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I accidentally bought a 7.4 volt servo when I should have bought a 5. Seeing that it is already integrated into the project it is being used for, what is the best way to drop 11 volts to 7 or 8 in a lightweight manner?

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If by "lightweight," you mean "quick and dirty," then the best way is probably just to use a standard 3-terminal voltage regulator:

schematic

simulate this circuit – Schematic created using CircuitLab

The LM317 adjusts the out terminal to 1.25V above the adjust terminal (which would be called ground in a traditional 7805 regulator). The purpose of the resistor network (now a set of fixed resistors in response to comments) is to generate a reference voltage 1.25 V below the desired output voltage (= 6.15V above ground): Approximately 5 mA of current will flow through R2 when when OUT is 1.25V above ADJ. The ADJ terminal sources less than 100 uA, so the voltage across R3 is about 6.4V. This is a couple hundred millivolts above the desired 7.4V, which is ok for this application. Originally, this schematic used a potentiometer, so that 6.15V could be easily generated without resorting to non-standard resistors or combinations of standard resistors.

The datasheet example circuit shows both input and output capacitors, however the text has this to say:

CI [C1] is required when the regulator is located an appreciable distance from power supply filter. CO [C2] is not needed for stability; however, it does improve transient response.

The output capacitor, C2, is there to supply charge between the time a large load is added to the output and when the reg's error amplifier responds. The servo is already designed to work over a fairly broad voltage range (because they normally run unregulated from a battery), so the transient response is not very important in this application.

If this circuit is meant to be powered by a battery, which has a very low output impedance, or the regulator is a short (wire) distance from the power supply, then C1 is also optional.

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    \$\begingroup\$ I'd put the potmeter between adjust and ground, and a fixed resistor between out and adj. \$\endgroup\$ – Wouter van Ooijen Aug 17 '13 at 7:37
  • \$\begingroup\$ It's not obvious to me why that's better... give me a hint? Is it to have a constant current through the voltage divider? \$\endgroup\$ – nick g Aug 17 '13 at 8:50
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    \$\begingroup\$ With the wiper at the upper position there is 0 resistance between out and adj. The poor 317 will try to maintain a 1.2 V drop across this 0 resistance. Though probably not fatal, this is not a mode the chip was designed for. \$\endgroup\$ – Wouter van Ooijen Aug 17 '13 at 9:00
  • \$\begingroup\$ Ah, that's a good point. I've added a resistor to the schematic to ensure a minimum resistance between the out and ground terminals of the regulator. It this case, I think we can get away without it, though, because the output is meant to be fixed at 7V, or so. \$\endgroup\$ – nick g Aug 17 '13 at 9:26
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    \$\begingroup\$ When I made the comment, you used a 100R fixed and 10k variable resistor. To get the 7V the variable resistor would have to be set to 460R. That was 0.046 of its full value, which is very difficult to get right. 1k would have been a better choice. \$\endgroup\$ – Wouter van Ooijen Aug 17 '13 at 21:48
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How about a couple of diodes in series. Make sure they can handle the current. Each silicon diode drops .6 V.

Cheap, easy to salvage from all kinds of equipment, you don't even have to go to RadioShack to get a lm317.

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