Take the 2-minute tour ×
Electrical Engineering Stack Exchange is a question and answer site for electronics and electrical engineering professionals, students, and enthusiasts. It's 100% free, no registration required.

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?

share|improve this question
2  
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). –  Nick Alexeev Apr 16 at 18:20
    
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. –  Spehro Pefhany Apr 16 at 18:49

2 Answers 2

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.

share|improve this answer
    
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. –  Scott Seidman Apr 16 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).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.