The aim of the design was to provide a cheap compact power supply for Cmos logic circuits. The circuits were used to control mains equipment (fans, lights, heaters etc.) through an optically isolated triac such as the MOC 3020.
Apart from the 5 to 10mA required to power the diode in the optical isolator, the Cmos control circuits themselves used very little current.
Formula for Current in an AC Circuit
The resistors in the AC portion of the circuit amount to 1100 ohms in total. This is made up of R1 plus the 100-ohm fusible resistor. R1 limits the peak current through the capacitor; but it has very little effect on the RMS current flowing through the circuit - because the contribution of R to the denominator is relatively insignificant.
There's no inductance to worry about, so
Thus the main control on the current is C1.
Its capacitive reactance is given by the formula: -
At 50 Hz the capacitive reactance of a 0.47uF capacitor is about 6k8.
The power dissipated by the 0.47uF capacitor is lower than the power dissipated by a 6k8 resistor; because the current through the capacitor is out of phase with the voltage across it.
Watts = Current x Volts
In the case of a resistor - as the voltage across it rises, so also does the current flowing through it. When both the voltage and the current are at maximum, the power reaches its peak.
But the opposite happens with a capacitor. When the voltage across the capacitor is zero, the current through it is at its maximum. At this point the power dissipated by the capacitor is:
Watts = Current x Zero = Zero
Similarly, when the voltage across the capacitor is at its maximum, the capacitor is fully charged; so the current flowing through it is zero. At this point the power dissipated by the capacitor is:
Watts = Zero x Voltage = Zero
It's only somewhere between these two extremes - when both the voltage and the current are below their maximum - that the power reaches its peak. Consequently, this peak has to be lower than that of the resistor.
Looking at the
Transformerless Power Supply - the AC current flowing through the 16v Zener diodes is approximately
The 33mA current flowing through ZD1 & ZD2 is available for rectification. You can think of these Zeners as the secondary windings of a mains transformer. However - unlike a transformer -under "no-load conditions" the Zeners will be required to dissipate the whole of the energy available. Consequently, if your circuit is to be powered up without R2 & ZD3 or the Output Load attached, ZD1 & ZD2 will need to be at least 1-watt.
The Cmos control circuits did not need a particularly smooth supply; and the choice of 47uF for C2 gave a good compromise between physical size and the degree of smoothing. If you have room - and you want more smoothing - then you can use a larger value capacitor.
The output from BR1 is about 15-volts. If we want to reduce this to 12-volts using ZD3 then there must be a drop of 3-volts across R2. There is approximately 33mA flowing through ZD1 & ZD2. We cannot try to take more than this from BR1 because it would simply cause a drop in voltage. So we choose R2 to pass a current of say 30mA.
The 30 mA flows through ZD3; and is - in theory - available to power your circuit. If your circuit only needs about 20mA then the remaining 10 mA continues to flow through ZD3 so that the voltage drop across R2 remains constant and the output stays at 12-volts.
If you try to take more than 30mA from the circuit then the voltage drop across R2 will increase beyond 3-volts; and the output will fall below 12-volts. In practice, up to 20mA at 12-volts is available; and another electrolytic capacitor across the output would give additional smoothing if required.
C1 MUST be a "suppressor" capacitor. They are made to be connected directly across the incoming mains supply. They are generally covered with the logos of many different Safety Standards Authorities.
The larger the value of C1, the lower will be its capacitive reactance; so the higher will be the current flowing through it. You can use a single capacitor with a higher value, or you can connect two or more smaller capacitors in parallel. For example, two 0.47uF capacitors connected in parallel will give the equivalent of a 1uF capacitor - and almost double the available current.
However, increasing the value of the capacitor also increases the significance of the contribution R makes to the denominator
- R1 has to work harder; and the extra current flowing in the circuit means that the resistors and Zener diodes will have to dissipate more energy (Watts). If C1 equals 1uF then R1 needs to be 7 watts, and the 16-volt Zeners need to be 2 watts.
This brings us to the circuit's main limitation. It's really at its best when it's used to provide up to about 20mA DC. If you try to produce any more current, the components start to get very big. There comes a point when it makes more sense to use a mains transformer.
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