Half Wave Rectifier Analysis

[teej]

http://www.docircuits.com/my-circuit/21 ... ave-actual



How to I go about finding (an expression for) the voltage across the capacitor in this circuit. I'm looking for a method of analysis that will produce an output like that provided in the results tab of the link above. The output appears to be in the form of a solution to a simple RC circuit (V(1-e^x)), plus a modulation term for the ripple.

I run in to trouble attempting to apply KCL to the capacitor node because of the sign in the capacitor leg. KVL works for the left half of the circuit without the diode and a supply of 170abs(sin(x)) (just to investigate full wave rectification), but then current flows back in to the supply when its voltage is below the capacitor.

Are iterative/numerical methods the only way to go about this?

[pebe]

I am curious. Is this an academic exercise or does it have a practical application?

[teej]

I am curious. Is this an academic exercise or does it have a practical application?

Well, I've already designed, constructed, tested and used a similar circuit for a real project, so I suppose you can say this is an academic exercise. Imagine D1 is an optocoupler.

I had to design it with the help of the tool at docircuits, because I do not know of a way to predict the current in the R1 branch (or voltage and ripple at the capacitor node) with my current techniques. I think this comes down to a failure to model the diode correctly.

[pebe]

I think it would be difficult to arrive at a single expression for obtaining the curve you display. For the purposes of the following, the voltage of the 170VAC supply is shown as Vs, the charge on the capacitor as Vc, and the forward voltage drop of each diode as Vd.

At switch-on, no charging will occur until Vs>Vd, and no discharging will occur until Vc>Vd. From there, I think the mean voltage across C will be an exponential curve, but for each cycle until C reaches its max voltage, this will apply:

Thevenin shows that when D0 conducts(Vs-Vd>Vc), C will charge from Vs/2 - Vd through (R0+R1)/2. Because the time constant of 100Ω and 47µF is very long compared to the rise time 0v to +Vmax of the supply, the charge will be exponential. When D0 turns off then C will discharge exponentially from Vc-Vd to 0V.

As Vc changes with each successive cycle of the AC supply, I don’t know how to arrive at an expression for that part of the curve.

Once Vc is at max the circuit stabilises, and as D0 cuts off, C will discharge exponentially through 2000Ω towards 0V+Vd.

When D0 conducts again, C will charge through 1000Ω towards Vs/2-Vd.

I hope that may have helped.

[larry80]

teej, Its a simple RC-network in the area where you don’t have any discontinuity. The diode in this case will be the source of discontinuity. An RC-network has a simple curve as you’ve showed in the figure above. At times we make things more complicated in our minds than they actually are in the practical world so it’s important to take a step back and realize that this is a simple RC charging curve with a bit of discontinuity.

pcb assembling

[teej]

teej, Its a simple RC-network in the area where you don’t have any discontinuity. The diode in this case will be the source of discontinuity. An RC-network has a simple curve as you’ve showed in the figure above. At times we make things more complicated in our minds than they actually are in the practical world so it’s important to take a step back and realize that this is a simple RC charging curve with a bit of discontinuity.

Larry,

I think I tried to model what you're describing. The problem is the discontinuity due to diode D0 is a function of the capacitor voltage. My initial attempt was to forget the diode existed, and solve for Vc assuming a source of 170abs(sin(t)). That ended up being foolish because it resulted in a first order, non-homogenous linear differential equation whose solution didn't make any sense. (because it didn't model the diode cutoff correctly)