I have a small differential drive vehicle that has two 24V 400W DC motors, powered by two 12V batteries connected in series. I am planning to attach a UST-10LX scanning range finder at the front of the vehicle, so I can implement some obstacle avoidance algorithms. The voltage operation range of the sensor is from 10 to 30V, ripple within 10%.

I would like to know if it is safe to connect the sensor to the batteries without any protection circuit, or if the current drained by the motors could possibly damage the sensor in any way.


2 Answers 2


The problem isn't different devices demanding different amounts of current. Mainly, the issues you may encounter are:

  • Voltage/current drops caused by the motors that may affect sensitive devices powered from the same source. For this you must make sure using batteries rated for currents (with some margin), required by the motors. Another measure is installing bulk capacitors near the sensor, it will guarantee correct voltage/current to the sensors in case of short voltage drops caused by the motors. For this, the bigger the capacitor the bigger will be protection. Of course using too large ones for the application will be a waste of money and space.
  • Noise: Inductive loads causes lots of noise on the power rails. For this you can protect the sensor adding capacitors with the right values across motor input poles. Like Methods of filtering noise caused by DC brush motors e.g. It looks coherent that you add also noise protection near the sensor.
  • Inductive fly-back: This is very important. Commuting DC motors OFF generates voltages spikes that are far away greater than other parts connected can take, like sensors. You can waste your sensors easily without protection to this. Refs: https://www.westfloridacomponents.com/blog/inductor-need-fly-back-diode/ https://en.wikipedia.org/wiki/Flyback_diode
  • It's very important that you check the range finder datasheet for recommendations, absolute maximum rates or other precautions not listed here required by the sensor.

How to measure surge load regulated supply ripple.

Short answer

  • Supply Load Regulation ~ ESR source / DCR load *100%
    • e.g. Load Ripple = +/-10% max, source in this case is ESR of battery
    • DCR = DC resistance is the ESR of a motor at DC= winding resistance
  • Effective series resistance, ESR = ΔV/ΔI

  • DCR = V /I surge , I surge = startup same as locked rotor but T=L/ESR for total ESR loop resistance.

Full answer

  • If the motor consumes 400W @ 24V then it draws 400W/24V = 16.7Amps at full speed and rated load.
  • But we know from efficient DC motor design that inrush current is 8~10x rated current,
  • thus assuming ideal battery and MOSFET switch;
    • Inrush 10*16.7 = 167A, Since Inrush is from DC resistance, therefore;
    • Motor DCR = 24V/168A= 144 mΩ
  • since the Laser tracker is rated for +/-10% ripple rejection, with Ohm's Law and neglecting for now ripple in cable ESR and ESL, inductance,( both very important;)
    • Ripple Vp / Surge Current = 2.4Vp/167A = 14 mΩ max total battery ESR
  • This defines the CCA rating for each battery based on std test of 5V drop from 12.5V fully charged, therefore; - CCA = ΔV(=-5) / ESR =7 mΩ (per batt.) = 714 A CCA

  • Normally cars cannot sustain this strain on batteries or has a short charge life with 55Ah typical. So battery and MOSFET selection are critical for performance and reliability. High battery ESR from low charge or aging will result in high back EMF voltage. (I^2*ESR)

  • The MOSFET RdsOn should be <2% of the DCR of the motor for thermal reasons.

    • Thus the RdsOn <= 2% of 144 mΩ = 2.9 mΩ
    • This also requires high quality batteries with rugged plates and deep discharge rated and well maintained for specific gravity and a heatsink for the MOSFET. Pd=I^2RdsOn=168A^2/2.9 mΩ = 10W

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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