Input voltage control by resistor, is this method is correct or not?

Question

In my application I have a requirement to switch 40V to 5V with the help of a linear voltage regulator. My load circuit consumes approx 7 to 8 mA.

Due to the unavailability of any specific model of a linear voltage regulator, I am using 7805 for this purpose. But the input voltage limit of 7805 is 35V.

I already tried a voltage divider method put before 7805 to control the input voltage level, but that method has other issues because of the internal resistance of the 7805.

Here I applied one input resistor to control input voltage flow.

Circuit:

Now due to 7mA of load, and (2.2Kohm, 1-watt) resistor, the voltage drop of the resistor is (7/1000)*2200 = 15.4V.

The input voltage of 7805 is ( 40 - 15.4 ) = 24.6 V in this scenario which is less than the safe limit of 7805.

Is this solution is suitable for this type of application, or there is another efficient way to solve this kind of thing?

Answer 1

Here I applied one Input resistor to control Input voltage flow.

This solution won't work if there's a possibility for the load to be disconnected or for its current to drop significantly because then the 7805 input will be subjected to ~40 V.

If you know the load will always be there and always draw at least 3 mA, then this is viable.

If you want to be safer about it, you could add a 30 or 35-V zener in parallel with C1.

or there is another efficient way to solve this kind of thing.

You could choose another regulator that allows more than 40 V on its input.

Answer 2

Use a voltage divider in front of the 7805 for safety.

^{simulate this circuit – Schematic created using CircuitLab}

or a zener as Photon says (preferable method) I put 10V zener, as 7805 recommended operating voltage is 25V.

^{simulate this circuit}

Answer 3

If you really care about efficiency, use a Buck converter like MC34063 - it can handle input voltage up to 40V.

Using a regulator for converting 40v to 5v is not an efficient way. But since your load is 8mA max, you can find a regulator with higher input voltage or use 7805 with a voltage divider or with a Zener diode as told by @ThePhoton.

Answer 4

Just use a single shunt zener or shunt regulator like this: -

Maximum current into zener is 35 volts / 2200 = 16 mA and power dissipation is 80 mW.