audioguru said:
The datasheet for the TIP120 shows it saturating well at 500mA with a base current of only 2mA.
The datasheet for a PIC says its max output current is 25mA.
So why not feed the TIP120 base 10ma?
The output of the PIC is about 4.0V with a 10mA load.
The input of the TIP120 is about 1.3V at our currents.
So the current -limiting resistor is (4V - 1.3V)/10mA= 270 ohms.
I have been unable to locate the pic output voltage/current relationship in the PIC data sheet. Audioguru states 4.0V at 10mA, where do I look in the pic Data Sheet to find this information?
If I understand the gain of the TIP120 = 1000, if the base current is set to 10mA, then .01A * 1000 would mean I could drive a theoretical load of 10 amps, is this correct?
1.3 or 1.4VDCVDC I believe is the voltage drop across the darlington pair, is this correct?
ULN2803A
I also want to to layout a prototype board with the ULN2803A (8 outputs). If I understand the ULN2803A the +12VDC is pulled to ground through the darlington pair. Is this ground connection the actual ground pin (9) on the IC itself? The reason I ask is that I would like to socket the ULN2803A, but most DIP socket pins have a maximum current ratings of 1 Amp. If 4 solenoids are ON at the same time then this would be 2AMPS of current through the ground pin, not good.
I have been using another formula for my calculations, which produces the same result.
((Drive Voltage - 1.4) / Current Required) * 1000
The main issues for me right now is where do I find the TIP120 information "saturates well at 500mA with base current of 2mA" and the "output Voltage/current relationship" for the PIC. The TIP120 data sheet I found has no graphs in it, and I looked in the PIC data sheet but could not locate the output voltage/current relationship.
Thank you for the learning experience!