Click here to purchase the entire book in PDF format. Rectifiers and Power Supplies What use are diodes to us? Well, what happens if we replace the battery from last chapter's circuit with an AC source as shown in the Figure 2.53? Figure 2.53: Circuit with AC voltage source, 1 diode and 1 resistor. Now, when the voltage output of the function generator is positive relative to ground, it is pushing current through the forward-biased diode and we see current flowing through the resistor to ground. There's just the small issue of the 0.6 V drop across the diode, so until the voltage of the function generator reaches 0.6 V, there is no current, after that, the voltage drop across the resistor is 0.6 V less than the function generator's voltage level until we get back to 0.6 V on the way down... When the voltage of the function generator is on the negative half of the wave, the diode is reverse-biased and no current flows, therefore there is no voltage drop across the resistor. Figure 2.54: Comparison of voltage of function generator in blue and voltage across the resistor in red for the circuit shown in Figure 2.53. This circuit shown in Figure 2.53 is called a half-wave rectifier because it takes a wave that is alternating between positive and negative voltages and turns it into a wave that has only positive voltages - but it throws away half of the wave... If we instead connect 4 diodes as shown in Figure 2.55, we can use our AC signal more efficiently. Figure 2.55: A smarter circuit for rectifying an AC waveform. Now, when the output at the top of the function generator is positive, the current is pushed through to the diodes and sees two ways to go - one diode (the green one) will allow current through, while the other (red) one, which is reverse biased, will not. The current flows through the green diode to a junction where it chooses between a resistor and another reverse-biased diode (the blue one) ... so it goes through the resistor (note the direction of the current) and on to another junction between two diodes. Again, one of these diodes is reverse-biased (red) so it goes through the other one (yellow) back to the ground of the function generator. Figure 2.56: The portion of the circuit that has current flow for the positive portion of the waveform. Note that the current is flowing downwards through the resistor, therefore $V_{R}$ will be positive. When the function generator is outputting a negative voltage, the current follows a different path. The current flows from ground through the blue diode, through the resistor (note that the direction of the current flow is the same - therefore the voltage drop is of the same polarity) through the red diode back to the function generator. Figure 2.57: The portion of the circuit that has current flow for the positive portion of the waveform. Note that the current is flowing downwards through the resistor, therefore $V_{R}$ will still be positive. The important thing to notice after all that tracing of signal is that the voltage drop across the resistor was positive whether the output of the function generator was positive or negative. Therefore, we are using this circuit to fold the negative half of the original AC waveform up into the positive side of the fence. This circuit is therefore called a full-wave rectifier (actually, this particular arrangement of diodes has a specific name - a bridge rectifier) Remember that at any given time, the current is flowing through two diodes and the resistor, therefore the voltage drop across the resistor will be 1.2 V less than the input voltage (0.6 V per diode - we're assuming silicon...) Figure 2.58: The input voltage and the output voltage of the bridge rectifier. Note that the output voltage is 0 V when the absolute value of $V_{IN}$ is less than 1.2 V, or 1.2 V below the input voltage when the absolute value of $V_{IN}$ is greater than 1.2 V. Now, we have this weird bumpy wave - what do we do with it? Easy... if we run it through a type of low-pass filter to get rid of the spiky bits at the bottom of the waveform, we can turn this thing into something smoother. We won't use a ``normal'' low-pass filter from Section 2.5, however. We'll just put a capacitor in parallel with the resistor. What will this do? Well, when the voltage potential of the capacitor is less than the output of the bridge rectifier, the current will flow into the capacitor to charge it up to the same voltage as the output of the rectifier. This charging current will be quite high, but that's okay for now... trust me... When the voltage of the bridge rectifer drops down, the capacitor can't discharge back into it, because the diodes are now reverse-biased, so the capacitor discharges through the resistor according to their time constant (remember?). Hopefully, before it gets time to discharge, the voltage of the bridge rectifier comes back up and charges up the capacitor again and the whole cycle repeats itself. The end result is that the voltage across the resistor is now a slightly weird AC with a DC offset, as is shown in Figure 2.59. Figure 2.59: ``DC'' output of filtered bridge rectifier output showing ripple caused by the capacitor slowly discharging between cycles. The width of the AC of this wave is given as a peak-peak measurement which is a percentage of the DC content of the wave. The smaller the percentage, the smoother and therefore better, the waveform. If we know the value of the capacitor and the resistor, we can calculate the ripple using the Equation 2.94 : $$\textrm{Peak-to-peak Ripple} = \frac{100 \%}{4 f R C \sqrt{3}}$$ (3.94) where $f$ is the frequency of the original waveforem in Hz, R is the value of the resistor in $\Omega$ and $C$ is the value of the capacitor. All we need to do, therefore, to make the ripple smaller, is to make the capacitor bigger (the resistor is really not a resistor in a real power supply, its actually something like a lightbulb or a portable CD player). Generally, in a real power supply, we'd add one more thing called a voltage regulator as is shown in Figures 2.60 and 2.61. This is a magic little device which, when fed a voltage above what you want, will give you what you want, burning off the excess as heat. They come in two flavours, negative and positive, the positive ones are designated 78XX where XX is the voltage (for example, a 7812 is a + 12 V regulator) the negative ones are designated 79XX (ditto... 7918 is a -18 V regulator.) These chips have 3 pins, one input, one ground and one output. You feed too much voltage into the input (i.e. 8.5 V into a 7805) and the chip looks at its ground, gives you exactly the right voltage at the output and gets toasty. If you use these things (you will) you'll have to bolt it to a little radiator or to the chassis of whatever you're building so that the heat will dissipate. A couple of things about regulators : if you reverse-bias them (i.e. try and send voltage in its output) you'll break it - probably gonna see a bit of smoke too. Also, they get cranky if you demand too much current from their output. Be nice. (This is why you won't see regulators in a power supply for a power amp which needs lots-o-current.) So, now you know how to build a real-live AC to DC power supply just like the pros. Just use an appropriate transformer instead of a function generator, plug the thing into the wall (fuses are your friend) and throw away your batteries. The schematic below is a typical power supply of any device built before switching power supplies were invented (we're not going to even try to figure out how they work). Figure 2.60: Unipolar power supply. Below is another variation of the same power supply, but this one uses the centre-tap as the ground, so we get symmetrical negative and positive DC voltages output from the regulators. Figure 2.61: Bipolar power supply.