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I came across this post on TI E2E forum.

In this post user wants to drive LMK04832 PLL IC clock inputs using a sine wave. User has a source of sine wave with -1.53dBm or 0.528Vpkk which they will use to drive LMK clock inputs.

LMK04832 input clock level requirements are:

enter image description here

Now since user has 0.528vppk input signal so to increase the peak to peak voltage they are planning to use 1:2 TC2-1TX+ balun and drive the LMK clock inputs differentially.

This balun has around 0.5dB insertion loss, so taking this into account, they have calculated that the voltage at differential(balanced) side will be 1.33 times of voltage at unbalance side.

So, peak to peak voltage at balanced side will be 1.33*0.528 = 0.7vppk. They have assumed that LMK clock inputs are high impedance so they will not draw too much current.

But on the forum TI engineer said the following :

"Ideally, if the input to a 1:2 balun is X dBm, then the power on each of the differential output pin will be X-3 dBm. (We cannot get more energy from a passive device!)"

But I think a balun is a kind of transformer which idealy keeps the power level on input and output same and voltage/current changes accordingly.

So, my question is :

Can we increase the voltage using 1:2 balun and in doing so balun will keep the power level same at balance and unbalance side?

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  • \$\begingroup\$ Hi Andy, so it means if we measure peak to peak voltage at differential side (secondary)of balun by probing P and N(not P/N wrt ground), for the selected balun with 1:2 balun, the peak to peak voltage will be 1.33 times of single ended(primary)? \$\endgroup\$ – Alan Nov 20 '19 at 13:53
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Two lots of (X - 3 dBm) is X dBm. They are talking about both wires on the output of the balun and, it’s likely they are referencing each output to a (imaginary or actual) centre tap on the secondary.

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