| Electronics > Beginners |
| Power disipation through Opto PC817 |
| (1/1) |
| rscott:
Hey Guys Just teaching myself a bit more about opto's Question is, I know how to bias one and have gotten good results that match the CRT ratios and such but curious how to calculate the power dissipated. I don't know what the internal resistance of the opto is so how do I see how many watts or more likely mWatts I am running? thanks |
| magic:
As always, voltage times current. How much voltage is dropped for given current or how much current flows for given voltage you will find in the datasheet (forward voltage vs forward current plot). It typically is 1~1.5V for reasonable currents, IIIRC. |
| rscott:
However there is no voltage drop the way this is setup. So there is no voltage there but there is current flowing through it. So if you don't know the internal impedance of the BJT side of the opto how do you know the power dissipated The load is connected above the collector and the emitter is connected to ground |
| magic:
Okay, I thought you mean the primary. For the secondary, it's of course current times voltage again, as usual. And this time you need the plots showing how secondary side voltage drop (Vce) varies depending on secondary side collector current (Ic) and primary side LED current (If). My PC817 daatsheet from Sharp has two plots, one showing Vce vs If for a few different values of Ic and another showing Vce vs Ic for a few chosen If. The specifications table also shows "collector to emitter saturation voltage", which is guaranteed never to be exceeded as long as If is high enough and Ic is low enough. It is specified as 0.2V maximum for If=20mA and Ic=1mA in my datasheet. |
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