DAQ analog output signal becomes distorted when connected to resistor. Why?

Question

I am an undergraduate student in electrical engineering. I just started working with a DAQ (PCIe 6341 connected to SCB68A) which can output analog signals (sine, triangular, square.)

When I output 2 Vpp, 1 kHz sine wave and connect it directly to the oscilloscope, I get this:

(FYI, my DAQ has an update rate of 900 kHz, so the above signal is composed of 900 samples per period.)

When I connect a 470 ohm resistor to the DAQ signal and measure the voltage across the resistor, I get this:

I noticed that the voltage is almost halved, which seems to imply to me that the output impedance of the DAQ is close 470 ohm, which seems way too high to me. More importantly, I have no idea why my sine wave is distorted like this. This also happens at different frequencies and amplitudes.

Can someone explain why this is happening and how to fix this issue?

Answer 1

According to the datasheet for the PCIe 6341, it has an output impedance of 0.2 ohms, but a drive current of only 5mA.

Your 470 ohm resistor draws too much current when the voltage rises above a certain level. When the current through the resistor is more than 5mA, the output voltage can't rise any more - the signal is "clipped."

Answer 2

On page 3 of the PCIe 6341 spec sheet, in the section "Analogue Output". it states that the maximum current drive for the device is 15mA (but it shouldn't really be asked for more than 5mA).

Given what's on the scope screen, it seems like you aren't asking for that amount of current (from your first image):

$$ I = \frac{V}{R} = \frac{2V}{470\Omega} = 4.3mA $$

In the bottom image, though, I see that clipping happens at 1V peak, so I'm wondering if you haven't got your probe set to x10, and in reality the signal is being clipped at 10V:

$$ I = \frac{10V}{470\Omega} = 21mA $$

It's above the stated 15mA maximum, but it's close, and I imagine that NI's specifications are somewhat conservative.