ADL5382 LO Leakage and Baseband Loading Clarification

Question

Hello, 
 
 The ADL5382 datasheet states in multiple locations the LO to baseband leakage in terms of dBm. However, the baseband ports are not loaded in 50 Ohms. Can someone clarify what the roughly -40 dBm leakage figure means? How can I convert this figure to, say, LO voltage amplitude at the baseband ports? 
 
 If I treat it as an actual power measurement, I would do sqrt(10^((-40 - 30)/10) * 2 * 450) * 2 = 19mVp-p, which assumes a 450 Ohm baseband load. Is this the correct assumption? Or should 50 Ohms be used here? 
 
 Thanks

dmailloux · Accepted Answer

Hi David, 
 The datasheet page 21 has an explanation of the characterization setups. The -40dBm LO leakage is into a 50 ohm system by means of a 9:1 impedance transformer. The load presented to the ADL5382 is 450 ohms. 
 Your calculation seems correct, but I'm trying to follow the quantities, particularly the -30 within the exponent. Is that to account for mW to Watts conversion? 
 Here's my hand calculations: 
 
 PL dBm = -40dBm 
 PL = 10^(PL dBm / 10) / 1000 = 100e-9 Watts 
 Z1 = 450 ohms 
 Z2 = 50 ohms 
 V2 = sqrt(PL * Z2) = sqrt(100e-9 * 50) = 2.24e-3 Vrms 
 N1 / N2 = sqrt (Z1 / Z2) = sqrt(450 / 50) = 3 
 V1 = N1 / N2 * V2 = 3 * 2.24e-3 = 6.71e-3 Vrms 
 V1 PP = 2 * sqrt(2) * V1 = 19e-3 Vpp 
 
 Of course this assumes the power in to the transformer is equal to the power out. There may be some loss in the transformer (maybe 1 to 2dB depending upon frequency), which means the voltage on the baseband side is a little higher. 
 Best Regards, 
 David)

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ADL5382 LO Leakage and Baseband Loading Clarification