Date: 2/19/2004 11:16 AM Central Standard Time
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The 6mA (continuous DC) per output pin is in the ballpark for HC logic. So
is 6mA adequate for driving the 4N26? Yes it is, but it depends upon how
much output current you want the 4N26 output stage to provide.
For this post I'll refer to this datasheet for the 4N26:
http://www.fairchildsemi.com/ds/4N/4N26.pdf
Under the transfer characteristics the CTR for this device is specified as a
minimum of 20% with a test LED current of 10mA. This 10mA rating is just a
simple test condition. You can drive the LED with quite a bit less or quite
a bit more than this (from 0-100mA DC, more for pulsed operation) and the
optocoupler will still work.
So what does the CTR mean? It stands for Current Transfer Ratio. If you
multiply the CTR by the forward LED current you can get a rough idea of how
much current the output transistor of the optocoupler can provide. For
example: suppose you drive the LED with a forward current of 6mA. Then the
minimum guaranteed current the optocoupler output transistor can sink (or
source depending upon circuit configuration) is theoretically something like
6mA * 0.20 = 1.2mA. Basically this behaves very much like a bipolar
junction transistor with a beta of about 1/5 or 0.2. Other optocouplers
with CTRs greater than 100% are available and are not necessarily more
expensive than the 4N26. Take a look at the large array offered by your
favorite distributor such as Digikey or Mouser.
In practice the above formula (LED current * CTR = output current) isn't
quite perfect. The problem is the CTR is not a perfect constant property of
the optocoupler. Instead the CTR is somewhat a function of LED current.
Look at figure 4 of the datasheet for instance. For small LED currents the
CTR is somewhat lower than at 10mA. For instance it appears with a LED
current of about 1.8mA the CTR drops to a mere 0.8*0.20 = 16%. On the other
hand with an LED current of about 6mA the CTR is slightly higher than the
10mA test current conditions at around 1.08 * 0.20 = 22%. At very high and
very low currents relative to the 10mA test current we can see the CTR drops
quite a bit.
So the bottom line is with the 4N26 you can drive it with 6mA, but you will
be limited to something in the range of 1.2mA of output current. So this
means in order to drive your 150mA relay you would need a PNP transistor
with a minimum DC gain of around 150ma/1.2mA = 125. Generally you want to
drive both the optocoupler and PNP transistor with a bit more current to
insure good saturation in all conditions.
If this problem were up to me I would certainly not use the 4N26. It is a
rather pitiful device, and doesn't even really command much (any?) price
advantage over other superior optocouplers. Some device like the NEC PS2501
for instance provides more bang for the buck.
Datasheet at:
http://www.cel.com/pdf/datasheets/ps2501.pdf
Notice this part comes in a variety of flavors. One type comes with two
optocouplers in a single package while another type has four inside. It
sounds like your project may need more than one, so this might be a good
choice.
For pricing and looking for other possible choices use your favorite
distributor's catalog such as:
http://dkc3.digikey.com/pdf/T041/1200-1205.pdf
