Why don't people tend to use voltage dividers or zeners in front of linear regulators

Question

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?

Answer 1

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.

Answer 2

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).

Answer 3

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.

Answer 4

Having taken apart a number of 5V wall-warts, I can attest that all of the ones I have examined have a resistor/zener divider to bootstrap the low voltage regulator. The current requirement for this bootstrap is miniscule, so efficiency is not a problem, and it is a very cheap solution for the manufacturers.

Answer 5

I realize this is an old question. I have used a dropper resistor in series with an linear regulator in designs powered from higher voltage (say 48 V) battery packs, and I have seen it done for the BMS of such a battery pack.

Typically the BMS draws its power from the top of the battery stack. It is typical to use a linear regulator rather than a switcher for this application because the Iq of a good linear regulator is much less than the IQ of a switcher. Or, said another way, most switchers are not very efficient at micro-power levels.

Even though the average current on this regulator is very low, so dissipation is not an issue, there may be times when it is called upon to deliver 20 or 50mA. Putting a resistor in series with the regulator lowers its dissipation during these high current events.

A series Zener could accomplish pretty much the same thing, but it is usually cheaper and more reliable to dissipate power in a resistor when you have the option (rather than in silicon).

Care must be taken to insure that the regulator will not drop out under the heaviest anticipated load and the lowest expected battery voltage.