0
\$\begingroup\$

I'm studying a popular motor driver board based on the ULN2803a Darlington Array chip (top and bottom pictured below).

A capacitor is between the common diode/flyback trace and GND.

Does the capacitor collect the flyback energy and return it into the supply current? I presume this happens whenever the differential between the cap's stored voltage and the supply voltage reaches certain level?

enter image description here

enter image description here

\$\endgroup\$
0
\$\begingroup\$

The idea is to reduce external large current loops with dV/dt by re-directing the motor current in the same direction but from diode cap instead of U1. This also means less surge EMI from the battery wire loop area. The dI/dt radiated noise increase with loop area. So having Cap and diode close to U1 keeps this loop small.

It is assumed you can choose Cap with lower ESR than the battery ESR and wire inductance at frequency equivalent to the decay rate of T=L/R = 0.35/f-3dB where R include motor, diode and cap or battery.

Using twisted pairs is useful to reduce interference for all current wire ( supply and return)

schematic

simulate this circuit – Schematic created using CircuitLab The ceramic cap has much lower ESR than the battery or power supply after inductive wires.

The flyback energy is small but can still create interference with sensors as it supplies only the power last used ( motor current) for xxx ns.

Depending on actual values of RLC for load, wire and circuit, it can still cause EMI problems if badly tuned. ( high Q MHz ringing)

\$\endgroup\$
  • \$\begingroup\$ Thanks Sunnyshyguy, The concept that flyback current travels in a loop was new to me. Is that why the EMI current isn't put straight to ground... because the current can travel through the ground wires back into the circuit? \$\endgroup\$ – H P Ladds Jan 14 at 15:17
  • \$\begingroup\$ The current is dumped thru the C1 ground path which is why it is kept close to the transistor and maintains the same current direction in L yet the voltage opposes as polarity depends on the polarity of dI/dt \$\endgroup\$ – Sunnyskyguy EE75 Jan 14 at 17:31
  • \$\begingroup\$ drive.google.com/file/d/1xEr21DaeMLnnTu_GRDkL7QSrf1RwBRQO/… \$\endgroup\$ – H P Ladds Jan 14 at 20:54
  • \$\begingroup\$ Man! I'm having a hard time getting an image in the comments. I'm wondering if this is an effective implementation of a capacitor. \$\endgroup\$ – H P Ladds Jan 14 at 20:55
  • \$\begingroup\$ Correct pins but use non-polarized ceramic \$\endgroup\$ – Sunnyskyguy EE75 Jan 14 at 20:57

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.