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I'm wondering what \$C_B\$ attached to the bottom of \$R_{C2}\$ does. This is a screenshot from a textbook called RF Circuit Design: Theory and Applications by R. Ludwig And P. Bretchko. Looking at 8-36a and 8-36b, \$C_B\$ doesn't seem to ever get used (the one at the base of the \$Q_2\$ does though).

The book does say somewhere else, though, that often the RFC is replaced with a \$\frac{1}{4}\lambda\$ line which converts the short created by \$C_B\$ to an open circuit, so would that imply that that cap is redundant if an RFC is used?

bias

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It provides a low impedance point for RF for the voltage divider formed by the RFC and Cb, thereby making the decoupling more effective at keeping RF out of the power supply.

Without CB the RF appearing on the power supply is simplistically
Vout_RF x Zsupply / (Zsupply + Z(RFC >~~< RFC2)
(Where >~~< is the complex real world Murphy transform operator whose operation is often summarised as "good luck with that one")

I say simplistically as there are all sorts of real world considerations to reduce the level of reduction (and at RF >~~< probably features in most of them).

Add Cb and you (again simplistically) get a divider formed by the RFC and Cb and then this is in turn divided by Rc2 & Z supply. Cb will usually be able to be placed close to Q2 in such a way that impedance at frequencies concerned is able to be minimised. Simplistic as above plus of course the 2nd stage filter loads the first and there is an impedance mismatch or few and ... . Grab a Smith chart, work out the real world effects of what is present where Cb is terminated (>~~<) and lay to.
Ends up all very arcane and uncertain very rapidly BUT adding a low impedance point half way through the decoupling path has got to be good for you.


Entirely consistent with RF and Murphy being at work, if you type >~~< without a space between the > and the preceding letter the 2nd half of the operator and a semi random portion of the following text does not appear in the output copy.
Ask me how I know.

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