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I'm doing some starter circuit design and assembly. I used a linear voltage regulator to provide 6V to an RC servo, which is nominally 6V driven. This was done on a PCB using a mix of SMT and hand solder (hand soldered regulator, SMT capacitors,) taking a 12V input and regulating it to 6V. I'm confident with digital design and with resistive loads, but this is my first foray into servos/motors.

The regulator I used was a TO-220 MC7806CTG, and it worked for a while, after soldering the regulator, and testing it 6V was correctly output.

However, whilst I was tuning the servo calibration, the servo stopped working, and I found that the voltage of the regulator was outputting 8.8V, rather than 6. I have capacitors on both input and output to ground, per the datasheet.

The servo datasheet (an acutonix PQ12-100-6-R) says it has a maximum draw of 500mA, but the regulator datasheet says that it should be good for 1A.

I know I've made a mistake (or multiple,) but I don't want to repeat the error - the servo isn't cheap, so I don't want to blow it.

I did not heat-sink the regulator. I thought at worst it would fail open, I also thought that the power draw I had was low, using a simple power calculation for the regulator (12-6)*0.5= 3W. I think this was a mistake, but am unsure if it was the sole problem.

How should I do the re-design? Should I add a large capacitor to the regulator output to suppress any inrush? Is just redoing the board and heat-sinking the regulator enough?

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  • \$\begingroup\$ If you ignore the warnings about overloading the servo motor, and stalled with 8V , both parts failed. Use the TO220 with a large heatsink and current limit to 250mA with 10V with a 20 Ohm resistor or better use an array of 3V Xmas mini lights as the current limiting R to give you foldback stall Current of 250mA with peak cold lamp currents of up to 1A. Cold resistance should be 5 Ohms and hot 20 Ohms on stall with a 12 Ohm 10W resistor load before using it. \$\endgroup\$
    – Tony Stewart EE75
    Jun 6, 2021 at 11:22
  • \$\begingroup\$ Then learn to control static load and acceleration to prevent overloading the motor. \$\endgroup\$
    – Tony Stewart EE75
    Jun 6, 2021 at 11:23
  • \$\begingroup\$ What heatsinking arrangements did you have on the 7806? \$\endgroup\$
    – user_1818839
    Jun 6, 2021 at 11:24
  • \$\begingroup\$ @TonyStewartEE75 : I'm a bit confused about your stall statement - the servo was being calibrated unloaded, and the total load is only a few grams in service. The only way I can think that it could have stalled is if the servo can be driven to stall unloaded (which Id have assumed not). I've considered using a polyfuse for stall protection, but did not implement it. I'm unclear why I would use a light rather than a normal resistor?? The motor is already acceleration controlled in software to not exceed the max-drive rate listed. \$\endgroup\$
    – User218311
    Jun 6, 2021 at 11:45
  • \$\begingroup\$ Light has PTC like characteristics but a smaller range so PTC is OK. It seems like your circuit may have been oscillating and needs a low ESR cap on output or else the noise was getting somewhere \$\endgroup\$
    – Tony Stewart EE75
    Jun 6, 2021 at 11:54

2 Answers 2

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The regulator can't irradiate 6 or even 3W power without a heatsink.

65C/W, so 3W means delta T = 195C, now add the air temperature of 25C you get that junction temperature would be 220C, so far too much.

enter image description here

EDIT:

Instead of using a linear voltage regulator, use a DC/DC step down converter- the servo motor won't suffer as it is not a low noise amplifier or a device that needs a linear PSU. There are DC/DC converters of the same size as TO-220 case, so nothing to change.

enter image description here

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  • \$\begingroup\$ And assuming that it would fail to open circuit was also a mistake. Given that a linear regulator will waste half of your input power you might consider a switch-mode device. It might seem a lot more complex but they aren’t that scary when you get down to it. \$\endgroup\$
    – Frog
    Jun 6, 2021 at 10:55
  • \$\begingroup\$ I originally considered the murata device,but had trouble sourcing it at the time, and redesigned with the linear regulator. If that's a good plan then I will switch back, thanks! As an aside, is there a good rule-of-thumb for the change in junction power dissipation with a heatsink (*C/W)? \$\endgroup\$
    – User218311
    Jun 6, 2021 at 11:05
  • \$\begingroup\$ @User218311 Look for some tutorial, for example: cuidevices.com/blog/how-to-select-a-heat-sink . The Junction to Case Resistance is 5C/W, then add the heatsink resistance (with a thermal paste) let's say 20C/W, then calculate 5+20= 25C/W. For a 3W disssipated power you get 75C temp. rise from ambient T. Larger heatsink has lower resistance, of course. \$\endgroup\$
    – Marko Buršič
    Jun 6, 2021 at 11:20
  • \$\begingroup\$ @MarkoBuršič : Thanks, I'll check it out. I was looking for linear heat transfer coefficients, which I could find; im happy doing heat transfer stuff (I've done this for fluid devices), but finding reference values is always a bit tricky. \$\endgroup\$
    – User218311
    Jun 6, 2021 at 11:48
  • \$\begingroup\$ After looking I can only find the 3-pin DC-DC convertors in 5V; it looks like a new board is required for the 5-pin trimmable versions. \$\endgroup\$
    – User218311
    Jun 6, 2021 at 12:37
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The power calculation is correct. At 500mA the regulator does dissipate 3W. The thermal resistance is 65°C/W so it would try to heat up by 195 degrees, which it surely cannot handle. The device does have built-in thermal overload protection, but it may have heated up too fast locally with the 3W before the protection would shut it down.

And when it fails, there is no evidence how it will fail, so assuming that it simply fails by turning off will not work, it might just as well fail with a short circuit.

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