But, if you're only using 15mA per segment, you'll only need about 105mA max collector current and about 10.5mA base current so use base resistors of 4.3/10mA = 430 Ohms, which is a standard resistance value. With 20mA coming in on the base, you'll be able to sink ~ 200mA total on the collector. 61mA/segment, and you're multiplexing them for a 1/4 duty cycle. If you divide 4.3mA between 7 segments, they will be awfully dim about. That will saturate the transistors up to only ~4.3mA collector current. Right now you are using 10k Ohms for the base resistors. The minimum resistance you should use from a PIC output to control an NPN base when the Vcc is 5v is 4.2/20mA = 210 Ohms 220 is the closest standard value. The base resistors for the NPN's are a big problem. Then re-calculate your current limiting resistors. Any idea? I suggest you try running 15mA through one of the segments to the common cathode connected to ground, and see what the Vf is. I don't know what the Vf of the LEDs in your displays are at what current. You also need a 0.33uF cap to ground on the input, and 0.1uF on the output to ground. However, a 7805 needs ~ 2v more input voltage than the output voltage so to get 5v out, you need 7v in. Admittedly, that's not the standard naming for those connections, but in schematics, the inputs usually come from the left, outputs flow towards the right. Looks to me like you have the 7805 connected backwards you're feeding 6v in to the VREG and expecting the output to come from LINE VOLTAGE. You willdesign, simulate, and create a Sixty Second Timer.EquipmentCircuit Design Software (CDS)Digital MiniSystem (DMS)omyDAQomyDigital ProtoboardoCmod.
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