03-14-2014, 10:58 PM
I don't see any reason to exactly duplicate the current drain of the bulb. Back when Fisher designed the circuit, they used a bulb which needed 35 mA, which was about as low as was available. If they were to design the same circuit today they would use a LED and drive it with 10 mA. So there is no real reason to burn off the excess current as heat with a very low value shunt resistor. Even if you exactly matched the current drain, the LED will still behave differently because it turns on at a much lower voltage than the incandescent bulb.
If you reduce the value of C11 the 100 uF cap, it will decrease the delay in the LED turnoff. With the lower drain of the LED, the cap value could be reduced proportionally.
Experiment with the shunt resistor and see how it behaves, but I certainly would not go below 1.5 K or so.
If you reduce the value of C11 the 100 uF cap, it will decrease the delay in the LED turnoff. With the lower drain of the LED, the cap value could be reduced proportionally.
Experiment with the shunt resistor and see how it behaves, but I certainly would not go below 1.5 K or so.
