Tips for triggering i_lim? (NFC WLC)

[OSilv.1]

Hello,

I've changed the impedance matching network on the NFC06A1 such that the system is resonant at 13.56MHz with an impedance of 2.6 ohms. The reader is in 5V mode with the regulator display register measuring 4.3V at VDD_RF. I have set the output resistance to 3.4 ohms.

I expect that it would output 1.6 Arms (4.3 V / 2.6 ohm) or possibly 0.72 Arms (4.3 V / (2.6+3.4) ohms) with the VDD_RF regulator limiting the current to 350Arms. However, the i_lim flag remains zero.

Is there a trouble shooting procedure I should follow for determining why the device will not output more than 350Arms?

Thank you!

Best,

Oliver

[Travis Palmer]

Hello Oliver,

The X-NUCLEO-NFC06A1 is in a configuration where the internal regulator is bypassed.

Therefore it will not be possible to trigger the i_lim bit.

This mode is described in the datasheet in chapter 4.2.11 Power supply system:

"If a transmitter output current higher than 350 mArms is required the VDD_RF regulator cannot be used to supply the transmitter. VDD_RF and VDD_DR have to be externally connected to VDD_TX (connection of VDD_RF to supply voltage higher than VDD_TX is not allowed)."

BR Travis

[OSilv.1]

Hi Travis,

Thank you for your reply. I have two questions:

(1) On the x-nucleo schematic, I cannot see VDD_RF connected to VDD_TX. It is just decoupled to ground. It is implied that VDD_DR is connected to VDD_RF with a flag, though I cannot find an explicit connection. Am I missing something on the schematic?

https://www.st.com/resource/en/data_brief/x-nucleo-nfc06a1.pdf

(2) If the VDD_RF is not regulating to 350mA, then the max current from the Nucleo board is about 500mA. To maximize power output (and therefore received power), the impedance of the antenna should be R = V/I = 5V / 0.5A = 10 ohms, leading to a maximum power output of P = I^2 R = 0.5*0.5*10 = 2.5W. I've been told that the highest output power occurs at the lowest possible impedance, however I believe this assumes a voltage source with infinite current capability. Is this thinking done correctly? Would I expect higher output power at 10 ohms impedance instead of 3 ohms? The impedance is purely reactance, as I have removed the damping resistors to increase Q.

Best,

Oliver

[Travis Palmer]

Hello Oliver,

unfortunately i looked at wrong/outdated altium files.

The power scheme is correct. VDD_DR is sourced via the LDO.

br Travis.

[Travis Palmer]

Hello Oliver,

I modified one of our ST25R3916-DISCO board to ~8.4Ohm which gives a driver current around 350mA.

The supply is done via USB (~4.8 to 5V).

If selecting an in-proper setting Reg: 2Ch = 0xf8 (too less drop) and then turning on the field, the Regulator display register is showing 0xf1 => i_lim bit is "1".

When selecting an proper setting (Reg: 2Ch = 0xd0 => 4.6V), then the i_lim bit is "0".

Values above trigger the i_lim bit to "1".

(Regulator display register is showing 0xa0).

The Procedure is always: Go to "ISO 14443A" tab, press configure and modify the registers via "Register map". Read back / update the register map via "Ctrl + R"

BR Travis

[OSilv.1]

Hi Travis,

Thank you for your response. I am glad for the procedure, which I will follow when I am in the lab. I have a few more questions:

(1) How did you calculate 8.4 Ohms as the impedance that would produce 350mA of output current? R = V/I, and if V = 4.8V and I = 350mA, then R should be 13.7 Ohms?

(2) I have modified the NFC06A1 to have a lower impedance than yours (~3 ohms) and set the regulator to manual 5V (2Ch = 0xf8), and still the ilim is not triggered. I would have figured that, while 3 ohms is not an optimal impedance, it would certainly draw over 350mA. Is it known why this is not the case?

(3) I have set the mode to ISO15693 as I will be wirelessly charging and communicating with a st25dvi2c chip. Does the ISO standard influence output current?

Best,

Oliver

[Travis Palmer]

Hello Oliver,

can you prepare your setup to measure the supply current?

We cannot simply do a DC operating point calculation to estimate the current consumption.

The typical NFC application has a matching network attached to the RFO pins. This matching network transforms the inductive impedance to a real ohmic load. The EMI filter is the first part of this matching network.

The EMI filter is transforming the rectangular voltage, which is outputted by the driver into a sinus shaped voltage. As a result, also the current going through the regulator is approximately sinus shaped. This is because the drivers produce a square wave voltage, but the EMC filter only passes the first harmonic.

below an example of calculating the current and also the dissipated power:

Input parameters:

VDD = 5V

Mathing Impedance = 8.4Ohm

LDO drop = 0.3V

RDRV = 1.75Ohm

IAL = 0.021A

Calculation:

PAL = VDD * IAL = 5 * 0.021 = 0.105 W

PREG = (VDD – VDDDR)*(VDDdr*1.27*0.637)/(2*RDS_on + ZLoad) = (5-4.7)*(4.7*1.27*0.637)/(2*1.75+8.4)

PREG = 0.0959W

VREG = (VDD-0.3)/2*1.27 = (5-0.3)/2*1.27 = 2.9845 V

IPEAK = (2*VREG)/(2*RDS_on+Zload) = (2*2.9845)/(2*1,75+8,4) = 0,501A

IAVG = IPEAK*0,637 = 0.501*0.637= 0,32A

PDRV = (IPEAK²/2)/(2*RDS_on) = (0,501^2/2)*(2*1,75)=0.439W

The total dissipated power of the device is then:

PTOT = PAL + PREG + PDRV = 0.105 + 0.0959 + 0.439 = 0.64W

The DC current (IAVG) is in this case around 320mA.

For triggering the i_lim bit, it is important to turn on the field.

BR Travis

[OSilv.1]

Hi Travis,

On my NFC06A1 I have swapped out the EMI filter such that it filters at 20300kHz and uses a low DCR inductor (60mOhm). I have replaced the serial capacitors in the impedance matching network with 56pF. I have removed all parallel capacitors to allow for more turns and higher parasitic capacitance in a custom antenna. The antenna itself has an inductance of 1600nH with a parallel resistance of 9000 ohms and a self-resonance of 33MHz. My goal is to transmit as much power as possible to the listener in a wireless charging and communications application, therefore I've removed the damping resistors to maximize Q.

I have been told that to maximize power output, I should set the impedance matching network (EMI filter, serial capacitors, and custom antenna) such that it is minimal (as close to 0+0j ohms as possible). However, with this setup, I do not see the Ilim flag triggering.

Thank you for the breakdown of your calculation. I have some questions on some of your steps.

Why choose 8.4Ohms as the matching impedance?

Is "RDRV" the driver output resistance? I ask because the datasheet specifies 1.7 ohms.

What is IAL? Is this a leakage current? I cannot find this value on the datasheet.

If I want maximum output power (1.75W), how should I set my impedance matching network?

Thank you for your help, and I am looking forward to your response.

Best,

Oliver

[Travis Palmer]

Hello Oliver,

can you measure the current consumption of your board in default configuration?

The setup mentioned in AN5584 - Figure 10 could be used to adjust the current on the fly.

br Travis