DW1000 Power Consumption

According to the PDF file above provided by DecaWave Inc, rule 5 said that:
" To minimize power consumption always use a high efficiency external DC/DC
converter to supply VDDLDOD and VDDLDOA and not 3.3 V. "

I did an experiment to test the total current flow through STM and DW1000 in two cases.
In each case, I let DW1000 worked in “Only TX, Only RX and Sleep”.
Here are the data I get after 180minutes.
1st Case: I provided 3.3V to VDDLDOD and VDDLDOA.
Only RX. 0.1958 A
Only TX. 0.0592 A
Sleep . 0.0103 A

2nd Case: I provided 1.8V to VDDLDOD and VDDLDOA.
Only RX. 0.1958 A
Only TX. 0.0591 A
Sleep . 0.0103 A

Here are my DW1000 Configuration.
.LoadCode = DW_LOAD_UCODE,
.Channel = DW_CHANNEL_4,
.PulseRepFreq = DW_PRF_64M,
.TxPreambLen = DW_TX_PLEN_1024,
.PreambleAcqChunk = DW_PAC_32,
.TxCode = 20,
.RxCode = 20,
.NonStandardSFD = DISABLE,
.DataRate = DW_DATARATE_110K,
.SFDTimeout = (1024 + 1 + 64 - 32)
.PGdelay = 0x95,
.power = 0x39393939

Apparently, I did not see any changes of current consumption in two cases which is contradict to the information provided in the PDF file. Does anyone have any ideas?

Hi Andy,

Keep in mind that the main thing you generally care about is energy consumption, i.e the power drawn by the IC, not just the current. power = current * voltage, so if your current draw is the same, a higher voltage means a higher power draw.

When supplying the DW1000 with 3.3V the external LDO will be used. An LDO will drop the voltage in a very inefficient way, basically “burning up” the excess. The higher the voltage difference and the higher the current, the more inefficient it is. The DW1000 will draw short pulses of high currents, it might average out to be about the same, but the losses during these peaks will be a lot bigger. Measuring with a higher bandwidth current measurement device can show this. See e.g section 6.3.2 in the DW1000 user manual.

The longer you average for, the closer the results will look since VDDLDOA is pretty much only used during TX and RX.

Where are you measuring the current? Are you measuring it before or after the DC/DC? To compare apples to apples (i.e current including losses in the regulator), you should measure before the DC/DC.

Here is my setup, I am pretty sure that current is measured before DC/DC.
Here the power is drawn from 3.3V source, so even if 1.8V is presented, I assume that the main voltage still remain the same.