Zener Regulator supply to IC

Question

In below Design , working ok with Vcc was 5V, but due to some modification I have only 15V power supply available to use this circuit. So i change as per 15V requirement ( see below image), Pin 1 is open Drain so did not removed R34 Removed R43 WDI_IO is drive form micro controller Reset of Pin 1 goes to Trigger input of 555 IC ( which has max working voltage of 18V , not shown in SCH)

Since U3 max voltage range is 5.5V , So I am thinking to change FB28 with 1K resistor and C40 with 5.0V Zener Diode.

Zener Spec:- 5.0Vz, Test Current 5mA Max PD 500mW So Max current = 500mW/5V = 100mA Min Rs = (15V-5V)/100mA = 100oHM

I choose 1K as Series Resistor so Current will be

I= (15-5)/1K= 10mA, This can supply both zener and load whch is my IC I as per data sheet IC Current suppy need in approx 200nA).

Does this look ok ,to use zener in this case as my load current in not much.

Answer 1

Using a zener diode as a voltage dropper is unsafe and inefficient — at best you'll end up dissipating heat in both the diode and limiting resistor. As the load current is low enough you could (and should) use a voltage regulator such as the good old LM7805 in a TO220 package. For a 5mA load current the regulator would dissipate 50mW — i.e. \$ (15V - 5V) \times 0.005A = 50mW \$ — which is way better than any zener + diode contraption. As such there's no more need to hack your circuitry, much simpler, much safer.