I'm studying optocouplers in general, and in special the 4N25.

I'm applying the optocoupler in a zero crossing circuit, like this one:


I know the CTR is minimum of 20%, but checking the datasheet of 4N25, I'm wondering how to get the aprox. CTR based on IF and ambient temperature:


So, my question is, how to determine if is in saturation mode or not? Which curve could I look?


1 Answer 1


You don't look at any curve - you learn what saturated means. The general definition is that the collector-emitter voltage is less than the base-emitter voltage. Since the base of a 4N25 is available, you measure that. For most NPN silicon transistors, you can assume a base-emitter voltage at currents of 1 mA or more to be about 0.7 volts.

Your graph is useful, but not for the curve. See the conditions at the upper left? It shows that saturation (in this case) was defined as collector-emitter voltage of 0.4. This is a decent rough number, but could quite possibly get as low as 0.1 or less for high LED currents and large load resistors.

  • \$\begingroup\$ I got your point @WhatRoughBeast, but what I was trying to do is like, assuming the IF = 10mA, can I consider the CTR = 0.7 (sat) or 0.8 (non-sat)? Can I say this? remind the If is in 120Hz. \$\endgroup\$ Jun 8, 2016 at 18:15
  • \$\begingroup\$ @BrunoAraujo - No, but you can say that the CTR is 0.7 or 0.8 of the nonsaturated CTR at If = 10 mA. And you cannot speak of "the If is 120 Hz". The current value is an instantaneous reading (or a DC level, if you prefer). That's at 25C. At 50 C, as your figure 3 suggests, the CTR will be lower. \$\endgroup\$ Jun 8, 2016 at 18:25
  • \$\begingroup\$ Yes, I took the picture at 50C because 25C is without the scale. Sorry to insist, but is not clear to me, how do I know when to use the nonsaturated or the saturated CTR? \$\endgroup\$ Jun 8, 2016 at 18:34
  • \$\begingroup\$ @BrunoAraujo - Start by assuming the output is not saturated. From If and the CTR curve, calculate the output current. Multiply this by the load resistor. If the resulting voltage is greater than the supply voltage, you know you're saturated. \$\endgroup\$ Jun 8, 2016 at 19:12

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