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I have a H-bridge driving a 12V motor that at 100% duty cycle draws around 15A. To start the motor I am applying a soft-start, the duty cycle of PWM to H-bridge goes from around 60% to 100% and PWM frequency is around 500Hz.

How can I calculate the needed capacitor value on 12V for 10% ripple? At such high currents and low PWM frequency would it be better to go for more than 10% ripple (this 12V line isnt driving anything else than motor)?

I am aware of: $$I=C\times\frac{dV}{dt}$$ But plugging the numbers in I get: $$C=\frac{I\times dt}{dV}=\frac{15*\frac{1}{500}}{1.2}=25mF$$

Which seems quite high to me, is this correct?

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  • \$\begingroup\$ The calculation you did does not include the current delivered by the power supply. You need to specify what the 12 V power supply is (current, ripple), and what drop in the 12 V supply do you see currently on starting? \$\endgroup\$ Dec 3 '16 at 21:03
  • \$\begingroup\$ @JackCreasey Its night here, so I will do this tommorow. But so you dont wait, what would be the calculation then? \$\endgroup\$
    – Golaž
    Dec 3 '16 at 21:05
  • \$\begingroup\$ Essentially the same, though this time you will actually know what the ripple value is. I doubt your power supply is dropping by 10% (1.2 V), and if it is, then you probably need to be working on it first. I'd expect that you might expect droop values of around only 1% maximum from your power supply. \$\endgroup\$ Dec 3 '16 at 21:17
  • \$\begingroup\$ May I ask why would you reduce voltage ripple out of your H-bridge? And exact schematic you are planning for? \$\endgroup\$
    – carloc
    Dec 3 '16 at 23:12
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    \$\begingroup\$ Okay . I believe you over estimate capacitor. I'd rather try to analyze an RLC circuit including power supply and cabling stray L and R and capacitors with their ESR. A few runs on a simulator with different duty cycle are going to be the fastest way. \$\endgroup\$
    – carloc
    Dec 4 '16 at 23:46
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For similar problems, with 10% V ripple , I let RC= 8T. If motor equiv R=12V/15A and T= 1/500Hz=2ms then C=8*2ms*15A/12V= 20 mF

(assuming your current is RMS)

This is consistent with your calculation and you can achieve this with a power supply with Zout = <<10% of load or better with a 12V battery and trickle charger. Batteries have much larger capacitance and low ESR and would perform better but Caps will be cheaper.

However since startup current due to DCR of motor can be up to 10x Imax rated current, soft start is necessary or a much higher capacity such as from a battery.

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