I have a simple question about calculating resistor.
I have lots of leds driven by negative, wired with one resistor for each one and 2 leds wired together. I can't modify this as it's enclosed in a box. This is the schematic on the left.
I need to lower the current in the leds, to lower the light amount.
My only option is to add a resistor, like on the right schematic. One resistor for 2 leds+resistors.
How can I calculate the new current flowing through the leds or equivalent resistor? Everything I've found on internet is parallel or serial network, but never parallel + serial.
You know the voltage drop across the left hand network (5V) and you can measure the current flowing through it. That tells you its resistance (Ohm's law).
Now imagine the whole thing is just a big resistor with that value... the rest is easy.
If the LEDs are different colours you are likely to run into problems this way - if the same
colour then it will work better (but current distribution won't be as balanced - this may not matter)
Another approach is to PWM the supply to the existing LEDs - reducing the average brightness
without changing the resistors.
You know the voltage drop across the left hand network (5V) and you can measure the current flowing through it. That tells you its resistance (Ohm's law).
Sort of... Maybe, "Close enough for government work"
The resistance of an LED is not constant like a resistor... An LED is a non-linear device. Ohms Law is always true, but when you change the current through the LED, its effective resistance changes.
The voltage drop across the LED is (approximately) constant under normal operating conditions. When you increase current, its effective resistance decreases. If its resistance was constant (like a resistor), the voltage would increase proportionally with the current.
DVDdoug:
The voltage drop across the LED is (approximately) constant under normal operating conditions. When you increase current, its effective resistance decreases.
I know...but I think it's (fairly) safe to assume the LEDs are in the mostly-linear zone, not the exponential zone.