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• • • ## The diagram

Let's take a look on the diagram below. On the left, you see the aforementioned transformer. In this case, the transformer's output voltage is 15V AC. This AC voltage is rectified by bridge rectifier G1. The rectified voltage is smoothed by capacitor C1. Without C1, the output would be just a rectified sine wave signal. If you would power your walkman with this voltage, you would hear a terrible 100Hz humming. The result of a computer simulation below shows what C1 does. In this picture you see two signals: a rectified sine wave (the situation without C1), and the situation with C1. At t=0, C1 is discharged, so VC1=0V. G1 will then charge C1 until t=T1. The top value of the rectified signal is 15V∙√2 - 2∙0.6V = 20V. After t=T1 C1 will be discharged by the load until t=T2. Then, everything will start all over again.

Vr is called the ripple voltage. Especially in audio equipment it should be as small as possible, because voltage ripple in the power lines means voltage ripple in the sound signals! You can reduce the ripple voltage by using larger capacitors. You can calaculate the value of the capacitor, but you can also use the rule of thumb: 2000...5000uF (2...5mF) per ampere load current.

You might think that the maximum voltage across C1 will be 20V. But this is only true if the secondary voltage of the transformer is 15V. Unfortunately, this voltage depends on the load. The open line voltage may be 18V or even more! Take this into consideration when buying a capacitor for C1, since all capacitors have a maximum voltage they can sustain.

The load-dependency of the output voltage of the transformer also explains the presence of zener diode D1. Without D1 the output voltage would depend on the load, and that is something we don't want. Thanks to D1, the voltage across P1 and R2 is always 12V. So the voltage at the base of T1 only depends on the position of P1. With P1 turned to the maximum position, VB=12V. The output voltage VE will be 12V-0.6V=11.4V. With P1 turned all the way down, VB=VR2. VE=VR2-0.6V. Of course we want VE to be 0V, so VR2=0.6V. We can now calculate R2: R2:P1=VR2:VP1=0.6:11.4, so R2=(0.6/11.4)∙10k = 526Ω. 470Ω is a good choise.

If we want 5mA current flow in D1, IR1=5mA+VD1/(P1+R2)=5mA+12/10470=6.15mA. R1=(Vtop-VD1)/IR1=(20-12)/6.15mA=1.3k. 1.2k is a good choise. This calculation assumes that T1's base current is very small. And it should be, because a large current will cause a high voltage drop across the top part of P1. This will reduce the base voltage and this the output voltage. And since the base current depends on the output current, the output voltage would depend on the output current. And we don't want that. You may need to replace T1 with a darlington.

Zener diode D1 has another advantage: a small ripple voltage across C1 does not appear on the output terminals. Even if Vr=5V, the voltage across C1 will never drop below 15V and VD1 remains 12V. This means that you can create a ripple free power supply without coffee table size capacitors! However, don't make C1 too small. Charging C1 causes a high current flow in the transformer. If C1 is small, charging takes a relatively long time and might overheat the transformer.

When VC1=15V, IR1=(15-12)/1200 = 2.5mA. This leaves 2.5-1.15 = 1.35mA for D1. And this might not be enough for D1 to operate properly. If we want ID1 to be 5mA (and thus IR1 6.15mA), then R1=(15-12)/6.15m = 488Ω. So R1 should be replaced with a 470Ω resistor. Let's now calculate the current when VC1 reaches its top value (20V). IR1=(20-12)/470 = 17mA. So ID1=17-1.15 = 15.35mA. PD1=12∙17m=0.2W. So a 0.4W zener will survive.

Please keep in mind that the top value is only 20V when the transformer voltage is 15V. We already saw that this voltage depends on the load current, and that the open line voltage can be 18V or more. Always consider the worst-case scenario. Assume we measure an 18V open line voltage. Add 10% to this, just to be safe. So we assume a top value of 20V∙√2 - 2∙0.6V = 27V. IR1=(27-12)/470 = 32mA. PR1=(27-12)∙32m = 0.48W. R1 must therefore be a 1W resistor. ID1=32-1.15 = 30.85mA. PD1=12∙30.85m=0.37W. Although this is just below 0.4, D1 should better be replaced with a higher wattage zener.

Another method is enlarging C1, making the ripple voltage smaller. Assume that the ripple voltage has been reduced to 2V. VC1,min is now 20-2=18V. R1=(18-12)/6.15m = 976Ω. We'll take an 820Ω resistor. When VC1=27V, IR1=(27-12)/820 = 18.3mA. PR1=(27-12)∙18.3m = 0.27W. An 1/3 of 1/2W resistor will now be sufficient. ID1=18.3-1.15 = 17.15mA. PD1=12∙17.15m=0.21W. A 0.4W zener can handle this very easily.

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