs can get hot, so it is safer to add an overcurrent protection fuse
to flasher bulb circuit (or inside of LED bulb that you bought).
Disable the 30mΩ resistor.
Cut the 30mΩ resistor and remove the vertical metal rod.
Install a 0.18Ω register here.
best regards.
K1100RS/LT - R1200RT - R1100RS - Cagiva SST 350 Ala Verde - K75LT project - K75 Schurgers - K75S - K1100RS - K75RT - K75C
