0

I'm currently using a breadboard Arduino Uno clone. Basically a Atmega328p powered using an L7805 linear regulator with a 10uF capacitor on both the input and output. The Atmega328p is connected with a 433Mhz receiver (WL101-541) and two SG90 servos (PWM pins). Basically the signal from the receiver is used to control the servos. The servos need to be driven by 5V(4.8V) as well so their power and ground come from the L7805 as well.

I have encountered a situation where the receiver stops functioning when the servos are under load. I suspect that it is very sensitive to the input voltage and the additional load on the L7805 from the servos causes a slight drop in the voltage.

I have temporarily worked around this by adding a second L7805 and attached the servo power to it. Both L7805 are attached to the same battery. So input voltage will start around 9V then drop to maybe 6.5 or 7 over time.

The circuit power will be a standard 9V battery (perhaps a lithium) as I have weight considerations and am only aiming for around 1.5 hr of runtime.

1) I am looking for information on why one L7805 does not cut it as I believe it is rated for providing up to 1 amp, and the entire circuit appears to be drawing around 0.2A at the time the servos are under load and the receiver is not functioning.

2) Am I missing something, can the circuit be improved? Perhaps through changing capacitors or their placement.

  • 1
    7805s are a very very poor choice for battery operation. You are better off with a UBEC (buck regulator). – Majenko Jan 11 '18 at 19:09
  • 1
    What is the input voltage going into the 7805? – Gerben Jan 11 '18 at 19:15
  • 2
    Linear voltage regulators are wasteful of power. You haven't specified your battery, but the efficiency of a linear regulator is approximately (Vout/Vin)*100. At 9V Vin that's 5/9*100 = 55.6% efficient. You'd be losing 44.4% of your battery power as heat. – Majenko Jan 11 '18 at 19:17
  • 2
    Also you may need far more than the steady-state current for when the servos start moving, when they will need a peak of current to get them going (the stall current). – Majenko Jan 11 '18 at 19:18
  • 2
    They're simple: two wires connect to your battery, and two wires connect to your circuit. Far simpler than building your own regulator with 7805 and capacitors. – Majenko Jan 11 '18 at 19:21

Your Answer

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

Browse other questions tagged or ask your own question.