On 10-Jan-02 Michael Vinson wrote:
> James Paul wrote, in part:
>>[...]
>>  Lets say we have an NPN transistor and that transistor has a Beta of
>>  100.  Lets further say that the circuit we are powering uses 300 mA of
>>  current.  To drive this transistor full on, we'll make the Base current
>>  about 10% of the Collector current.  So in this case, the Base current
>>  will be ~30 mA.   Now lets say that the PIC pin will output a worst case
>>  logic high level of 4.5 volts.  So, if we divide the 4.5 volts by 30 mA,
>>  we get a value of 150 ohms.  So, if we use a 150 ohm resistor in the
>>  Base, and connect a PIC pin to the other end of the resistor, and make
>>  the PIC output a logic high, [...]
>
> The problem with this configuration is that you are asking the PIC
> i/o pin to source 30mA of current. According to the 16F84 data sheet,
> for example, under "Absolute Maximum Ratings", the most current that
> can be sourced by any i/o pin is 25mA. So it sounds to me as though
> you are asking for more current than the PIC can comfortably supply,
> as well as much more than is actually needed (you only need a few
> mA going into the base).

Michael

The problem with this is that James' calculations above are wrong IMHO.

If the transistor  has a beta (Ic/Ib) of 100, and the collector current is
300mA, then the base current is 3mA, NOT 30mA.   I'm not sure where the "10% of
the collector current" rule came from, but I think it's wrong.

The only thing I can think of is the rule that a common emitter bias network
should be designed with a standing current of ~10 times the base current.

Peter.


----------------------------------
E-Mail: Peter Onion <ponion@srd.bt.co.uk>
Date: 10-Jan-02
Time: 16:00:58

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