I'm trying to figure out why the output voltage waveform of this circuit would not include a DC offset of E. My understanding is that D1 becomes forward biased when Vs > E, and so that is when it should conduct. Every waveform drawing I've seen, on the other hand, begins conduction at 0 volts.
Also, what would be the difference between this circuit and one that replaces two diodes with thyristors? I've seen the latter referred to as half-controlled.
Thanks!
It may seem that conduction begins at 0 volts, but it does not; it depends on the LR time constant and the frequency of the AC source.
The LTspice circuit file for these two circuits is here, so you can simulate them simultaneously and see the differences between the start of conduction.
Notice particularly that for the circuit on the left, conduction begins at a voltage more negative than zero volts.
Your intuition is right for the first cycle. Imagine at first there is \$V_z\$ positive, say 2V and E is 1V, so there is a voltage drop in forward direction of 1V and the current \$i_0\$ starts to increase (due to the inductor) from 0 until the voltage on the inductor is 0 (it is basically a conductor, or a piece of wire if you will). Then 1V is left for the resistor which limits the current to 1V/R (say R = 1 Ohm, current will be 1A). If there was no resistor the current would increase linearly to infinity. But to return to your question, the diodes start conducting as soon as \$V_z\$ becomes greater than \$E\$ (provided we neglect the voltage drop at the diode!).
But now something happens: Imagine after that said above, you switch off \$V_z\$, i.e. it becomes 0V. Then the conductor can't just switch off, but it will still have a current flowing through it, since the current can't change abruptly. So there will be current flowing through D1/D4 as well as D2/D3, which are now in forward bias (note that the current does not change directions). After switching of \$V_z\$ the current is for a short moment still 1A (with values assumed above) which makes the voltage dropping on R remain 1V (from the upper side to the lower side of R). Furthermore we have \$E\$ and this sums up to 2V. The Diodes would be reverse biased at first sight, so the inductor has to make up for it to reverse its voltage to the same amount, namely -2V (counted as with R in direction of the current). As the current decreases voltage of R decreases accordingly until it becomes zero. The inductor is then free of current and no voltage drop is across it (here the voltage \$E\$ makes it a tough one... I don't get it from here...)
Anyway, if we apply negative voltage during the current decreasing phase the current will go through D4 into the upper side of \$V_z\$ and then through D3 (in the same direction as before). Lets now assume that \$V_z\$ is -0.5V (absolute value lower than 1V, means \$E\$). When \$V_z\$ was zero \$V_o\$ was zero (or slightly above), \$V_R\$ was 1V, \$E\$ was 1V and \$V_L\$ was -2V. When now \$V_z\$ goes to -0.5V the voltage on the inductor will go from -2V to -1.5V abruptly and the voltage \$V_o\$ will be 0.5V! This is so, since again the current does not change abruptly and has to be maintained, hence inductor voltage drops from -2 to -1.5 and \$V_o\$ increases to 0.5V. There we are. After starting the cycles the \$V_o\$ will follow the absolute value of \$V_z\$, provided high enough frequency/high enough inductor/high enough resistor. Man this is complicated!
But without the inductor it is basically as you said, Vs has to be > E. Anyone find a better explanation without breaking your brain? Anyway, hope that helps.
