![]() Although they should be a lot cheaper than the IRF520s. The 2N7000 MOSFET will handle up to 200mA and a TO-92 package, but kids being kids The IRF520 MOSFET is a TO-220 package, and slight overkill being able to pass 9-Amps or so, having said that, it's unlikely to be blown up on a small battery supply. A total of 3.6 Watts power required for 3 Watts of light. So (10.5v*300mA) = 3.15 Watts dissipated in the LEDs and (1.5v*300mA) = 0.45 Watts dissipated in R. Using LED1+LED2+LED3+R setup, each LED drops a total of 10.5v. And for all three parallel circuits that’s a total of 10.8 Watts of power required to get 3 Watts of light. That’s (3.5v*300mA) = 1.05 Watts dissipated by the LED and (8.5v*300mA) = 2.55 Watts dissipated by the resistor. Using LED1+R1||LED2+R2||LED3+R3 setup, each individual LED+R circuit will use 300mA and will have 3.5v dropped across the LED and 8.5v dropped across the R. So the closer you can get to the supply voltage with your series fV’s, the less power is lost in the resistor for any given LED current.ġ2v supply and 3 white power LEDs, each needing 300mA with a fV of 3.5v. The higher the voltage dropped by the resistor, the more power is wasted by it as heat. Each LED drops it’s fV and the remaining voltage is dropped by the resistor. The advantage to having LEDs in series is obvious. Same reason why you can’t use normal Diodes in parallel. ![]() It relies on an exactly matched forward voltage drop for each LED, and even a tiny difference will significantly bias most of the current to the LED with the smallest fV, thereby heating it more than the other which reduces it’s fV even more. “( LED || LED ) + R” should never be done. I’m sure many of you know this, but for those who don’t… Yes, one package would replace 8 discrete transistors – but you still need a series resistor for each LED. Is the benefit of the ULN2803 in reducing the number of components?Ī. Others have in fact reduced the current to as little as 5mA and still retained visible LED indication but a higher current may still be worthwhile in the classroom environment when ambient light levels and viewing angles are not necessarily ideal. If they are “normal red LED’s with a maximum rating of 20mA, then the 330 Ohm resistor will limit the current to avbout 12 mA which is still very readable. That will help to achieve equal current and therefore equal light output level from each LED.Ī starting point would be say a 330 Ohm resistor in series with each LED. What you do need to do is install a separate series resistor for each LED rather than one (100 Ohm) resistor for the entire 8 LED’s. I suggest that you stay with the 4.5/5.90 Volt supply. There is no current reduction for the same illumination level. Raising the voltage from 4.5 volts to 9 volts requires that you drop a greater voltage across the series resistors which equates to greater losses.
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