ISO 14443 - Timing and AN6065

[TBaue.1]

Hello to the support community, 

 

i'm currently working on some custom coils used with the ST25R3916. I've studied the ISO/IEC 14443-2 for quite a while and recently had a look into the AppNote AN6065 from you guys. 

 

There is quite some confusion on the topic stated in the AN around Figure 15.

First of: the minimum timing values stated in Table 3 are somewhat different from the ones in the actual ISO norm. That is not really the point of my question, but i just wanted you to know. 

 

The real question arises from the following fact:

In the AN the Q factor of the coil is calcaulated to be Q<= 13.56 MHz * 3µs = 41. (The very bottom line on Page 18).

The 3µs represents the time that the unloaded coil would need to rise/ fall to the min/max amplitiude. So far so good. 

 

Looking at this timing diagram: 

I would expect that those 3µs would represent t3 or t1-t2 as those are the rise and fall times of the symbol. Looking in the table:

* t3 = 16/fc = 1.18µs

* t1-t2 = 2.98µs - steady_state_time

 

Obviously, t1 was used as the T for the T=1/B and thus the Q related calculation, which in my opinion does not make any sense as it is at least three times bigger, then the actual required rise/ fall times (t3, t1-t2). 

 

Could you somehow explain why you only care about the circuit being damped to 5% at the very end of t1 rather then t1-t2? 

 

Thank you and kind regards,

Tony

 

[Travis Palmer]

Hello Tony,

 

many thanks for highlighting. I assume this numbers are based on an old standard and have been propagated to newer versions of the antenna design application note. We will consider it for upcoming changes on the AN.

In the 2020 standard the maximum t1 is 40.5/fc. The minimum t2 (in case t1> 34/fc) is 7/fc. Resulting in a falling slope of 33.5/fc.

Due to the active driving of the antenna, the rise time is much faster than the falling slope.

You can see this for example in Figure 29 of AN6065 Rev4. 

The fall time is 35-27.3 = 7.7/fc, the rise time is  5.2/fc.

This is also reflected in the standard. The maximum fall time (as calculated above) is 33.5/fc, the maximum rise time is limited to 16/fc.

BR Travis

 

[TBaue.1]

Hello Travis, 

 

thanks for the quick and detailed reply. 
This describes and explains quite a lot as it's in line with what I understand. But still: in the AN you calculate the Q factor by 3µs*13.56MHz which would result in a maximum feasible Q of 41. As you described, the fall time must be in the order of t1-t2 = 40.5-7 = 33.5/fc which would equal 2.47µs and thus, a max Q-factor of 33.5 which is ~18% different than what is stated in the AN. 

 

Did I get something wrong here or is that an error in the AN? 

 

Kind regards,

Tony  

[Travis Palmer]

Hello Tony,

 

No, it is in line with what I stated before. 

For ISO14443-A 106kbps the Q-Factor of the system should be a maximum of ~33.5.

 

BR Travis