Path: utzoo!utgpu!jarvis.csri.toronto.edu!rutgers!phri!roy From: roy@phri.UUCP (Roy Smith) Newsgroups: sci.electronics Subject: Tx with no collector connection? Message-ID: <3834@phri.UUCP> Date: 5 Jul 89 00:40:22 GMT Organization: Public Health Research Institute, NYC, NY Lines: 48 Sigh, it's amazing how much you can forget after college. I was browsing through one of those TAB books (The Humungus Book of Electronic Circuits, or something like that) looking for RF mixer ideas. I found one that looked useful and sat down to analyize it. First, the DC model: open all the capacitors, short the inductors, and see what's left. For one of the stages, I got: o +6V | [key to ascii art: < and > are > resistors, V are ground points, < 10k C and the transistor is NPN] > / | B |/ +-------| | |\ > \ E < 2.2k | > < | > 1k V < | V There are, of course, connections to the collecter, but all of them use a capacitor as a series element, so there is no DC collector connection at all. How does one go about analysing a DC circuit like this? After taking the thevinin equivalent of the 10k/22k voltage divider you get 1.1V through 1.8 k. Assuming Q is active, you use KVL to write 1.8V = 1.8k*Ib + 0.7V + 1k*Beta*Ib and solve for Ib. This gives you an open-circuited current source of (Beta+1)*Ib at the collector. Obviously, with Ic=0, Q is cutoff, but that's not very useful. Am I missing something? If it helps, the AC model sees the modulated signal fed in via a series capacitor at the base terminal, and the Local Oscillator fed in at the emitter, above the 1k resistor. The down-converted signal is picked off the collector. Presumably this uses product detection, but I don't see where the product comes from either. -- Roy Smith, Public Health Research Institute 455 First Avenue, New York, NY 10016 {allegra,philabs,cmcl2,rutgers,hombre}!phri!roy -or- roy@alanine.phri.nyu.edu "The connector is the network"