Do you have a question? Post it now! No Registration Necessary
- Posted on
- Pic Currrent
Re: Pic Currrent
I run a PIC16F628, four EEPROMs, and one LED (LED is ON for only 100ms
ever sec.) from a 9v (actually 7+ volt) NiCd battery. LED anode is
connected to the 78L05 output through several hundred ohms resistor, LED
cathode is connected to a PIC pin which pulls it low to turn the LED
on. I have used this circuit for about a year and it works fine.
As long as the total current of your circuit never exceeds the
current-supplying capability of the battery or the 78L05, your circuit
should work fine too.
"Just do it."
Re: Pic Currrent
and ma as required by the circuit otherwise.
The following is a reply to the same problem discussed in another newsgroup.
pick a resistor that you believe will drop a tolerable amount of voltage at
the average current. For example, if you believe the average current is
150uA, you might pick 1K 1% which will drop 150mV at that current. Then put
a capacitor across it to smooth the reading at the repetition rate of
whatever pulses etc. are being drawn from the supply. Maybe you pick a time
constant of 1 second, so 1000uF electrolytic. This is a single pole LPF. You
use the meter in voltage mode, of course. In stubborn cases, you may wish to
add a second pole, say a 100K resistor and 10uF tantalum. That won't affect
the reading with a >10M input Z, the average current is still 1mV/uA within
a couple of percent.
If the voltage turns out to be too large or small - so it drops too much
voltage to the circuit, or the voltage is too small to measure easily, then
just change it (and the capacitors)
Note that if current is drawn in short pulses you have to make sure that the
instantaneous decrease in the voltage is not too high or the reading may be
inaccurate or the circuit may malfunction. A 'scope is useful in this
regard. You can also calculate it to bracket the problem.
Eg. Supply the 150uA is essentially all composed of pulses: 150mA for
200usec with a 200msec period. A 100uF capacitor would see a voltage change
at the terminals of delta-V = I * Ton/C = 300mV, which is a bit high. The
1000uF would yield a more reasonable 30mV, so the troughs would be 180mV
down from nominal, not 450mV down. The 30mV/300mV p-p of ripple will yield
jumpy readings as the 5Hz frequency "beats" with the measurement cycle of
the meter. This is where the second pole filter comes in handy.
- » OT: "Your" money and Duped by the Soviets
- — The site's Newest Thread. Posted in » Electronics Design