Every type of electrical or electronic apparatus needs a source of electrical energy to function. The source of electrical energy is called the **power supply**. The two main classifications of power supplies are line-operated power supplies (operated from a standard 120-volt AC wall outlet) and battery supplies (electrical energy is provided through a chemical reaction). It is relatively safe to say that any device capable of functioning properly from a battery power source, can function equally well from a properly designed “raw” DC power supply, receiving its energy from a wall outlet. This is important because you will probably run into many situations where you will want to test or operate a battery-powered device from standard household power. As an exercise to test all you have learned thus far, here is a hypothetical exercise in designing a raw DC power supply for a practical application.

Assume that you own an automobile CB (citizen’s band) radio that you would occasionally like to bring into your home and operate as a **“base station.”** In addition to installing an external stationary antenna (which is irrelevant to our present topic of discussion), you would have to provide a substitute for the automobile battery as a power source. The CB radio is specified as needing “**12 to 14 volts DC at 1.5 amps**” for proper operation.

The CB radio power supply will have three primary parts: the transformer, a rectifier network, and a filter. You should choose a transformer with a secondary “peak” (not rms) voltage rating close to the maximum desired DC output of the power supply. In this case, a 10-volt AC secondary would do nicely, and they are commonly available.

**The peak voltage output of a 10-volt secondary would be:**

**Peak = 1.414(rms) = 1.414(10 volts) = 14.14 volts**

The rectifier network will drop about 1 volt, so that would leave about 13 volts (peak) to apply to the **filter capacitor**. The current rating of the transformer secondary could be as low as 1.5 amps, but a 2-amp secondary current rating is more common, and the transformer would operate at a lower temperature.

A transformer with a 10-volt AC at 2-amp secondary rating can also be specified as a 10-volt 20-volt-amp transformer. The volt-amp (V A) rating is simply the current rating multiplied by the voltage rating (10 volts * 2 amps = 20 VA).

A full-wave bridge rectifier can be constructed using four separate diodes, or it can be purchased in a module form. Bridge rectifier modules are often less expensive, and are easier to mount. The average forward current rating should be at least 2 amps, to match the transformer’s secondary rating. The peak reverse-voltage rating, or PIV, would have to be at least 15 volts (the peak output voltage of the transformer is 14.14 volts), but it is usually prudent to double the minimum PIV as a safety margin.

However, a 30-volt PIV rating is uncommon, so a good choice would be diodes (or a rectifier module) with at least a 2-amp, 50-volt PIV rating. If you purchase these rectifiers from your local electronic parts store don’t be surprised if they don’t have an associated “peak forward surge current” rating. Most modern semiconductor diodes will easily handle the surge current if the average forward current rating has been properly observed. This is especially true of smaller DC power supplies, such as the hypothetical one presently being discussed. If it is desirable to estimate the peak forward **surge current**, measure the DC resistance of the transformer secondary, and follow the procedure given earlier in this chapter. (The secondary DC resistance can be difficult to measure with some DVMs because of its very low value.)

In order to choose a proper value of filter capacitance, you can equate the CB radio to a resistor. Its power requirement is 12 to 14 volts at 1.5 amps.

**Using Ohm’s law, you can calculate its apparent resistance:**

**R = E / I = 12 volts / 1.5 amperes = 8 ohms** (worst case)

As far as the power supply is concerned, the CB radio will look like an 8-ohm load. Note that the 8-ohm calculation is also the worst-case condition. If the upper voltage limit (14 volts) had been used in the calculation, the answer would have been a little over 9.3 ohms. Eight ohms is a greater current load to a power supply than 9.3 ohms (as the load resistance decreases, the current flow from the power supply must increase).

You now know two variables in the load time constant equation: the apparent load resistance (Rload) and the desired time constant (83 milliseconds with a full-wave rectifier). To solve for the capacitance value, the time constant equation must be rearranged.

**Divide both sides by R:**

**Tc / R = R(C)/R**

**The R values on the right side of the equation cancel each other, leaving**

**Tc / R = C or C = Tc / R**

**By plugging our known variables into the equation, it becomes**

**C = 83 milliseconds / 8 ohms = 0.010375 farad or 10,375 micro-Farad**

According to the previous calculation, the filter capacitor needed for the CB radio power supply should be about 10,000 micro-Farad. A capacitor of this size will always be electrolytic, so polarity must be observed. The voltage rating should be about 20 to 25 WVDC (this provides a little safety margin over the actual DC output voltage), depending on availability.

If this calculation had been based on a half-wave rectifier circuit, the required capacitance value for the same performance would have been about 20,000 micro-Farad.

**RESOURCE:** McGrow Hill. TAB Electronics Guide to Understqanding Electricity and Electronics

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