Hi,

in my adl5380 application BB I/Q max freq is 190MHz.

ADL5380 should be connected with either a 2¨1 balun (100Ω -->50Ω) or a 4:1 balun (200Ω-->50Ω)

A microstrip 100Ω differential line of aproax 12cm long connects al5380 with the balun.

Balun could not be placed at adl5380 side.

adl5380--> 100Ω microstrip 12cm long diff line --> balun --> passive network

passive network Zin=50Ω

Balun 4:1 or 2:1

In order for adl5380 to see a load impedance >=200Ω, I'm considering the following interface options:

1. Use a resistive impedance transformation network near adl5380 to implement a 200Ω (adl5380 load)-->100Ω transformation, driving differential line with 100Ω source impedance and use a 2:1 balun at the end of the line. That solution will yield to the optimum matching , adl5380 driving scheme, but will have a rather large pwr loss.

2. place two 75Ω resistors in series with I,I', Q,Q' outputs , and a 4:1 balun at the and of diff. line. There will be an impedance mismatch at the adl5380-diff line side , adl5380 load impedance will be >200Ω, pwr loss will be less than choice 1.

3. As choice 2 , but using 25Ω resistors , not creating impedance mismatch at adl5380 side.

Could ADL5380 drive directly a 100Ω diff line terminated at its end with a 200Ω impedance?

Which is the recomended interface option?

Which are the pros & cons for each 3 solutions?

would be any problem selecting choices 2 or 3 as far as adl5380 is concerned?

Thank you

Hello limni,

I would agree with your analysis concerning properly terminating the ADL5380. Option 1 would be the best in the sense that the source and load impedance would be matched, however the tradeoff is power loss. Option 2 and 3 will also generate power loss where option 3 will provide the least loss out of all three options. There is no one answer but you will need to way your options.

The ADL5380 can drive directly a 100 ohm differential line terminated by 200 ohms from the 4:1 transformer however the load as seen by the ADL5380 may be too low. If my understanding is correct, the 100 ohm microstrip line would be in parallel with the 200 ohm load impedance and the resulting load as seen by the ADL5380 would be 67 ohms. This load may be too low for the proper signal swings. This is where option 2 and 3 comes in because by placing the series resistor you're increasing the load as seen by the ADL5380. However, any time you put in series resistors you're forming a voltage divider and this results in power loss. With the 25 ohm series resistor, you'll get about 2.4 dB in power loss and with the 75 ohms the power loss increases to 5 dB.

The load as seen by the ADL5380 is 117 ohms for option 3 and 217 ohms for option 2. The load as seen by the ADL5380 affects the voltage conversion gain of the device. The baseband output of the ADL5380 behave like voltage sources with 50 ohm differential output impedance, 25 ohms on each side, as described on page 6 of the datasheet. It's the voltage drop across the internal impedances that causes the voltage gain to change with termination.

Vo (load) = Vo (open) * RL/(RL + RS)

where RL is the differential load impedance,

and RS is the differential output impedance (50-ohm).

With this said, option 3 will have approx 0.7 dB less conversion gain than option 2.

I hope I didn't confuse you with all the different scenarios. As I had mentioned earlier, there is no one answer and you will have to weigh your options to determine the best solution for your application.

Qui