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I am trying to design and improve the power input part of my circuit to supply power to a motor driver that controls 2 DC motors. I did some research and i found this circuit but i cannot fully understand why these components have been used. I am using DC batteries to supply 12 V at VS and VIN goes to the VIN input pin of the Arduino UNO.

UNO power supply

I read about power supply decoupling capacitors; electrolytic capacitors filters out low frequency signals and ceramic capacitors act as a low pass filter.

I have the following questions:

  1. Why use 2 electrolytic capacitors instead of just one?

  2. Why no ceramic capacitors has been used? Has this been omitted intentionally?

  3. Why is a high value of 470uF used? I searched and found that many power supply decoupling capacitors are less than 100uF for my application where the supply is only about 12 V DC.
  4. The Schottky diode is the most intriguing to me. It is a rectifier, so is it also helping to smooth out the DC signal? Shouldn't it be connected in series instead of in parallel?

Thanks for any help.

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  • \$\begingroup\$ Please link to the source of the circuit. \$\endgroup\$
    – Andy aka
    Commented Jun 18, 2019 at 11:25
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    \$\begingroup\$ "Why is a high value of 470uF used?" - This just depends on a) the 'ripple' present on V[in] and b) the current drawn from V[s]. Look at the classic transformer-based PSU. They tend to have those big caps after the bridge rectifier because the 'ripple' just after the rectifier is 100% (the full rectified sine wave, briefly reaching 0V and V[max] once every 10ms). The caps must provide all of the required voltage and current even at those times where V[in] becomes temporarily 0! \$\endgroup\$
    – JimmyB
    Commented Jun 18, 2019 at 11:59
  • \$\begingroup\$ Don’t forget that when you are tying to slow down the motor, it is likely acting as a generator and pumping power back into your driver power rail. The capacitors absorb some of this for re use, and the zener makes sure the voltage doesn’t get too big. \$\endgroup\$
    – user69795
    Commented Jun 18, 2019 at 23:56

2 Answers 2

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Why use 2 electrolytic capacitors instead of just one?

That's often a choice, maybe 2 capacitors of 470 uF are cheaper / smaller / fit better than one 1000 uF capacitor.

In high frequency switching power supplies a certain low value series resistance (ESR) is needed, that might be more easily achieved when using two capacitors

Why no ceramic capacitors has been used? Has this been omitted intentionally?

The designer decided that these were not needed. Perhaps the low ESR could be met without using ceramic capacitors.

You might want to explain why you think that ceramic capacitors are needed!

Why is a high value of 470uF used? I searched and found that many power supply decoupling capacitors are less than 100uF for my application where the supply is only about 12 V DC.

The designer decided that this was needed to achieve low enough ripple.

Saying "the supply is only 12 V" does not mean anything. The value of smoothing capacitors scales with the current (and switching or AC frequency), not with the voltage.

The Schottky diode is the most intriguing to me. It is a rectifier, so is it also helping to smooth out the DC signal?

No it is simply a protection against reverse voltage. If for some reason the voltage becomes negative the diode will conduct and that way tries to limit the voltage. What happens to electrolytic capacitors which are subjected to a reversed DC voltage ? They go BANG!

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  • \$\begingroup\$ First, thanks for the answer. I really appreciate your effort and time to help. To consolidate my understanding: 1. I want to add a switching buck-boost voltage regulator that will decrease the VIN voltage from ~12 V to fixed 6 V. Is it recommended to add a similar setup (2x470uF and Schottky diode) after the voltage regulator to filter noise generated at the regulator output? 2. I mentioned ceramic capacitors as i read that it filters out high frequency noise. But i suppose it's not needed in this case? 3. If reverse voltage, the diode will conduct but won't this damage the battery? \$\endgroup\$
    – DryRun
    Commented Jun 18, 2019 at 12:09
  • \$\begingroup\$ 1) you add what the buck converter needs (see datasheet) so don't add 2x 470uF because you've seen that being used elsewhere. Also you need a buck converter, you want 12 V in, 6 V out so a buck-boost isn't needed. 2) If high frequency noise isn't present and/or not an issue it does not need to be filtered out. Again, follow the datasheet of the buck converter. Use electrolytic caps when that is suggested. On a PCB you could reserve the space for adding ceramic caps and then add them if you find out they're needed. If your application has no RF you probably don't need them. \$\endgroup\$
    – Bimpelrekkie
    Commented Jun 18, 2019 at 12:14
  • \$\begingroup\$ 3) Indeed, too much current can flow when the battery is reversed connected. If you want to protect against that then add a fuse in series with the battery. Then when the battery is reversed, the fuse will blow and no harm is done. \$\endgroup\$
    – Bimpelrekkie
    Commented Jun 18, 2019 at 12:16
  • \$\begingroup\$ To fix the reverse voltage battery damage and avoid using a fuse, could the Schottky diode be simply placed in series at VIN instead? \$\endgroup\$
    – DryRun
    Commented Jun 18, 2019 at 12:49
  • \$\begingroup\$ could the Schottky diode be simply placed in series at VIN instead? Yes, that is also an option but has the disadvantage that there will be some voltage drop across that diode. \$\endgroup\$
    – Bimpelrekkie
    Commented Jun 18, 2019 at 12:50
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The Schottky diodes are essential to protect the drivers from reverse voltage when power is abruptly disconnected with an inductive load. (Been There Done That with my OEM Buck driver on a few strings of LEDs THAT DID NOT SUCH PROTECTION and the drivers failed., so ordered a new one from Amazon and received in 2 days) I had an intermittent load connection between source and Buck input.

The Battery ESR*C time constant of say 50 mOhms * 10kF for one 18650 Li Ion cell is equal to 50 seconds.

The 1mF ultra low ESR caps combined may have an ESR*C time constant as low as 56 mOhms * 470uF = 26 us and two in tandem give 26 mOhms and 6.6us time constant. The demand load on startup will be the DCR of the motor and impedance rises with inductance and BEMF as frequency rotates the magnetic flux. However the decay rate of these caps increase with the DCR of the motor. They are a big help to stepper motors but BLDC ones with large inertia may not be as much help.

This results in the caps taking more of the start surge and ripple current than your battery or possibly **much better for your SMPS design ** IF NOT just rated just above max motor current rather than ~ 10x surge load current. !!! THis is why they invented constant V/F VFD motor controllers.

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