Power dissipated through zener

[bauerjack]

Hello,
I need some assistance calculating the power dissipated by CR24?

- 200VDC is applied across V5+ and V5-
- no jumpers are installed on J11.
- CR22 has a regulator current of 2mA



Thanks

 

[schmitt trigger]

82volt times the current.

The current is the sum of the 2mA regulator, plus 3.5mA contributed by the transistor (which is configured as a constant current source).

So it will be (0.002 + 0.0035) * 82 = 0.451 watt

If you ask me, this is an excessively complicated and expensive circuit to drive the optocouplers.

 

[bauerjack]

Hi,
I agree Could you please explain how you calculated the 3.5mA

Thank you.

 

[betwixt]

Keeping the Bto E voltage constant means the transistor runs as a constant current device. The TL1431 stabilizes 2.5V on the base with respect to the top of R9, the Vbe drop in the transistor is about 0.7V (at max can be 0.9V, slightly higher than most transistors) so the voltage across R9 must be 2.5 - 0.7 = 1.8V. That means the current through R9 can only be 1.8/523 = 3.44mA. The voltage across the bias network (R10,R11,CR22) is 200 - 18 - 51 - 82 - 2.5 = 46.5V which adds an additional 1.5mA. With all the links removed CR22 will be unable to stabilize 2mA so I have ignored it. The total should therefore be 3.44mA + 1.5mA = 4.94mA. The power dissipation in CR24 is therefore 82 * 0.049 = 0.4W

There will be a degree of error in the calculation due to the natural variation in Vbe.

Brian.