balun, and balun circuitry

Question

As I understand it a balun is something that converts a line with a signal into 2 lines with mirrored signal to reject the noise like showed in this picture : I am trying to understand the balun of the CC1101 circuit : . As I understand it, the balun is usefull to reject common mode noise picked up by the RF cable.

1)I don't understand how the capacitors and inductors showed in the yellow circle create the signal and a mirror copy of it like showed in the first picture.

2) Why is the balun locate just before entering the IC? It will receive alrady plenty of RF noise before no? I(In meander antenna, in the matching network,...)?

Answer 1

According to the data sheet, \$\small C_{124}\$ is a relatively high value decoupling capacitor and \$\small L_{131}C_{131}= L_{121}C_{121}=LC =1/{{\omega_{n}}^2}\$. Component values are: \$\small L=33nH;\:C=6.8pF\$, giving a natural frequency of \$\small\omega_n=2.11\times 10^9 rad/sec\$, or \$\small f_n= 336MHz\$, which will lead to a slightly lower resonant frequency, \$\small f_r\$, in a practical circuit where resistance is taken into account.

Assume now some small series resistance, \$\small R\$, in each of the branches leading to pins 12 and 13, hence we have two series L/C/R circuits with the \$\small V_n\$ and \$\small V_p\$ signals taken across the capacitor and inductor respectively. Using the voltage divider rule, \$\small V_n\$ and \$\small V_p\$ may be expressed in terms of the input voltage to the balun, \$\small V_i\$ (i.e. the signal at the junction of \$\small L_{122}/L_{131}/C_{121}\$) as: $$\small V_n=\frac{\frac{1}{LC}}{s^2+\frac{R}{L}s+\frac{1}{LC}}V_i $$ $$\small V_p=\frac{s^s}{s^2+\frac{R}{L}s+\frac{1}{LC}}V_i$$

Writing these in standard form:

$$\small V_n=\frac{\omega_n^2}{s^2+2\zeta\omega_n s+\omega_n^2}V_i $$ $$\small V_p=\frac{s^2}{s^2+2\zeta\omega_n s+\omega_n^2}V_i$$

where \$\small \zeta=\frac{R}{2}\sqrt{\frac{C}{L}}\$

Notice that \$\small V_n\$ and \$\small V_p \$ are low-pass and high-pass filtered versions of \$\small V_i \$, respectively.

Moving now from Laplace domain to frequency domain (\$\small s\rightarrow j\omega \$):

$$\small V_n=\frac{\omega_n^2}{(\omega_n^2-\omega^2)+j2\zeta\omega\omega_n}V_i $$ $$\small V_p=\frac{-\omega^2}{(\omega_n^2-\omega^2)+j2\zeta\omega\omega_n}V_i$$

Now let \$\small \omega=\omega_n\$:

$$\small V_n=\frac{\omega_n^2}{j\:2\zeta \omega_n^2}V_i\:=\:\frac{-j}{2\zeta}V_i =\frac{V_i}{2\zeta}\large\angle\small-90^o$$ $$\small V_p=\frac{-\omega_n^2}{j\:2\zeta \omega_n^2}V_i\:=\:\frac{j}{2\zeta}V_i =\frac{V_i}{2\zeta}\large\angle\small+90^o $$

Hence \$\small V_p\$ and \$\small V_n\$ are the same magnitude but phase shifted by \$\small 180^o\$, i.e. one is the inverse on the other, so when they are subtracted: $$\small V_p-V_n=\frac{V_i}{\zeta}\large\angle\small+90^o$$

Using the values for \$\small L\$ and \$\small C\$ given in the data sheet, \$\small \zeta =7.2\times 10^{-3}R\$, and the voltage gain at \$\small 336 \:MHz\$ will be large (around \$\small 60\:dB\$ for \$\small R=0.1\Omega\$). This gain reduces to around \$\small 15\:dB\$ at the operational frequency of \$\small 315 \:MHz\$.

Common mode noise signals are rejected due to the subtraction performed on the \$\small V_p\$ and \$\small V_n\$ signals.

Answer 2

• Answer to (2):
  You wrote

  "As I understand it, the balun is usefull to reject common mode noise picked up by the RF cable"

  That's true but that's not what it does in general; that's also not why it's here in this circuit.

  In general (and also here in this circuit) it just interfaces a balanced line (IC side) to an unbalanced line (particular antenna).
  Therefore the name: BALanced-UNbalanced.

• Answer to (1):
  The impendance of the two subnetworks (composed of C121, L121, C124 and composed of C131, L131) should create such phase shifts at the particlular frequency that signals of both paths superpose constructively at the unbalanced node or vs. versa that the power from the unbalanced line is split into two signals of opposite phase relation (depending on whether this is a transmitter or receiver).

Answer 3

At series resonance L131+C121 create +90, -90 deg signals to create the 180 deg difference over the Q range.

The reference design forms a balun that converts the differential RF signal on CC1101 to a single-ended RF signal to the antenna. Together with an appropriate LC network, the balun components also transform the impedance to match a 50R load.

Answer 4

The differential behavior is very important for the successful, non-spurious, operation of that RF IC.

Single-ended output of a transmitter, with load current looping back into the silicon by any and all GND and RTN pins and even VDD pins what with bypass capacitors injected GND ripple, is a huge oscillation risk.

Using this formula: Foscillation = Rload/(2*piAvLgnd), we find with 50_ohm load, gain of ONE, and 10nH Lgnd, the lowest frequency of oscillation is

50/(6.28 * 1 * 1e-8) or approximately 900MHz.

Yet inside a transmitter, there is lots of amplification, even if just the final few stages of RF power amplification. Thus Av = 1 is way too low.

Summary: single-ended transmitters are big risk; the IC of your question uses differential output pins, for stability and thus avoidance of spurious energy.

The same risks occur in the receiver.