led
I thought of making a 12 LED light with input power of 48v and 20 mA. Is there any circuit using only transistors to act as a switch running on the above inputs. I intend to use the above circuit with LDR which should activate the above 12 LEDs when it is dark.
Kindly help with the circuit diagram. I am already using the above circuit with a switch and now, out of curiosity, I intend to use it in my bedroom for lighting purpose.
A 555 isn't really required for a simple task like this. You can make the LED latch with another transistor and a few extra components:
When the voltage at the junction of the photoresistor and R4 rises high enough, Q1 turns on, turning on the LED. Q1 pulling low also turns on Q2, which charges C1 and latches Q1 in its ON state. Discharging C1 with the switch turns the LED off, and assuming it is sufficiently light (the resistance of the photoresistor has decreased), Q1 and the LED will remain off.
Here is how you can wire your LEDs. The resistors used are 150 ohms and 1/4 watt. This is assuming that you have 20 ma flowing through each LED.
Wiring Diagram:
LED Polarity:
Best Answer
It can easily be done with just a transistor, like this:
If your LEDs are common indicator LEDs (which the 20 mA suggests) they may drop for instance 2 V. Then 12 of them in series is 24 V, the remaining 24 V will be across the series resistor R. For 20 mA R can be calculated as
\$ R = \dfrac{V_+ - V_{LEDs}}{20 mA} = \dfrac{48 V - 24 V}{20 mA} = 1.2 k\Omega \$
A simple solution like this has the disadvantage that it's not efficient because the supply voltage doesn't match the required 24 V. You'll lose half of the total 1 W, i.e. 500 mW in the series resistor, so take a 1 W type for that.
Also make sure your transistor has a high enough \$V_{CE}\$ specified. The BC546B can have 65 V, so that's OK, and it also has an \$h_{FE}\$ of minimum 200, so you won't need too much base current.
If you're vexed by the low 50 % efficiency then there's a solution for this, but you'll need more than a transistor. A switching LED driver can very efficiently take care of the 24 V difference between power supply and LED voltage.
The LED driver doesn't require many external parts, but it isn't cheap. This one is the cheapest I found at Digikey and it's 2.50 dollar in 1s. But it has a great efficiency of up to 95 %, which means you lose only 25 mW instead of 500 mW.