How to find Total resistance in an parallel circuit including three resistances.

Discussion in 'General Electronics Chat' started by abbas_6007, Sep 9, 2009.

  1. abbas_6007

    Thread Starter New Member

    Sep 7, 2009
    I want to know about Clipper/limiter of we can find out the output voltage waveform of fullwave rectifier diode?
  2. steinar96

    Active Member

    Apr 18, 2009
    Can you post a schematic of the circuit you are trying to analyse ?.

    But usually when calculating circuits involving diodes you simply assume a 0.7V drop across them (assuming they're silicon diodes). There are some waveforms of rectified waveforms in the AAC lectures on this site.
  3. abbas_6007

    Thread Starter New Member

    Sep 7, 2009
    Thanks Sir.
  4. kkazem

    Active Member

    Jul 23, 2009
    your posy said you wanted to know how to compute the total resistance of a 3 resistor parallel circuit. That is easy. The formula is as follows, no matter how many parallel resistors there are:
    Rtotal = 1/[(1/R1)+(1/R2)+(1/R3)], and if R1=R2=R3, then Rtotal=(R1)/3. Similarly, if ther are n resistors in parallel all the same value, Rtotal=(R1)/n, otherwise, if they are different values, the formula is: Rtotal = 1/[(1/R1)+(1/R2)+ ... +(1/Rn)].
    Regarding the Clipper-Limiter diode circuit, that's one thing. The waveform of a full-wave rectified AC signal is quite another. In short, a diode clipper or limiter circuit can work on AC or DC circuits, so I'll address the DC version here (meaning unipolar, and AC meaning bipolar). If one has a unipolar input signal feeding a diode limiter, the input signal will be clamped by the diode when it conducts. The Anode connects to the signal input and the cathode connects to the desired limiting voltage plus the diode drop of 0.3 for Germanium and Shottky, and about 0.65 to 0.7 for silicon. For the circuit to work, the input signal needs to be of relatively high impedance compared to the diode circuit. As an example, lets say the diode cathode is connected to +4.3 VDC, and the diode used is a 1N4001 (1 amp, 50V PIV) and the signal has a series impedance of 1K Ohm. As the signal starts from zero and climbs up, it will be hard clamped at about 5.0 VDC (4.3V + 0.7V). It's that simple. But for a full-wave bridge rectifier acting on an AC (bipolar) input, if the Amplitude of the AC input is high enough, the diode drops can be ignored and the full-wave rectified waveform will be unipolar with the positive half cycle untouched (except for the diode drop, if it's significant for your circuit) and the negative half cycle will be mirrored around the X-axis (time axis), or you could say that the negative half-cycle gets multiplied by -1, and the positive half-cycle gets multiplied by +1 (again, except for the diode drops, and only if they are significant). Also regarding Full-wave diode rectification, there are two ways to do this if you have an input from a transformer winding (which is often the case) and both have their advantages and disadvantages. First, the full 4-diode bridge. this works on any AC (bipolar) signal, only needs a transformer winding with a single, 2-wire output. The disadvantage is that the input always goes thru 2 diode drops, which doubles the dissipation in the rectification circuit compared to the next type. The second type is a 2 diode, common-cathode full-wave rectifier and requires a 3-wire transformer output, the 3rd wire is a center-tap. So, for the same output voltage, it needs twice the secondary output winding area compared to the first type. Its advantage is that the voltage to be rectified only goes thru one diode drop and therefore, needs only half the diodes and has half the diode dissipation compared to a 4-diode full-wave bridge.

    That's the whole story. Post a reply if you have any questions for me.

    Kamran Kazem