# current limiting in phototransistors

J

#### Jeff Lawlis

Jan 1, 1970
0
I am using 5 infrared phototransistors in parallel (Radio Shack part #
276-142). The emitters have a rating of 5V reverse voltage and 1.7V
max forward voltage. How can I calculate which resistance to use (in
series with the battery I presume) to limit the current when using a
9V battery? Do I wire the emitters so that the cathode is connected
to the positive end of the battery? I realize that these are novice
questions, but I would appreciate the help. (I am not using the
detectors included with the emitters).

Jeff

J

#### John Popelish

Jan 1, 1970
0
Jeff said:
I am using 5 infrared phototransistors in parallel (Radio Shack part #
276-142). The emitters have a rating of 5V reverse voltage and 1.7V
max forward voltage. How can I calculate which resistance to use (in
series with the battery I presume) to limit the current when using a
9V battery? Do I wire the emitters so that the cathode is connected
to the positive end of the battery? I realize that these are novice
questions, but I would appreciate the help. (I am not using the
detectors included with the emitters).

Jeff

You will be using the emitters in the forward direction (that is the
only way that emit anything). You can wire all 5 in series and put a
small resistor in series with that to limit the current to the desired
amount, but the current will fall rapidly as the battery voltage
sags. But this is definitely the most efficient way to drive them,
because you get to use the same battery current to run all of them.
The 1.7 volt spec may be a maximum value, instead of typical, so a
resistor between 50 and 100 ohms may work okay.

Are you at all interested in battery life, or is maximum output your
only concern? If life is no concern, you could drive them separately
or in smaller series groups with larger resistances in series with
each (or each group.) For instance, you might put a pair in series
with a 560 ohm and a triplet in series with 390 ohms to drive all 5
with about .01 amps (10 milliamps).

J

#### John Popelish

Jan 1, 1970
0
Jeff said:
I am using 5 infrared phototransistors in parallel (Radio Shack part #
276-142). The emitters have a rating of 5V reverse voltage and 1.7V
max forward voltage. How can I calculate which resistance to use (in
series with the battery I presume) to limit the current when using a
9V battery? Do I wire the emitters so that the cathode is connected
to the positive end of the battery?
(snip)

The cathode of the emitter is the more negative terminal. You add up
all the forward voltages (1.7 volts, each, in this case) of the
devices you will operate in series, and subtract that total from the
available voltage (9 volts, in this case). Then you divide that
excess voltage by the current you want through the while series string
to solve for the resistance needed to limit the current ot that amount
while it burns up the excess voltage.

For instance, to drive a single emitter, the excess voltage is
9-1.7=7.3 volts. If you want ot drive the emitter with .01 amp (a
conservative amount for many emitters), you divide that voltage by .01
to get a resistance of 730 ohms. Standard 5% values near that are 750
ohms and 680. If you drive all 5 emitters this way, the total battery
drain will be 5 times .01 amp or 50 milliamps.