amplifierbjttransformertransistors
The DC and AC load lines of a transformer coupled Class A amplifier is shown below.
My question:
How can Vce vary from 0 to 2Vcc? In class, I learned that it was due to Lenz Law working in the transformer and a voltage of Vcc being induced to reverse the change in Ic. However, I cannot picture this. Can someone please explain with a diagram of how this comes about in the transformer windings.
This is an AC analysis of signals and all the clutter of DC components is removed - input and output capacitors are replaced with short circuits and power supplies are also replaced with short circuits. It's not a DC analysis or any other analysis - it's an analysis of signal amplification and AC impedances with a minimalist approach diagramatically. This method is meant to help beginners but I've never thought it was helpful - show me the full circuit any day and I'll figure it out from that has always been my approach!
And for your point 3 it is assumed that base current is zero in a lot of AC analyses.
Point 4 - this circuit can be used with the collector effectively seeing a load resistance of under 100 ohms. Try researching an emitter fed cascode amplifier - two transistors are used in common base and the collector of the 1st transistor feeds the emitter of the 2nd transistor. Most cascode amplifier circuits feed the base of the first transistor from the external input but it can work with both transistors in common-base configuration.
The input resistance is determined by the forward conduction dynamic resistance of the emitter-base junction - it's forward biased and has a dynamic (AC) impedance of 20 ohms in your question - this is presumed to be the slope of the emitter-base diode. Load resistance will not affect this.
What you may be getting confused about is the "ideal transformer" in the equivalent circuit. You should not regard it as having any magnetic qualities at all. Try and see it like this: -
Whatever voltage you have on the input to the ideal transformer, \$V_P\$ is converted to \$V_S\$ on the output of this "theoretical" and perfect device. It converts power in to power out without loss or degradation such that: -
\$V_P\cdot I_P = V_S\cdot I_S\$
The ratio of \$V_P\$ to \$V_S\$ happens to be also called the turns ratio and almost quite literally it is on an unloaded transformer because there will be no volt-drop across R3 and L3 and only the tiniest of volt-drops on R1 and L1.
This means you now have a relatively easy way of constructing scenarios of load effects and recognizing the volt drops across the leakage components that are present.
The equivalent circuit of the transformer is really quite good once you accept that the ideal power converter is "untouchable" and should just be regarded as a black box. For instance you can measure both R1 and R3 and, by shorting the secondary you can get a pretty good idea what L3 and L1 are. With open circuit secondary you can measure the current into the primary and get a pretty good idea what \$L_M\$ is too.
Best Answer
OK, here is a simple schematic of a transformer coupled amplifier;
Here are the voltages;
Notice before the sin starts the collector is at Vcc, 12V in this case. This is because for DC the inductor is a short. When the sin starts, it can only go as low as VE, which I decided to make about 2V.
The input voltage is multiplied by 100 so it can be seen, otherwise it would look like a flat line at 0V.
Here is the collector current;
I was trying to get the bias current to be 50mA. This is close enough.