Is it possible to make rectified LC tank like this?
Let's see EXACTLY what happens.
STEP1: Let's say top side of the cap chargest first. Ok, current flows the only way it can through the diode, inductor, another diode and comes to point X.
STEP2: Now, it has to chose, to the right there is a diode with certain resistance, to the left there is nothing but empty cap, logically it goes to the cap and charges it.
STEP3: Cap is fully charged and current starts to flow the only way it can through the diode, inductor, another diode and again comes to point X, STEP2 repeats and oscillation continues between bottom cap plate and the inductor.
Think before answering, please.
What if AC is connected to x and dot diagonally from it.
Doesnt seem right, i don't see a way to make this work. Say if you see one.
I'm not sure, that's why i ask. I just wanna know if LC tank can be made to produce pulsed DC on the inductor.
Are you sure you went to college, you dont seem to know what you're talking about. Here are few formulas for you to study..
1F = 1C / 1V - amount of electric charge in coulombs that is stored per 1 volt
C = Q / V i C = kA/d
energy in a cap E = 1/2 QV i E = 1/2 CV²
qrk I did follow where current flows.
To answer my own question, after in-depth analysis, it is clear this circuit would have to be fed pulsed DC to produce pulsed DC which of course defies the purpose. Fed with AC no matter how connected will not produce continous oscillation.
- amania_rLv 76 months ago
Assuming the capacitor is charged and then connected to the circuit...
The capacitor will discharge through D1 and D3.
The current will continue to flow through the inductor and through D3 and D2
The currents will fade and nothing else will happen.
The other poster is correct "Let's say top side of the cap charges first" is meaningless. A capacitor either has a charge (of either polarity) or no charge.
- 6 months ago
In all the EE courses I took in college, they never talked about the "top side of the cap charges first" That doesn't really have any meaning. A capacitor has two contacts. The voltage across the capacitor multiplied by the capacitance will tell you how much charge the capacitor has on it. According the internet, a "tank" circuit IS DEFINED as a capacitor in parallel with an inductor.
- qrkLv 76 months ago
You show a diagram and don't say how you are energizing the circuit. There are 2 points that are diagonal from "x". What are you trying to accomplish with this circuit?
Why not model this in LTspice and see what happens. LTspice is free and a very good simulator, rivaling $5000+ simulators.
You can spout all sorts of formulas, but one thing you really need to understand is how current flows. Amazingly, most graduates out of college lack this skill. I will suggest you follow the path of current and see where your circuit gets you. As previously mentioned, LTspice is a wonderful tool for examining a circuit.
- billrussell42Lv 76 months ago
where is the source of current? what is it referenced to? Is the source AC or DC?
Your description seems only to apply to a positive signal.
First guess, see below. AC source.
During a positive cycle, cap charges and L in in parallel. D2 is reverse biased and the other two forward biases.
During a negative cycle, the inductor is disconnected, both D1 and D2 are reverse biased.
So you get a +DC voltage on the inductor. But the whole thing would take a very rigorous analysis....
edit: just give us the full circuit, with applied voltage points, and exactly what you expect this strange circuit to do.