new one Posted May 10, 2020 Report Share Posted May 10, 2020 i want to design a power supply circuit from 220v ac to 5v with ripple 100mv Quote Link to comment Share on other sites More sharing options...

HarryA Posted May 10, 2020 Report Share Posted May 10, 2020 Are you interesting in designing a circuit using mathematical equations or are you interested in building a power supply using proportional engineering (as a 6.3 volt filament transformer should work)? Proportional engineering is the way I build most things. Are you looking for help here? Quote Link to comment Share on other sites More sharing options...

new one Posted May 12, 2020 Author Report Share Posted May 12, 2020 hi Harry i want to know the circuit design with also mathematical equations i bought a transformer 12 volts 1 ampere Quote Link to comment Share on other sites More sharing options...

HarryA Posted May 13, 2020 Report Share Posted May 13, 2020 The 12 volt transformer supplies a peak voltage of: Vpeak = 12 * square root of 2 = 12 * 1.4 = 16.8 With 16.8 volts into the bridge we subtract 0.7 volts for each diode. As the current passes through 2 diodes for 1.4 volts. Vdc = 16.8 - 1.4 = 15.4 without a load. The actual loaded voltage will be more likely around 14 volts as we will see later with a capacitor and load. I am using 20ma LED here. For the capacitor: ------------------ ripple voltage Vr: Vr = (0.0024) * I/C = 100 mv C = 0.0024 * I / Vr = 0.0024 * (0.120A / 0.10V) = 0.00288 F 0r: C = 2880 micro-farad from Schertz and Monk "Practial Electronics for Inventors" section 11.6 For a Zener diode ------------------ Lets try the 5.1 volt 1N4733A. As 5.0 volt Zeners are hard to find. For the 1N4733A Zener diode the value of Vz is 5.1V and Pz is 500 milliWatts now with a Vdc (dc supply voltage) of 14V the value of R2 will be: Rs = (Vdc - Vz)/Iz Rs = (14 - 5.1)/Iz Iz = Pz/Vz = 500mW / 5.1V = 98mA max. Thus Rs = (14 - 5.1)/98ma = 90.8 ohms Rs = 91 ohms (approx) 91 ohms is a standard. The power/watts into R2 will be: Pr2 = V^2/Rs = ((14 - 5.1)^2)/Rs = (8.9 * 8.9)/91 = 0.87 watts or 2 watts in practice. So with a load of 120 ma the voltage drop across Rs: Vrs = 91 * 0.12 = 10.92 volts ........oops! Looks we only have 14 - 10.92 = 3.08 for the load and the Zener is out of business! Lets try a 1 watt Zener and 5.1 volts: Iz = 1.0 watt / 5.1 volts = 0.196 A or 196 ma. Rs = (14 - 5.1)/ 0.196A = 45.4 ohms or 47 ohms standard Voltage drop across Rs at full load: Vrs = 47 * 0.12 = 5.4 volts; Or 14v - 5.4v = 8.6 volts for the Zener to chew on. The power/watts into R2 will be: Pr2 = V^2/Rs = ((14 - 5.1)^2)/Rs = (8.9 * 8.9)/47 = 1.685 watts or 2 watts or better in practice. Next we can run this through the simulator and see what it looks like. (to be continued). Quote Link to comment Share on other sites More sharing options...

HarryA Posted May 15, 2020 Report Share Posted May 15, 2020 Doing a prototype of the circuit as shown below using a voltage regulator as I have no 5 volt Zeners: I used 1N007 diodes and a LM7805 regulator. Transformer output 13.85 Vac 17.3 Vdc at the capacitors. What I got for ripple voltage at the capacitor was: 536 mvpp with 1523 mfd cap. 140 mv ac with multimeter 288 mvpp with 3005 mfd cap 71 mv 200 mvpp with 4454 mfd cap. 48 mv 140 mvpp with 6050 mfd cap 36 mv Quote Link to comment Share on other sites More sharing options...

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