# Ac voltage regulator to keep output voltage constant about 6v using diodes

I'm trying to design a circuit to keep the output voltage constant at 6 VDC for output load resistances between 0.5 to 3 kiloohm. It doesn't matter to have errors about 0.25 VDC on the output.

My input voltage is a sine wave : $$\v(t)=9\sqrt2\ \sin(100\pi t)\$$

I have only simple elements such as capacitors, diodes and Zener diodes (4.7 and 6.8 break down voltage).

I think it's better to use Zener diode to keep output voltage constant as it's shown in below circuit but in this way the voltage is kept constant in about break down voltage of Zener but not 6 VDC.

• Your worst case load is $12 \:\text{mA}$. Which is pretty light. What zeners do you have? The one you show there? If so, that one works better with an NPN BJT added to it to deliver $6\:\text{V}$. And that would be a good idea if you can manage it.
– jonk
May 5, 2022 at 4:46
• I have zeners with 6.8 and 4.7 break down voltage, and unfortunately i can't use transistors May 5, 2022 at 4:50
• You ought to define your "expectations" or Specs for voltage error including ripple, before you design anything. May 5, 2022 at 4:51

The current for a resistive load = 500 to 3k will depend on the output voltage. The load draws current normally fed into the Zener limited by a current limiting resistor and the voltage mean and Vpp ripple of the rectifier filter.

The choice of rectifier makes some difference in its series resistance and thus voltage drop with current above the standard Si diode 0.6V @ ~ 1mA. The bulk cap does not need to be ripple free with 2mF and may be reduced significantly with a secondary cap on the Zener for more ripple filtering. Use dV/dt=Ic/C for dt = 10 ms and Ic = max load + Zener.

I=V/R   6V    6.8V  Load Current [mA]
-------------------
500     12    3.4  mA
3k      0.2   0.34 mA
----   -----  --------
Diff.   11.8  3.1  mA

Then choose 20 to 30 mA for Zener current without load so with load Iz= {8 to 17} to {18 to 27}

Since 9Vrms = 12.7Vp choose the peak voltage drop for rectifier 2 diode drops 1.5 to 2V depending on diodes at peak charge current which depends on % ripple.

Choose between < 10% ripple voltage or < 1 = dV.

Now remember if 30 mA is steady discharge current max at 10% voltage ripple then the charge time d.f. is also 10% but partly triangular shaped current so the rectifier current will be about 5x or 150 mA pk.

Thus $$\C = Ic *dt / dV \$$= 150 mA * 10 ms / 1Vpp = 150 uF min or choose 220 uF

If you use Si signal diodes for the rectifier like 1N4188 rate at 100 mA that should suffice and add filtering and reduce surge start currents.

If you choose the 4.7 V Zener and add another of the same in the forward direction (1.5V @ 200mA) then you can achieve 6.0 V with a stead max current of 30 mA from 12.7-2V = 10.7 minus 6V = 4.7V / 30 mA = 160 ohms and ~ 180 mW so use 1/4 W resistor

Compare your test results with my quick and dirty calculations for voltage error max. under both conditions 500 to 3k ohms.

using 1N4148 (4)
1N750A, 4.7V (2)
220uF (1)
160 Ohm (to 200) (1)