1
\$\begingroup\$

I'm having some problems with an a LT1301 step up DC/DC converter. I'm using the circuit that appears on the datasheet to convert 5V to 12V, but I'm getting periodic spikes on the output when loading with a LED. The spikes are high, +- 5 Volts around the 12 V signal.

enter image description here enter image description here

I don't understand why it doesn't behave like the datasheet says, if this is the same circuit.

Thanks in advance.

EDIT1:I haven't enough reputation to post more than 2 links, but the spike duration is around 60 ns

\$\endgroup\$
7
  • \$\begingroup\$ Please show your measurement setup and that you have watched m.youtube.com/watch?v=Edel3eduRj4. Also, show layout. \$\endgroup\$
    – winny
    Aug 7, 2017 at 15:04
  • \$\begingroup\$ Hello, I've watched the video this morning looking for information. I measure directly with the probe connected to the ground in the protoboard I've the design. Sorry if this is little information, I'm starting on that. I try to do a differential measure with 2 probes (one substract the other), and it continues appearing. \$\endgroup\$
    – gutiale
    Aug 7, 2017 at 15:32
  • \$\begingroup\$ "directly with the probe connected to the ground in the protoboard" as in a 20 cm loop? The sawtooth shaped waveform is probably true, but the very shoes spikes could easily be a measurement issue. See Davids answer below also. \$\endgroup\$
    – winny
    Aug 7, 2017 at 15:41
  • 1
    \$\begingroup\$ Can show us the protoboard? \$\endgroup\$
    – Jason Han
    Aug 8, 2017 at 3:42
  • 1
    \$\begingroup\$ Hello, I add this cap, and measuring with the ground clip now I get a signal with less than 400 mV peak to peak. In a protoboard, so I think in a PCB it will be better. Thanks \$\endgroup\$
    – gutiale
    Aug 9, 2017 at 7:05

3 Answers 3

Reset to default
2
\$\begingroup\$

There are two likely explanations:

  1. The spikes aren't really there (mostly). They are scope artifacts, possibly make worse by poor grounding of the scope probe. Connect the ground clip of the scope probe directly to the negative side of C2, and the tip directly to the positive side.

  2. The high ESR of the output cap (C2) is contributing to the spike size. It's OK to have a electrolytic on the output for bulk storage, but there should also be something there with low impedance at high frequency. At only 33 µF, C2 could have been ceramic. Leave what you have there, but add at least 10 µF of ceramic across it.

\$\endgroup\$
3
  • 1
    \$\begingroup\$ Whoever downvoted this, please explain what you think is wrong. \$\endgroup\$
    – Olin Lathrop
    Aug 7, 2017 at 22:04
  • \$\begingroup\$ I have a 10 uF ceramic capacitor here, but using a ground clip and measuring directly in C2 spikes have gone down to just 1 V peak to peak. I think that using a 10 uF ceramic capacitor in parallel with C2, good PCB design it will go lower, but do you think is worthy to add an snubber? \$\endgroup\$
    – gutiale
    Aug 8, 2017 at 9:05
  • 1
    \$\begingroup\$ @gut: A snubber makes no sense in this context. \$\endgroup\$
    – Olin Lathrop
    Aug 8, 2017 at 11:37
3
\$\begingroup\$

The "spikes" are the ringing in the inductor (self-resonance) associated with the switch-off transient of the regulator's switching transistor, coupled through the capacitance of the rectifier diode. There is actually very little energy associated with them, and they can be reduced to negligible levels by good circuit layout and some judicious bypassing of the output with a ceramic capacitor. This keeps the ringing from affecting other circuits via either conduction or radiation.

In extreme cases, a "snubber" (resistor and capacitor in series) connected between the switch node and ground is required to suppress the ringing. The component values required in the snubber depend on the specific characteristics of the inductor.

\$\endgroup\$
4
  • \$\begingroup\$ Thanks, but why there is little energy associated with them if they get almost 5 volts? Ok I will try what you recommend \$\endgroup\$
    – gutiale
    Aug 7, 2017 at 15:41
  • \$\begingroup\$ Because their duration is so short. If you decrease the time/div on your oscilloscope, you'll see the characteristic decaying sine wave. \$\endgroup\$
    – Dave Tweed
    Aug 7, 2017 at 15:41
  • \$\begingroup\$ The duration of the spike is around 60 ns. I can see the sine wave as you said. \$\endgroup\$
    – gutiale
    Aug 8, 2017 at 7:18
  • \$\begingroup\$ How can I calculate the values of the R and C for the snubber? Using a ground clip I can reduce the spikes to just 1 v peak to peak, so I would like to get rid of the whole spike. The sawtooth wave is little enough to be in the specifications of the dc-dc converter. So it is almost solve. Thanks \$\endgroup\$
    – gutiale
    Aug 8, 2017 at 9:03
2
\$\begingroup\$

Like Dave Tweed mentioned, these spikes are likely very low energy. Zoom into them (in the time scale) and check the duration. If they are in the order of a few nanoseconds, you can easily solve it following Dave Tweed suggestions e.g. adding a snubber. I would just like to know how much load you are drawing from this converter. You mentioned you connected an LED but you didn't mention the series resistor, or if there is more than one LED. I would be more concerned with the sawtooth-like shape you have after the ringing; generally, the converter shouldn't have that jump in the output voltage.

\$\endgroup\$
2
  • \$\begingroup\$ There is only 1 LED, and as far as I could characterised the V-I curve, the impedance of it is around 130 ohm. \$\endgroup\$
    – gutiale
    Aug 8, 2017 at 7:27
  • \$\begingroup\$ So I believe that this doesn't have anything to do with the loading. I've read that the sawtooth isn't a problem for you, so if you are not looking to track down that issue, you could just add a ceramic capacitor in the output of about 100nF. That should help suppress the high-frequency spike.The problem is that if you are doing this on a breadboard, it will be very hard to know what the voltage waveform really looks like. That spike may not even be there when you have your circuit in a PCB with proper ceramic caps on the output. Also, try to see how your output reacts to different loads. \$\endgroup\$
    – Lucas
    Aug 9, 2017 at 19:53

Your Answer

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

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