This questions consists of two parts.


I'm designing a digitally controller class D amplifier. Main IC's are a TI TPA3116D2 Class D amplifier, a TI PGA2311 volume control and an ATMega328P (running at 5V with internal 8Mhz oscillator).

Problem 1: Brown out when relay engages

Power supply schematic.

I power my board with 24VDC. For now that's coming from a lab power supply. U303 provides 5V for the MCU.

The MCU in turn, controls relay K301, which will be used to power the other IC's. 24VDC will go directly to the TPA3116D2. U304 regulates 5V for other digital parts that are normally off, like a display. U301 and U302 create ±5V for the analog part of the PGA2311.

When the relay activates, the 5V from U303 drops enough to reset the MCU.

Somehow U303 can't keep up with either (or both) increased current draw from the relay coil or drop in input voltage because the rest of the circuit is powering up.

  • Would adding bigger caps on the input/output of U303 (or a big buffer cap on the 24VDC input) resolve this issue?
  • Is there a better way than using a bulky relay? Worst case my system would need 4-5A if the amplifier is at full volume. Would it be better to use a MOSFET and gradually bring up VCC to 24VDC?
  • I notice that U303 gets quite hot (about 60°C) when the relay is active, this might also be a reason to switch to a MOSFET based setup (or use a bigger regulator).

Problem 2: An unexpected voltage

When powered off, there seems to be about 1.3V on the switched 5V rails.

My PCB is 4 layer. Top and bottom are for signals, the inner layers are GND and power (VCC in most places, ±5V in others).

I've split the GND into two planes (red outline). Left is for the input/digital parts, right is for the analog parts.

  • Could this problem occur because I leave the VCC power input floating when powered off?
  • What is a good solution for this? Connect VCC to GND with a resistor? Ideally I'd use the relay to switch between 24VDC and a resistor to GND, but I've designed the thing the wrong way around. :-(

enter image description here


For now, I've by-passed the relay and power the entire circuit directly - so no soft power on/off. The amp functions correctly then (except for some fine-tuning in the software for volume control, etc.)

This is a hobby project, the first one on this scale, so feel free to point out any other obvious mistakes.


1 Answer 1

  1. You're much better off using a FET to switch the power than a relay. The current spikes from a relay coil can give you all kinds of supply problems. The FET is also far more efficient.

  2. U303 (and likely the other linear regulators) are heating up because each one is dissipating about four times as much power as they're delivering to the load. It would be far better to use a switcher to get into the 5-10V range, and run your linear supplies from there.

  3. You're probably seeing stored charge in the caps. In the 1.3V range, the devices may lose any capability of switching on, and present a high impedance, allowing the caps to stay charged. A bleeder resistor in the 10K-100K range would probably get rid of that.

  • \$\begingroup\$ Thanks! 1) Good idea. Not sure if I can retrofit it on this board, but the surely the next revision. 2) They heat up sure, and the 19V drop is quite large, but U303 only powers the MCU, a 10mA LED and the relay. That last one is probably the culprit. 3) I should be able to hack in some bleed resistors. I have a R + LED on each rail, but the Vf of the LED prevents a full discharge of caps. Thank you for your answers! \$\endgroup\$
    – Ariejan
    Jan 24, 2019 at 14:09
  • 1
    \$\begingroup\$ I agree with Chistobol Polychronopolis. If you want to keep the relay, you could switch it to a 24VDC type and led the 24V supply feed the relay coil. That way you wont load the +5VD supply. All it takes is a new relay, at cut PCB trace and a piece of wire. \$\endgroup\$ Jan 24, 2019 at 14:26
  • \$\begingroup\$ @PeterKarlsen Good idea! \$\endgroup\$
    – Ariejan
    Jan 24, 2019 at 15:01

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.