## Demodulating down to DC with HF2LI

Many of our customers use the possibility to measure DC voltage and current, together with an ac signal by setting the demodulation frequency ω=0. The measured value is seemingly incorrect as you need to divide the obtained rms value with the factor of √2 to get the actual signal amplitude. We can check this by applying AuxOut voltage to the HF2LI input either manually or by using the Sweeper tool as shown in the Fig. 1. The data taken with the Sweeper tool can be saved and fitted to get the exact coefficient and an offset for a selected Input range.

We note that for say, 1V applied at the input, the demodulator 1 in the Lock-in MF tab measures ~1.4 V (see graph). This looks like a factor of √2 too much, but it is actually fine and can be easily explained.

In the simple demodulation chain, a signal we want to measure Vs(t)= Vs*(cosωst+φ) is multiplied by the reference Vref(t)=Vref*cos(ωrt). Here |Vref(t)| =1. The measured rms signal value (before filtering) for ωs=ωr=ω is:

<Vs>= (1/√2) Vs*cos(φ)+(1/√2) Vs*cos(2ωt+φ). (1)

For arbitrary ω>0, we get two components of the signal, a DC and a 2ω component. The 2ω component is filtered out with appropriate filter setting of BW< ω/2 and we are left with just the DC component.

However, when ω=0, i.e DC measurement, equation (1) becomes,

<Vs,ω=0> = 2 (1/√2) Vs <cos(φ)>.

Here we note that the phase of a DC signal is ill defined with respect to a reference DC signal and for the sake of simplicity we will put it to φ=0, so finally we get measured DC voltage,

<Vs,ω=0> = √2 Vs.

We see that the measured DC voltage amplitude will be larger from the applied signal amplitude by a factor of √2. This can be explained by DC and 2ω components becoming equal and inseparable for ω=0. You can take care of this factor only by hand, i.e correct for it through separate calibration. In order to see the sign of the applied voltage/current, the phase can be zeroed and x-component of the demodulator used to measure further.