Electronic – Voltage regulator circuit, efficient, low necessary drop


I want to get a 4v supply from >=5v. (I.e. as little as 1v drop)

The load current will vary by a factor of 10,000. (From 5uA to 50mA).

I need the circuit to be really efficient when there's a small load.

I wanted a darlington pair set up as emitter-follower following a zener diode regulator. The problem being fairly obvious, the 1.4v drop exceeds the 1v I have available, so it's not usable.

Instead, using a single NPN follower following a ZD in isolation would give an hfe of about 500 say, meaning the current taken by the ZD with the supply at 5v would have to be 20 times greater than the load is using in standby, in order for it to be able to supply the load's peak current! And significantly greater when the supply is say 8v, the current taken through the ZD would be 200 times the load's standby current!

Is there a better way?

Best Answer

A zener diode is absolutely no good for very low power (for the 1N4732A the zener voltage is specified at 53mA), and even LDOs often have ground currents 10 times your 5\$\mu\$A load. You want an LDO with a < 1\$\mu\$A ground current, like Seiko S-812C40. Output voltage is 2.0 to 6.0 V, selectable in 0.1 V steps, so there's also a 4V type. You get 65mA out. Dropout as low as 120mV, and stable even without output cap. The S-812C is available in SOT-23.

I just discovered that there's also a version with a shutdown. That S-812C is even cooler than I thought! :-)