1
\$\begingroup\$

I am getting 7 volts from a power supply source which I need to step down to 3.3 V for powering up my 2.4 GHz zigbee MCU (TI CC2530 + CC2592 amplifier).

I am using a linear regulator to step it down. My schematic is as shown below:

linear regulator

Randomly I am seeing MCU resets and brief communication failures. I checked the power line waveforms and they look like these (with ac coupling):

Input (7 V line):

1

2

FFT looks like this:

3

Output (3.3 V line)

1

2

FFT looks like this:

3

I can do nothing about the input. What should I do to ensure a better power line for the MCU in order to get rid of resets and communication failures.

Note - same circuit works fine if I power it up from my bench power supply.

EDIT (solution posted below):

I followed @analog's solution but with different values (because I didn't have the values that he mentioned):

cktt

Setup:

setup

Waveforms with 180 ohm load on the 7 V line (ac coupling):

enter image description here

enter image description here

Waveforms with 33 ohm load on the 7 V line (First one is DC coupling and rest are ac coupling):

enter image description here

enter image description here

enter image description here

EDIT: Adding more waveforms.

Both probes connected at Vout. Tested at no load and 33 ohm load conditions. Yellow line is DC coupling. Blue line is ac coupling. No load voltage is roughly 7.5 volts. At 33 ohm load, the voltage drops down to roughly 5.7 volts.:

enter image description here

enter image description here

enter image description here

enter image description here

enter image description here

\$\endgroup\$
  • \$\begingroup\$ How much current are you drawing from the linear regulator? \$\endgroup\$ – HandyHowie Jun 5 at 9:24
  • 1
    \$\begingroup\$ The regulator datasheet recommends a 22uF solid tantalum on the output for stability (and that probably means there is a minimum ESR). Have you tried that? \$\endgroup\$ – Peter Smith Jun 5 at 9:30
  • \$\begingroup\$ You have your noise filtering caps, but that looks like your input voltage is dropping below the point your linear regulator can handle? I think you need to add more capacitance to the output. \$\endgroup\$ – hekete Jun 5 at 9:31
  • \$\begingroup\$ I have tried adding 100 uf electrolytic with 0.1 uf ceramic on output, input and both. Didn't help much. \$\endgroup\$ – Whiskeyjack Jun 5 at 9:34
  • 1
    \$\begingroup\$ Pretty much what I thought, you have no PCB and hence no layout, so it's not going to work properly. At least you should get breadboard so you can keep things tight and have a decent strip of copper for power/GND. \$\endgroup\$ – Barleyman Jun 7 at 13:24
3
\$\begingroup\$

Like this

schematic

simulate this circuit – Schematic created using CircuitLab

If your current demand will not allow the 10 ohm resistor, then use 1uH inductor in parallel with 100 ohms Rdampen.

\$\endgroup\$
  • \$\begingroup\$ Thanks. I'll try it. Hopefully it fixes my problem. \$\endgroup\$ – Whiskeyjack Jun 5 at 15:18
  • \$\begingroup\$ I didn't have the exact component values that you suggested. I used different values for a proof of concept and it worked. Thanks a lot man. How can I buy you a beer? \$\endgroup\$ – Whiskeyjack Jun 6 at 14:19
  • \$\begingroup\$ How clean is the output? Set scope on 1 microsecond/division, ACcouple, 10 millivolt/division, have only one scope probe in use with its ground clipped onto the right(quiet) end of the Ground Plane. \$\endgroup\$ – analogsystemsrf Jun 6 at 16:28
  • \$\begingroup\$ I will be able to do that after 13 hours from now. Will post the image for sure. Till then can you please help me understand the impact on my output when I replace 1.5 mH with 1 uH. I will be reducing the inductance by a factor of 1500. What can I do to compensate for this? Will changing 5 ohms to 10 ohms (R1 as per your ckt) improve the performance? \$\endgroup\$ – Whiskeyjack Jun 6 at 16:42
  • \$\begingroup\$ examining your photos, I see very slow ramps which the LDO should be able to servo-out (regulate-out, remove). I also see ringing at frequency near 4MegaHertz, which LDOs have trouble with. The 10 ohm resistor and 100uF cap form a 1 millisecond time constant, which is 1/(2 * pi * 0.001) = 160Hertz. Given the ground plane, to help the big-cap be more effective, you ideally would see 4,000,000 / 160 = 24,000X attenuation. But caps have ESR that limits the high-frequency attenuation. The inductor helps at high frequencies. Use 1uH or 10uH (or any inductance in that range). No magic here. \$\endgroup\$ – analogsystemsrf Jun 6 at 17:43
1
\$\begingroup\$

You've got extremely low switching speed in the PSU, 350µs works out to about 2.9kHz, it should generate a clearly audible whine at such a low speed. It might be caused by low load power saving feature kicking in, we don't know what the current drain is.

The output is ringing at about 2MHz (0.5µs). I'd hazard that the problem here is your wiring/PCB, try adding 4.7µF ceramic caps right next to both IC power pins. 100nF is not very effective at 2MHz, depending on the package it'd be around 0.5-1R.

\$\endgroup\$
  • \$\begingroup\$ The output caps that you see in my voltage regulator circuit is placed quite close to the pins of zigbee. Trace length = 15 mm. Trace width = 10 mils (0.254 mm) \$\endgroup\$ – Whiskeyjack Jun 5 at 11:14
  • \$\begingroup\$ I tried adding load on the 3.3 V line. 43 ohms resistor. Didn't help. I added 22 uF electrolytic + 0.1 uF ceramic on 7 V line and 100 uF electrolytic + 0.1 uF ceramic on the 3.3 V line. Still the waveform didn't change much. I replaced the linear regulator with a different part number (linear again). Still no difference. \$\endgroup\$ – Whiskeyjack Jun 5 at 11:16
  • \$\begingroup\$ Unless you got a low-z electrolytic, 22µF can's ESR will be quite high. And even with "low" impedance cans it's around 1R at 100kHz. Generally speaking, you'd need something like 220µF can to have a significant impact on switching noise and the "low" impedance would be in the ballpark of 0.2R or so. You'll definitely get more mileage out of your 4.7µF ceramic capacitor. Do you have a proper GND plane? You could try adding a small RC filter in the IC power feed e.g. 1R and 4.7µF would suppress the 2MHz noise to ~1/50th or 34dB attenuation. \$\endgroup\$ – Barleyman Jun 5 at 11:43
  • \$\begingroup\$ I will check the RC filter and revert back. I tried 470 uF electrolytic. It reduced the noise a bit but not up to satisfactory levels. \$\endgroup\$ – Whiskeyjack Jun 5 at 11:53
  • \$\begingroup\$ I tried 680 uf on input and 680 uf on output. No change. Am I doing something wrong? \$\endgroup\$ – Whiskeyjack Jun 5 at 12:23

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.