I want to replace the LED pilot light in one of my guitar amps. The power source in the amp is the 6.3VAC tap on the power transformer. The published specs for the LED I want to use are:
I have seen two different ways to connect this. One has the anode of a IN4002 diode connected to the negative lead of the LED and the resistor of appropriate value connected in line on the positive lead. The other way shows the diode connected across the pos and neg leads of the LED, with the resistor inline on the positive lead.
Advice please.
- operating range 5-14VDC
- voltage drop 2.8V (typical)
- @ 20mA
I have seen two different ways to connect this. One has the anode of a IN4002 diode connected to the negative lead of the LED and the resistor of appropriate value connected in line on the positive lead. The other way shows the diode connected across the pos and neg leads of the LED, with the resistor inline on the positive lead.
Advice please.
Thank you. By "correctly" do you refer to "reverse polarity" connection of the diode - i.e. cathode of diode to the positive lead of the LED and anode of the diode to the negative lead of the LED?
Circuits 1 and 2 are electrically identical in performance.
Circuits 3, 4, and 5 are also electrically identical in performance.
Circuits 3, 4, and 5 are also electrically identical in performance.
Nowadays small 3/5 mm LEDs are typically quite bright (even cheap ones) so they don't need 20mA current to be seen. It can be rather closer to 0.1-0.5 mA.
And, there is a concern of a blinking. If it blink too awfull with single-diode rectification I would suggest to use a diode bridge, or even possibly a filter capacitor too.
And, there is a concern of a blinking. If it blink too awfull with single-diode rectification I would suggest to use a diode bridge, or even possibly a filter capacitor too.