This came from another thread and I am trying to make sense of it...can anyone please explain?
Given the circuit below:
Assume that the relay coil needs 40 mA @ 5V to energize.
The data sheet for the 6n139 is here https://www.mouser.com/ds/2/239/6N138S-TA1-1149909.pdf
The data sheet lists High output current – 60mA (p1) and a MAX average output current of 50mA.
Based on those characteristics, my original thinking was that the circuit will work - the NPN can energize the coil. [There is about 3.3 mA to the LED and the chip has a CTR above 200].
Then, I looked at this characteristic:
Output Collector Power Dissipation Po 100 mW MAX (p 8), and I thought, no, this will not work because 5V X 40 mA=200 mW.
Apparently I am quite wrong - I don't mind that so much, but I really don't like not being able to calculate the Po - how is it done, specifically, given these components? That is my question.
From here https://www.allaboutcircuits.com/textbook/semiconductors/chpt-4/transistor-ratings-packages-bjt/
I can read.
"Power dissipation: When a transistor conducts current between collector and emitter, it also drops voltage between those two points. At any given time, the power dissipated by a transistor is equal to the product (multiplication) of collector current and collector-emitter voltage. Just like resistors, transistors are rated for how many watts each can safely dissipate without sustaining damage. High temperature is the mortal enemy of all semiconductor devices, and bipolar transistors tend to be more susceptible to thermal damage than most. Power ratings are always referenced to the temperature of ambient (surrounding) air. When transistors are to be used in hotter environments (>25o, their power ratings must be derated to avoid a shortened service life."
I guess, in my mind, I was think that drop is going to be 5V. How do I calculate that drop? What measures are available in the data sheet to make the calculation?
Can someone please explain?
Given the circuit below:
Assume that the relay coil needs 40 mA @ 5V to energize.
The data sheet for the 6n139 is here https://www.mouser.com/ds/2/239/6N138S-TA1-1149909.pdf
The data sheet lists High output current – 60mA (p1) and a MAX average output current of 50mA.
Based on those characteristics, my original thinking was that the circuit will work - the NPN can energize the coil. [There is about 3.3 mA to the LED and the chip has a CTR above 200].
Then, I looked at this characteristic:
Output Collector Power Dissipation Po 100 mW MAX (p 8), and I thought, no, this will not work because 5V X 40 mA=200 mW.
Apparently I am quite wrong - I don't mind that so much, but I really don't like not being able to calculate the Po - how is it done, specifically, given these components? That is my question.
From here https://www.allaboutcircuits.com/textbook/semiconductors/chpt-4/transistor-ratings-packages-bjt/
I can read.
"Power dissipation: When a transistor conducts current between collector and emitter, it also drops voltage between those two points. At any given time, the power dissipated by a transistor is equal to the product (multiplication) of collector current and collector-emitter voltage. Just like resistors, transistors are rated for how many watts each can safely dissipate without sustaining damage. High temperature is the mortal enemy of all semiconductor devices, and bipolar transistors tend to be more susceptible to thermal damage than most. Power ratings are always referenced to the temperature of ambient (surrounding) air. When transistors are to be used in hotter environments (>25o, their power ratings must be derated to avoid a shortened service life."
I guess, in my mind, I was think that drop is going to be 5V. How do I calculate that drop? What measures are available in the data sheet to make the calculation?
Can someone please explain?