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For one of my labs I need to design a power supply with the ripple less than 1% of the nominal 4.7 V value (since Vz = 4.7 V) and calculate its load regulation.

I know that the ripple will most likely be within the required 1%, but I am having problems calculating it's exact value that I will need for load regulation calculation.

Here is the circuit:

Power Supply

The load has to be \$50kΩ\$. R1 and C1 are adjustable. Since I will only have one \$100µF\$ and one \$10µF\$ capacitors in the lab, I am going to shunt them for better filtering.

Attempted solution:

Input:

\$V = 15 V_{p-p}\$ or \$7.5 V_{p}\$ @ 50 Hz

After passing through bridge rectifier (two diodes - two 0.7 V drops):

\$V = 6.1 V_{p}\$ @ 50 Hz

Here's where my problems start.

After passing through the filter:

I wanted to calculate the amplitude of the ripple using: \$ V_{r} = {{V_{p}} \over {2fCR}}\$, where f = 50 Hz, \${V_{p}} = 6.1 V\$, \$C = 110µF\$ and R is the resistance of the circuit parallel to the capacitor. I have no idea how to get \$R\$, even if I pick any \$R_{1}\$, since I suspect the resistance will vary because of the Zener diode.

So, how can I find the ripple amplitude at the input of the Zener regulator? I need this value for Zener regulator calculations and, consequently, precise output ripple value and load regulation.

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  • \$\begingroup\$ Why the zener and not something less archaic? \$\endgroup\$ – PlasmaHH Feb 15 '17 at 16:36
  • \$\begingroup\$ @PlasmaHH That's how this lab is. R1 and C are really the only things I can change in this circuit. \$\endgroup\$ – Max Lawnboy Feb 15 '17 at 16:38
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    \$\begingroup\$ So this is basically homework? \$\endgroup\$ – PlasmaHH Feb 15 '17 at 16:52
  • \$\begingroup\$ @PlasmaHH This is not required for the lab, but I want to prepare for it to understand what exactly is going on. So no, it's not homework (it's not marked). \$\endgroup\$ – Max Lawnboy Feb 15 '17 at 17:00
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You can pick R1 to produce an average current of Iz (from the zener datasheet). While you are there, look up the zener impedance at that current. You can think of the zener as an ideal voltage source with a resistor in series. The value of that resistor is highly dependent on the zener current, of course.

Calculate the ripple voltage at the capacitor from the current from the capacitor (assume it is constant and equal to (Vp-Vz)/R1. Symbolically (as a function of C).

Calculate the ripple voltage across the zener from the zener impedance, ripple voltage and source impedance (R1 || 50K which you may be able to say ~= R1). Again, as a function of C.

Find the minimum capacitor value to yield 1% ripple.

You can calculate the load regulation similarly.

Here's part of a table from a zener datasheet:

enter image description here

There are two currents at which the zener impedance is specified, but the zener voltage is only specified at Iz.

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  • \$\begingroup\$ The R1 is picked that way, so that Zener impedance is too low and load impedance is too high to matter in the capacitor ripple voltage calculation, am I correct? I am still confused a little about what you wrote for capacitor ripple voltage calculation. Is its amplitude going to be equal to Ic/2fC, where Ic = (Vp-Vz)/R1? \$\endgroup\$ – Max Lawnboy Feb 15 '17 at 17:54
  • \$\begingroup\$ Yes, that's correct. The capacitor charges in a spike-y manner at the peaks of the mains (2*f since you have a bridge rectifier) and spends most of the half-cycle discharging with current Ic, just as you wrote. \$\endgroup\$ – Spehro Pefhany Feb 15 '17 at 18:48
  • \$\begingroup\$ Thank you! Your post was really helpful. R1 = ~300Ω and C = 250µF or higher keeps the ripple under 1%. \$\endgroup\$ – Max Lawnboy Feb 15 '17 at 20:17
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Your objective is to make sure that the voltage across the capacitor doesn't fall below (say) 5 volts. This ensures that the zener is always operating. So if your capacitor's peak voltage is 6.1 volts, you need to pick the highest value resistor (R1) that satisfies getting at least 5 volts on the load if the zener were removed.

Current taken by the load is a miniscule 94 uA so you might as well assume maybe 10 mA for the zener diode and this means R1 is 1.1/0.01 = 110 ohms. The 1.1 is 6.1 volts minus 5 volts.

If (approximately) 10 mA were being taken from the capacitor over a period of 10 ms how much would the voltage fall? I = Cdv/dt therefore dV = 0.91 volts.

The discharge of the capacitor will be from about 6.1 volts to about 5.2 volts but of course the zener won't be perfect so there might be a slight sag in its voltage. However, I have assumed 10 mA discharge current and this is over-egging it. At 5.2 volts, and R1 = 110 ohms feeding a 4.7 volt zener, the current is about 4.5 mA, so I would expect the droop in capacitor voltage to be somewhere about 0.7 volts.

Also my calculation assumed 10 ms but it will be slightly less than this, possibly 8 ms. So I would choose R1 to be 110 ohms.

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  • \$\begingroup\$ Thank you. The other answer was more clear with theoretical calculations, although I ended up using capacitance higher than 110 µF. I will keep in mind your approach when I will be in the actual lab and I will have to use 110 µF. \$\endgroup\$ – Max Lawnboy Feb 15 '17 at 20:21
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you specify 4.7v only,but you should also specify the out put cerrent with 4.7 nominal voltage,and simply you have to calculate Vripple= Iout/4fc because it is bridge ckt. Vripple= Iout/2fc or Vp/2FCR isfor half wave rectification(you use). there fore R= Vp/Iout.

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  • \$\begingroup\$ Welcome to EE.SE. Please put some effort in posting your answer. Try to format formulas in a beautiful way and try to improve the semantics of your writing. Currently your post is hardly understandable. \$\endgroup\$ – Ariser Dec 6 '18 at 19:06

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