>The input impedance of the amp is very dependent on the emitter
resistance. It gets reflected back into the base circuit, i.e. the
input, as hfe*Re. As the emitter resistance is varied it will have an
impact on the input impedance of the amplifier and therefore on the
first mixer. <
Yes and no. And a whole lotta depends. Hint the device is 2n3904 with a FT of 300mhz
First if you take out the input transistor and then parallel all the resistors the highest
impedance is about 217 ohms for the base uBitx. So we have set the upper limit.
The active feedback lowers that further with the device in.
Now we are debating the difference between having Re set to near 0 and that is dependent
on the ac beta of the device and the emitter current. So that the low end and for the basse circuit
its set with a 10 ohm resistor.. for an assumed 50 ohms, never ran the number or put it in the
HP4191A to measure. Now if the diode is substituted to get about 10 ohms you need about
2-3 ma current (25/ma is a fair approximation for diode resistance). So we can set for say
2.5ma. Now lower that to zero. the diode is not an open circuit and the 100 ohm emitter
resistor is dominant. Since Re of the transistor is unchanged and that resistor is added to
it its still parallel loaded by all the bias and output coupling resistors around it. Worst case
is 4:1 by eyeball, actual will be lower.
And no the circuit is not 50 ohms at all likelihood as the ac beta of 2n3904 is barely 10 at
30mhz and at 45mhz less than 7. so then the feedback is near zero as the transistor
cannot achieve the gain and impedances are not holding well too as device gain is a
factor in the calculation. The second IF is better but to terminate the second mixer it
has to have a uniform input impedance to near 135mhz. A 2n3904 is marginally useful
device up there.
For the case presented look at resistors r27(47), r26(470), r105(1K), r10(1K), r11(2.2k), r14(220).
Do the nodal analysis to convert those from a spaghetti-net to a single parallel value.
That will be the load value if Q21 and Q10 are not even there. The input of the amplifier
needs to be higher than 50 ohms as there is about 217 ohms present in that net or about
56 ohms to present an exact 50 ohms to the IF port of the mixer.
Refer to ubitxV3 schematic.