Little transistor questions....
Hi guys, I'm busy with a small project useing a PIC to operate some relays....
I need a bit of help with the practical side of things now, I'm quite happy with my pcb layout, schematic etc... sadly not the code, but my problem right now is learning about switching transistors....
What I think I learned so far... (the tranistors will be NPN)
you need at least about 0.7V between the base and the emiter.
I read that "Ic = hFE × IB"
So lets say my relays need about 20mA, to switch on...
And I have a measured transistor with hFE of 100
Would IB needed then be 0.020 / 100 i.e. 0.0002mA (is this uA?)
I know I can copy popular component values etc. from existing circuits, but that won't be teaching me these basics...
You can keep the explanation pretty low level, as my knowledge of transistors is still in the cocoon stage.... :D
If you don't want any scientific explantion, a 10 kohms resistor to base from the MCU. Relay coil to the collector. Don't forget to add a diode 1N4148, 1N4001 or whatever across the relay coil. Cathode of the diode faced towards supply voltage. Without this diode you will kill the transistor.
Thanks man, lol was looking for slightly more scientific approach.... Should have said explain like you would to a smart kid....
found this on the big ol' interweb in the meantime... does it look about right?
OK so lets see what we have so far
VS: 9V supply (relay needs little over 8V to switch)
Chip supply 5V
Relay coil 270R
Max chip current I want to draw from each PIC's pins should be about 5mA, may go lower if I see it is possible...
load current Ic = supply voltage Vs / load resistance RL
Ic = 9V/270R = 0.033.mA
hFE(min) > 5 × ( load current Ic / max. chip current )
hFE(min) > 5 x (0.033/0.005)
hFE(min) > 33
RB = Vc × hFE where Vc = chip supply voltage
= 5 x 100hFE assuming you use 100hFE transistor
right so far??
Does this mean the higher the hFE the lower current you need to base to switch on the transistor?
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