Basic Power Supplies

The Power Supply converts the 120 Volts AC into various DC voltages that the game actually runs on. A transformer steps the voltage down to a voltage close to what is desired. We still have AC coming out of the transformer. To convert that to DC we use a Rectifier. Various schemes provide different means of rectification, each with benefits and draw backs. The output of the rectifier is Pulsating DC. These need to be filtered out to a smoother DC using a large filter capacitor. The output at this point is smoother DC, but unregulated. To regulate the voltage to a specific voltage we use a Voltage Regulator. Now that we know where we are heading, lets build one, shall we?

First we select a Transformer. The output voltage for our simple application must be higher than the voltage we want to end up with after we regulate it. Since most of our circuits work on +5 Volts, we need a voltage higher than 5V coming out of the transformer. 6.3 V is popular and available even at Radio Shack, so let’s select that. The current rating of the secondary must be able to supply our desired current. Most of our exercises can be done with ¼ Amp, or less, so let’s select one with at least that much current capability.

Next we need to Rectify this 6.3 V AC coming out of the transformer to a DC level. We have at least three basic designs to choose from. The Half Wave rectifier lets us use only one side of the AC. Since we need positive voltage we can use this circuit. An alternative would be a full wave rectifier. This allows us to use both sides of the AC signal. The advantage is that we can use a smaller filter capacitor. With a half wave rectifier we have a 8 ms gap between positive pulses. Our filter capacitor must hold a charge during this time, meaning we need a larger filter capacitor. With a full wave rectifier, we have a sequence of 8 ms pulses (our rectified AC signal) with no gaps. This means our filter cap can be half the size as with half wave rectification. The disadvantage is that our transformer must be twice as large. Since we are using half of our secondary on each half cycle, we need twice the voltage out, or 12.6 V, with a center tap. A third alternative is to use a circuit called a bridge rectifier.

This allows us to use both sides of our AC signal using only a 6.3 volt transformer. The disadvantage is that we loose twice as much voltage dropped across the rectifier diodes. Two of the four are conducting at any given time. Full wave requires less filtering, so that’s a good idea for us in our application. Cost of a transformer is less for a 6.3 volt transformer, compared to a 12.6 volt transformer, so it looks like a Bridge rectifier design is best for this application. The question to us is can we afford the loss of voltage across the two diodes at one time. Lets take a close look at the design we are talking about. The voltage at the rectified output will be up to the peak value of our rectified AC line. 6.3 (AC RMS) x 1.414 = almost 9 Volts. So we will have a maximum of 9 Volts once we rectify the AC. Less our loss across the diodes (0.6 V each x 2) equals 1.2 Volts lost in rectifying. 9 – 1.2 = 7.8 volts. We have a 2.8 Volt difference between the 5V we want at the end, and the 7.8 Volts coming out of the rectifier and filter capacitor. This gives us a DC voltage, but it is unregulated. As our line voltage varies, the output of our power supply will too. As current in the load varies, so will our voltage. We need a regulator. A Zener diode provides a steady voltage across it, but will only provide 10’s of milliamps of current. A transistor amplifies current, so let’s put the two together.

If we want 5.0 volts out we need to choose a Zener diode that is 0.6 Volts above that. Note the direction our current is flowing through the regulating transistor, from emitter to collector. The base voltage will be 0.6 Volts more positive than the emitter. We have +5 Volts at the emitter, so we want to set the base to 0.6 volts above the emitter, or 5.6 V. It just so happens that this is a standard value for a Zener diode, so we are in luck. What resistance do we need for R1? We will have a maximum of 9V at the collector. Actually only 7.2 Volts we calculated including loss across the diodes, but we have to make our calculations for “worst case” conditions. When the transistor is off we draw no current, the maximum we will see at the collector is 6.3 x 1.414, or 8.8 volts. We will calculate for 9 volts since our line voltage may vary. 9 V minus the 5.6 volts across the Zener makes 3.4 Volts, maximum, dropped across the resistor. How much current will be passing through R1? If we have our expected 250 mA we want to supply, our emitter current will be 250 mA. If the gain of the transistor is 100, we will have 1/100^th of that 250 mA, or 2.5 mA of base current. To this 2.5 mA we will have the 20 mA through the Zener, according to the design specs of the Zener. So we have 2.5 mA, plus 20 mA, or 22.5 mA through R1. With 3.4 volts across it, and 22.5 mA through it, we need a 150 ohm resistor. This calculates out to less than 100 mW, so a ¼ W resistor is sufficient. Selection of filter capacitors is loose. With a bridge rectifier we need about 1,000 uF per Amp of current. With ¼ Amp of current, we need about a 250 uF (or higher) capacitor, rated at 10 Volts or higher. Looking at our shelves, we find a drawer full of 1,000 uF, 25 V, capacitors. That should work just fine. No more elaborate calculations are needed. C1 is now chosen to be a 1,000 uF 25 V capacitor. This gives us a working design, but has the disadvantage of not limiting our current. If we accidently short the output to ground the transistor will conduct itself into destruction trying to keep the output at 5 Volts. We need to limit the output current to 250 mA. Beyond that we want the power supply to just shut down. Adding current limiting to our power supply is a simple task.

We only need to add R2 and Q2 to our design. When the current through R2 gets to the 250 mA level we develop 0.6 Volts across it, turning Q2 on, shorting out the emitter – base circuit of Q1, turning Q1 off. What resistance do we want for R2? 0.6 volts across it. 250 mA through it. That comes out to be 2.4 ohms. That’s a standard value. 0.6 V x 250 mA comes out to 150 mW, so it also can be a ¼ Watt resistor. This gives us the safety of current regulation with only a small sacrifice to voltage regulation. Our output voltage will be a little less since we have the loss across R2 to consider. At lower currents we may only drop tenths of a volt across R2. As we approach 250 mA we will get a maximum of 0.6 volts across it. Beyond that our current limiter should activate and shut down the power supply. When the power supply turns off the voltage at the collector will raise to the full 9V level. That’s why we used 9 V when making calculations for C1 and R1. The whole of our circuit may be shown in discrete components, as above, or as a block diagram, and just call it a Voltage Regulator. This is such a standard design, why don’t we add a circuit for thermal safety, and a few more bells and whistles and stuff it all inside a transistor case? Great idea. That bring us to our next lesson.

Regulated Power Supplies

The next step in developing our power supply is to turn the pulsating DC, unregulated, in to a specific, well regulated, current limited voltage. Most logic circuits run on +5 Volts. This must be within 5% of 5 Volts under varying load conditions. In order to get a stable, predictable, output voltage under any predictable current level we need to add a regulator. In today’s circuits, this is usually a simple three- leaded component. The voltage regulator puts out a specific voltage over a certain range of operating current. The LM7805T is a very popular device in the gaming industry. It provides a regulated +5 Volts at any current level from about 0.050 Amps to over 1 Amp. The part number can be broken down to make sense of the device. “LM” is the manufacturer’s prefix for analog devices, and is an industry standard. Other manufacturers may use another prefix. The 78xx family of devices are all positive voltage regulators. 7805 is a 5 Volt device. 7812 is a 12 volt device. 7815 is a 15 volt device. 7824 is a 24 volt device. The T suffix indicated the case the device is built in, and therefore the current that we may draw from the regulator. T indicates a TO-220 device, rated at 1 Amp. “K” indicates a TO-3, rated at 1.5 Amp. “L” indicates a TO-92 device, rated at 100 mA, (0.100 Amp). We can read the part number and tell what the device does, and its ratings. The 79xx family is a similar device made for negative voltages. These devices provide a regulated output voltage, as well as current limiting. If we try to draw more than the rated current value the device shuts itself off, preventing circuit damage and blown fuses. An excessive current drain just turns the device off. The internal operations of these devices will be demonstrated as we get into electronic components.