I am at the point 4.6 "The common-collector amplifier"
After figure 4.40 it says:
"...it would be reasonable to presume that this amplifier will have a very large current gain."
I don't see why an CE vs a CC amplifier will have less current gain. I have not tested it with SPICE, but with iCircuit instead, and find not significant differences.
But above all I cannot understand why does it say that. Current gain should be the same and given a certain Vbe (aprox. above 0.7V) it should reach the Beta ratio (not more under regular humidity and temperature conditions).
On the other hand, in page 205 after figure 4.43 it says:
"Given the voltage polarities across the base-emitter PN junction and the load resistor, we see that these must add together to equal the input voltage, in accordance with Kirchhoff's Voltage law."
I always analyze circuits according to the electron flow, not the conventional.
So I presume that electrons will firstly traversse the load, and then the PN junction base-emitter. If that's correct, I cannot see how possible it is that the load has a voltage 0.7V less than the input voltage. Which input voltage does it mean, the signal? (Vin) or the other DC source?, the main one.
As far as I understand, Voltage drop means the amount of voltage (energy to move electrons) that gets lost when current traverses a passive component. Which means that the voltage (according to electron flow) can't be dropped by 0.7V until it has traversed the emitter-base junction. But the load lies before the base-emitter junction.
So for this case I am completely lost.
Any guidance to clarify these concepts is greatly welcome.
Many thanks.
Chal-lo.
After figure 4.40 it says:
"...it would be reasonable to presume that this amplifier will have a very large current gain."
I don't see why an CE vs a CC amplifier will have less current gain. I have not tested it with SPICE, but with iCircuit instead, and find not significant differences.
But above all I cannot understand why does it say that. Current gain should be the same and given a certain Vbe (aprox. above 0.7V) it should reach the Beta ratio (not more under regular humidity and temperature conditions).
On the other hand, in page 205 after figure 4.43 it says:
"Given the voltage polarities across the base-emitter PN junction and the load resistor, we see that these must add together to equal the input voltage, in accordance with Kirchhoff's Voltage law."
I always analyze circuits according to the electron flow, not the conventional.
So I presume that electrons will firstly traversse the load, and then the PN junction base-emitter. If that's correct, I cannot see how possible it is that the load has a voltage 0.7V less than the input voltage. Which input voltage does it mean, the signal? (Vin) or the other DC source?, the main one.
As far as I understand, Voltage drop means the amount of voltage (energy to move electrons) that gets lost when current traverses a passive component. Which means that the voltage (according to electron flow) can't be dropped by 0.7V until it has traversed the emitter-base junction. But the load lies before the base-emitter junction.
So for this case I am completely lost.
Any guidance to clarify these concepts is greatly welcome.
Many thanks.
Chal-lo.