Full Bridge Rectifier with Filter Capacitor circuit

Question

I have a problem where I have a 60 Hz sinusoidal power signal with an amplitude of 14.14 V through a full bridge rectifier with constant voltage modelled diodes (\$V_D = 0.7V\$)

The load current is \$30 mA\$ on average. I need to find the ripple voltage \$V_r\$, average DC output \$v_{outavg}\$, peak inverse voltage experienced by the diodes \$PIV\$, the peak and average currents through the diodes \$i_{dmax}\$ and \$i_{davg}\$. The circuit uses a \$250 \mu F \$ capacitor as a filter capacitor.

Since we aren't given the ripple voltage and the resistor are unknown i'm not sure how to proceed.

I have the input: \$V_1 = 14.14 V\$

Peak output: \$ V_{peak} = V_1 - 2V_d = 14.14 - 1.4 = 12.24V\$

Peak inverse voltage :\$PIV = V_1 - V_D = 14.14 - .7 = 13.44 V\$

$$i_{davg} = i_{loadavg} (1 + \pi\sqrt\frac{V_{peak}}{(2V_r)}) = 30 mA (1 + \pi\sqrt\frac{14.14}{(2V_r)})$$

$$i_{dmax} = i_{loadavg} (1 + 2\pi\sqrt\frac{V_{peak}}{(2V_r)}) = 30 mA (1 + 2\pi\sqrt\frac{14.14}{(2V_r)})$$

$$V_r = \frac{V_1 - 2V_D}{2fRC} = \frac{14.14 - 1.4}{2 \cdot 60 \cdot R \cdot (250\times10^{-6})}$$

How do I solve for \$V_r\$ without knowing the resistor? And where else do I go from here? Is what I have correct so far?

Answer 1

Vmax = Vavg(dc) + Vpp(ac)/2 is an approximation that the ripple is a triangle wave, but actually it is closer to a sawtooth.

You have estimated Vmax = 12.24V

Although the charge time is faster than the discharge time, the decay time and load current yields the peak-peak voltage AC ripple ΔV=dV/dt*T [Vpp]

• for interval T=1/2f and f=60Hz
• for Ic=C*dV/dt in the capacitor discharge interval
• from above ΔV = dV/dt*T
• thus ΔV=Ic*T/C for T=1/120Hz [s], C=250uF and Ic = 30mA CC discharge only
  • ΔV = 30mA*(1/120Hz) * (1/250uF) = 1 Vpp
  • let's get back to that ripple waveform
• what ratio of each half cycle is used for charged/discharge time?

• this ratio is directly related to % ripple and charge/discharge current ratio

  • we know % ripple = ΔV /Vdc*100% is approx 1Vpp/12Vdc = 8.3%.

    • If we double that we can reduce the ripple from 1Vpp to ~ 86% of 1V or 860 mVpp.
    • This agrees well with simulation below. 405(+pk) + 445(-pk) = 850mV
    • With more proof I can show % V ripple is same as % load/charge current
    • I'll let you compare your proof to my estimate
      • Thus charge current pulse in Diode and Cap= Ipeak = 30mA*1/ΔV[%] = 361 mA
      • This agrees well with simulation below. (385 mA)

  • Note in the simulation below that the lower scope during power up is a couple Amps limited by the Diode ESR

  • Also consider that for ideal diodes and cap, if power is switched on at peak AC voltage is the diode and cap current is infinite !!

Food for thought

Answer 2

Seems like there must be a simpler expression for the average diode current if you know the average output current. Just sayin'.

Vr can be calculated approximately (for Vr small) by assuming constant current discharge of the filter capacitor for 1/120 second.

The peak diode current can be calculated from the dv/dt of the input voltage when the diodes first start to conduct, since the capacitor current will be C*dv/dt (and there's still output current).