enter image description here

I am following the course on control systems from edx. I have this circuit as part of practical exercise. I would like to understand how the capacitor values were determined in this circuit

  1. The 47uF is for power supply filtering
  2. The 47nf is for reduce the switching speed of PWM
  3. The two 2.2uf capacitors for filtering noise signals from motor

The circuit is powered by arduino and the motor is powered by an external 5v 3A power supply. The lessons say that the motor can draw up to 2 A, If i assume the ripple voltage is 2 v and main freq is 60 hz, then the ripple frequency is 120 hz, the current that the capacitor can take/give is

$$ I = C \frac{\Delta V}{\Delta T} $$

which gives 11.2 mA only

$$ I = 47 \times 10^{-6} \times \frac{2}{\frac{1}{120}} = 11.2 mA $$

However the lesson says the motor can draw up to 2A how is this capacitor sufficient?

  • \$\begingroup\$ The formula you used is for calculating value of a capacitor within a power source based on a transformer and full bridge rectifier. You don't need that formula. You are not building a power source, you already have one. And most likely it is smps type. \$\endgroup\$ Dec 9 '18 at 7:48
  • \$\begingroup\$ What is the role of 47uf capacitor here then? Why a value of 47uf was chosen \$\endgroup\$
    – vumaasha
    Dec 9 '18 at 8:10
  • \$\begingroup\$ It is a decoupling cap. Filters out some noise or minor glitches. \$\endgroup\$ Dec 9 '18 at 9:19

the current of 11.2mA you calculated from the equation I=C*ΔV/ΔT has no matter here because the ripple voltage and main freq are related to a rectifer circuit based on bridge diodes. Here the 47uF capacitor is a charge supply connected to Q2 emitter and the motor's current is directly controlled by Q2 wich is controlled by Q1, the collector current of Q2 is Ic2 = I1+I2 I1 flows through R4 while I2 wich flows hrough R3 is flowing through motor and R5. Hence the motor current is taken from the 5v supply and transistor Q2 is controlling its ammount.

  • \$\begingroup\$ The capacitor has no internal resistance unlike a battery whose internal resistance Ri can affect more or less the intensity of the current. If the capacitor is charged at 63% of Vcc at the time CRi (s), it takes 5*CRi (s) to reach Vcc (99%) and therefore 5*CR to discharge in a circuit having the resistance R.The capacitor value is not too sensitive, it depends on the power required and the circuit parameters. The main function of the capacitor is to smooth current by charging and discharging for a short time following an exponential curve in a way that prevents against noises and shocks... \$\endgroup\$
    – a Kamli
    Dec 9 '18 at 19:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.