roy@phri.UUCP (Roy Smith) (07/05/89)
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"abali@parts.eng.ohio-state.edu (Bulent Abali) (07/05/89)
In article <3834@phri.UUCP> roy@phri.UUCP (Roy Smith) writes: = First, the DC model: = = = 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? Ic=0, therefore Ib=Ie (Ie to be taken out of the emitter). Voltage division at the base gives 6*2.2/(10+2.2)=1.08V. Therefore, 1.08V = 1.8K*Ib + 0.7 + 1K*Ib You have nothing to do with Beta in the DC model, because the transistor is not in the active region. Base acts like a forward biased diode. -=- Bulent Abali Ohio State Univ., Dept.of Electrical Eng. 2015 Neil Av. Columbus, Ohio 43210 abali@baloo.eng.ohio-state.edu