After seeing some students yesterday that tried to use a voltage divider instead of a regulator to provide a sensor with a lower power supply, with predictable results, I started wondering about this question.

When picking a regulator, it seems many look at the required voltage drop and the power dissipation required. Efficiency aside for the moment, if a linear regulator can drop that power within thermal limits, linear regulators are an option, and if they can't, move on to switching regulators.

If one could figure out the range of current draws, and calculate out a voltage divider that would simultaneously keep the input to a linear regulator high enough to maintain regulation and low enough such that the regulator doesn't burn away too much power across the current draw range, is this a viable approach?

I can think of a number of reasons why this might not be the best approach: power supply rejection ratio may not be good enough on the regulator; the range of current draws that make this approach feasible might be very small, unless you use small resistors that are likely to exceed their own power ratings; its just more efficient to use a switching regulator; etc.

Also, it might be that people do this all the time, and I just haven't noticed it, or maybe a zener is used instead of the divider. It just seems that when the power drop is too big, people mostly run to switching regulators.

Anything I'm missing?

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    \$\begingroup\$ Another approach: add a power resistor in series with the linear regulator input (not a voltage divider). At high current, it will reduce the voltage to the linear regulator and dissipate some power (which otherwise the linear reg would have to dissipate). \$\endgroup\$ – Nick Alexeev Apr 16 '14 at 18:20
  • \$\begingroup\$ Similar to @NickAlexeev's suggestion, you can put a resistor in parallel with a linear regulator if there is a guaranteed minimum load and a guaranteed maximum input voltage. Same dissipation but it moves to to the resistor. \$\endgroup\$ – Spehro Pefhany Apr 16 '14 at 18:49
  • \$\begingroup\$ Some Japanese cordless phone manufacturers use one "beefy" (1W) transistor with a zener to get 6V at board entrance from 9V brick supply and then 2-3 100mA SOT89 5V regulators spread on the board. The 6V supply is used directly by only one of the xtal oscillators. \$\endgroup\$ – Fizz Sep 27 '15 at 13:50
  • \$\begingroup\$ Also ti.com/lit/an/slva119/slva119.pdf \$\endgroup\$ – Fizz Sep 28 '15 at 4:48

This is certainly a technique I have used a few times to overcome the limited power dissipation abilities of the diminutive 78L05. I've known the range of currents that the load is taking and placed a dropper resistor in series with the power feed to the device.

Why didn't I use a switching regulator?

I couldn't - I was sending power and data down a 50 m cable (phantom power) and the extreme complication of filtering out the switching regulator's current surges meant it just wasn't feasible.

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  • \$\begingroup\$ That's exactly why the question sprang to mind. Aside from efficiency, there are some real reasons to avoid switchers, and the noise generated is probably top on that list. \$\endgroup\$ – Scott Seidman Apr 16 '14 at 18:46

Voltage dividers are terrible for efficiency (if you think of output impedance vis-a-vis power consumption). I'd be hard put to think of a good place to put them in front of a regulator.

Series zener diode- if you put a 24V zener diode in to knock a 35V input down to 11V for a 9V regulator, you've increased the sensitivity to input variations- a 10% drop in the input means there's only 7.5V left and your regulator drops out.

I have used a shunt zener with a capacitive dropper in series with a linear regulator to get power from the mains, and I think that's fairly common. With capacitive droppers you don't suffer much loss.

Many of us will also put a shunt TVS that effectively acts as a regulator under unusual circumstances, so I'd count that too.

Series or shunt resistors around a linear regulator- I think I used the latter once, the former not so far. The shunt resistor would be more attractive if the linear regulator was capable of sinking current (some are, but most are not), then you could just set the resistor to handle the mean current and the regulator would tend to run very cool (downside is that some power would be wasted if the required current drops below the mean).

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  • \$\begingroup\$ Series resistors with a linear regulator may be attractive if there's a huge gap between the guaranteed minimum voltage at the input and the regulator's required input voltage; they'll have no effect on total power dissipation, but they can cut the regulator's worst-case power dissipation by almost a factor of four (worst-case dissipation for the regulator will be at about 50% load current, where it and the resistor dissipate 25% of worst-case total power; at higher currents, the resistor's share of the power will increase faster than the total power, so the regulator's power will decrease). \$\endgroup\$ – supercat Dec 2 '15 at 15:18

If one needs to convert 12V to 5V for a load which may vary from 0 to 1 amp, and the regulator needs a minimum of 6 volts on the input, connecting the supply directly to the regulator will cause it to dissipate 7 watts with a one amp load. Adding a 6-ohm resistor in series with the input would cut worst-case power dissipation in the regulator to about two watts over a wide range of load conditions (as current goes up, the amount of voltage dropped by the regulator [as opposed to the resistor] would go down). Series resistors don't help overall efficiency, but they can shift heat dissipation away from the regulator. A key point to note, though, is that the bottom half of a resistor divider wouldn't really help much of anything, since its purpose would be to waste power when the load isn't drawing current, but in that scenario regulator current will be low so there's no need to share it.

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