I am trying to sample a LED light driver signal which is DC signal with an approximate mean voltage of just over 50V and a 3V pk-pk ripple. I would like to sample the signal with a microcontroller which has a maximum 5V input on it's ADC.

I tried building a simple potential divider circuit with 1M and 100K resistors, and then opened up the lamp and connected my potential divider in parallel with the LEDs being driven in the lamp (there appear to be 9). This did the trick for lowering the voltage, however it completely changed the shape of the signal. I'm assuming this occurred because I'm changing the load on the driver?

Here's some images of my signal before and after the potential divider. The stepped down voltage not only changes shape, but also becomes slightly unstable. The peak to peak voltage of the wave fluctuates ocassionally getting bigger, as well as the signals frequency which decreases (so the wave spreads out and gets bigger).


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Does anyone have any advice as to how do I should go about stepping this voltage down, while still preserving the original signal shape? Thank you.

Edit: So this is my schematic.

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The LED lamp is a normal lamp for domestic use, which is powered of 240V AC mains. The black box is the driver circuity connected to 9 LEDS in series. I didn't design this, the friendly lamp makers did. (As I can't post more than 2 links per post with a reputation below 10, I have had to include the pictures as I have below, if you copy and past it into your URL bar it should work however)

FakeMoustache, the scope itself does not have a ground connection and is connected to mains via an isolating transformer so that the ground is floating. I don't entirely know what you mean by "switched from the positive side". They're all in series connected positive to negative though :).

When I measure the across the LEDs and add the potential divider the signal does not change. It does however change when I tap into into the potential divider. Also pretty strange, but probably down to me doing some rooky mistake with my scope, the signal shape changes with the attenuation factor on the probe.

Here's some measurements AC coupled. The first two are with x10 attenuation on the probe (with the oscilliscope set to x10). The second two are with the probe and scope set to x1.

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I will be sampling at 1Khz, but I have yet to actually try it, will do that today and plot the graph to see how it compares to what I'm getting out my scope.

Stray capacitance is a possibility as I'm using a couple of roughly 1.5-2m wires to connect the lamp to my potential divider. But wouldn't that affect the signal across the whole divider as well as from the middle?

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    \$\begingroup\$ I'd be willing to bet that the shape really hasn't changed much at all it's just that the scale is different. It's interesting that the pk-pk value is a higher percentage of the mean in the second capture vs the first. It just doesn't make sense that a 1.1M load would affect the regulator at all. \$\endgroup\$ Oct 31, 2016 at 14:19
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    \$\begingroup\$ While watching the original measurement, add a small load (like the divider) to that signal. Does it change just under that load? \$\endgroup\$ Oct 31, 2016 at 14:21
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    \$\begingroup\$ Is the lamp in fact connected to AC power or is it a true DC set-up you have created? \$\endgroup\$
    – Andy aka
    Oct 31, 2016 at 14:22
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    \$\begingroup\$ To take this discussion to the next level you really should include a schematic. For example, it is unclear how the ground of the scope is connected. Are the LEDs switched on the positive side etc.. \$\endgroup\$ Oct 31, 2016 at 14:25
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    \$\begingroup\$ Time base appears to be different for the two images. Have you tried switching to AC coupling on the scope to see just the ripple? Also, if that "Post" image is picked up from the voltage divider, then it is possible the the high impedance is either letting it pick up noise or that some capacitance is corrupting your signal. \$\endgroup\$
    – JRE
    Oct 31, 2016 at 14:26

2 Answers 2


You need to buffer the signal before dividing it down. Op-amps are the usual way of doing that. Depending on the precision you need, an instrumentation op-amp costs a little more but has a lower offset voltage and tends to be the most linear over range.

The usual configuration is to connect the signal to the positive input of an op-amp, perhaps with a high value resistor for protection. The input of the op-amp is very high impedance. The output is then divided with resistors, and optionally fed into another op-amp buffer to avoid problems stemming from the impedance of the ADC.

The following schematic demonstrates a divide by 2:

op-amp buffer circuit

  • \$\begingroup\$ Don't know why some idiot voted this down, it's correct and tested on actual hardware. \$\endgroup\$
    – user
    Jan 15, 2018 at 11:54
  • \$\begingroup\$ The signal has 50V and a 3V ripple. MAX44246 can input maximum 36V. Not me. \$\endgroup\$
    – Dorian
    Apr 19, 2018 at 7:58

You should compensate the oscilloscope 10x probe first. Looks overcompensated to me.

The led might have a open drain driver something like that, the falling edge being the floating one that changes to driven low when you put the divider.

That's why the shape looks different with and without the divider.

A common driver like this will leave the negative side of the LED floating between inductor discharge and transistor open.


simulate this circuit – Schematic created using CircuitLab

The jitter you see with the 10X probe is due the HF noise amplified by overcompensated probe disturbing the trigger.

100K impedance is not so low to be unsensitive to tapping.


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