“The Electronics Of Radio” NorCal 40B Transceiver Build Lab Notes: Problem 23
This continues a series of blog posts on David Rutledge’s text, “The Electronics of Radio”, that I am studying while building the NorCal 40B transceiver. This series of posts will not be a review of the book, nor is it a assembly manual. Rutledge presents a series of problems at the each chapter that aid in understanding electronics and building the 40M QRP CW transceiver. I am going to try to go through all of these problems and document them here. All of these are titled similarly, so search for them that way. For what its worth, most people will want to skip these posts, they are really for my own self-education on electronics and may not make a lot of sense unless you have Rutledge’s book.
[The links to all problem solutions as I go through them will be posted here.]
Note: In order to get this problem to work, I needed to add a jumper from the key jack terminal corresponding to the tip (of the TRS) to the cathode of D11.
I recognized this because it was painfully obvious that the voltage measurement of the source of JFET Q5 should not be 0.35mV.
Investigation revealed no continuity between the tip of the keyer jack and the cathode of D11. I think the scratch I showed in Problem 20 may be the culprit. It explains some (most?) of the trouble I was having with the circuit I suppose. You can see the scratch at the bottom left near the jumper on the picture below.
This is what the jumper looks like (and yikes! the back of this board looks super junky and needs serious cleaning!):
A.
VDC (at source of JFET) = 1.718 V
Drain Bias Current = 3.37mA
B.
C.
Input voltage of buffer amplifier (vin(measured)) = 50mV
Output voltage of buffer amplifier (vout(measured)) = 47.5mV
Gv = vo / vi = 47.5 / 50 = 0.95
————-
Gv = vo / vi = 1/(1 + (1/gmR)
0.95 = 1/(1 + (1/gmR)
0.95(1 + (1/gmR)) = 1
0.95 + 0.95(1/gmR) = 1
0.95(1/gmR) = 1 – 0.95
1/gmR = 0.05/0.95
0.053 = 1/gmR
0.053gmR = 1
0.053gmR = 1
R = 510Ω
0.053*510*gm = 1
gm = 1/27.03 = 0.037S = 37mS
D.
Calculated transconductance = 6 mS
E.
vinput (at filter end of R10) = 50 mV
vinput (at gate end of R10) = 36 mV
vin(filter) / vin(gate) = 50/36 = 1.38
————————
Note: This is not the correct model for this problem, as I used strictly resistive components without taking into account the entire reactances.
F.
GaindB = 10 * log10 (Pout / Pin) = 10 * log10 (0.553µ / 20.2µ) = -15.63dB