## 3.4 Diodes

As discussed in Chapter 1, diodes are semiconductor devices (2 layer) that allow current to flow in one direction, but not the other. This property is very useful in both analog and digital design. Figure 3.39 shows the schematic symbol for a generic diode along with some labels (a good common signal and power diode to use is the 1N4001). A diode is in either of one of two states: conducting or not conducting. If the diode is conducting then there is a current Id flowing through the diode from anode to cathode and a small voltage drop over the diode of 0.5–0.7V (more on this shortly). No matter how much current you drive through the diode, the forward voltage drop will remain 0.5–0.7V for the most part, so the diode can control the direction of current flow. Now, let’s get a little more specific: To make the diode function in the forward direction you have to apply a forward voltage to overcome the depletion zone (remember that); this voltage is around 0.5–0.7V depending on the diode type. Once you have that voltage met then you can drive current through the diode; the actual curve for a generic silicon rectifier diode is shown in Figure 3.40. Let’s take this graph apart.

Figure 3.39 A diode.

Figure 3.40 Diode current Vs. voltage graph.

Referring to the graph in Figure
3.40, there are a lot of interesting
elements. First, we see that as you forward bias the diode, at about .5V it
starts to conduct appreciably and then it’s free to conduct almost
infinite current at about .7V (500mA–1A are common values). But, what
happens if we reverse the voltage and reverse bias the diode? The negative
voltage axis shows what happens in this case. As you can see, up to about -100V
the diode current is measured in microamps (nearly nothing), but the moment we
cross the -100V threshold the diode conducts heavily. What’s happening
here? The diode is breaking down and conducting; this is called the
** breakdown voltage** and if you apply a reverse voltage
greater than or equal to it you will destroy the diode, and at very least it
will conduct like a piece of wire! These specs are typically shown in the
datasheet for the device. Common switching/rectification diodes are the 1N400X
series; the X refers to the amount of current they can handle, so a 1N4001 can
handle about 1A of current. You can find these diodes just about anywhere and if
you refer to the datasheet you will find all the details for forward voltage
drop for various currents along with the peak inverse voltage or breakdown
voltage.

Now that you have a little practical background on diodes, let’s see them in action.

### 3.4.1 Rectifiers

Diodes are used to switch circuit elements on or off, so the first class applications are rectification. There are two basic types of rectification: half-wave and full wave.

#### 3.4.1.1 Half-Wave Rectification

Figure
3.41 illustrates the most basic of circuits, the ** half-wave
rectifier**. An AC input voltage is applied to the circuit. During
the negative cycle the diode is not forward biased; therefore, the load resistor
Rl has 0V across it. However, during the positive half cycle. the diode conducts
(once the forward voltage reaches 0.7 approximately) and a voltage is developed
across Rl of Vl. The graph shown in Figure
3.41b illustrates this, so in essence
we have "clipped" the negative half cycle. This gives us a positive-only
voltage. Now, if we were building a power supply that had an AC input, this
would be a step in the right direction, but we can do better. One of the cool
aspects of diodes is that they can

**signals based on their polarity, let’s see if we can’t steer the negative going cycle to the output correctly.**

*"steer"*Figure 3.41 Half-wave rectification.

#### 3.4.1.2 Full Wave Rectification

What if we could invert the negative going signal and perform a full wave rectification; that is, during the positive half cycle of the AC input we get the output, and during the negative half cycle we invert and send the input to the load as well. This is totally possible and the circuit shown in Figure 3.42 solves this problem. Aside from the details of the circuit operation (we will get to momentarily), an AC input is applied, call it Vin, then the diode bridge steers the signals to the output at Vout and both the positive and negative half cycles are represented. This is DC technically, but obviously not steady DC; the next step will be to smooth it out, but we will get to that a little later. Next, let’s look at how the network works:

Figure 3.42 Full-wave rectification.

**Positive half cycle analysis**—During the positive half cycle, the upper node 1 at the junction of D3 and D1 is positive and the lower node 2 at the junction of D2 and D4 is negative; therefore, the diodes D1 and D2 conduct, outputting the positive half cycle to the + and – terminals.**Negative half cycle analysis**—During the negative half cycle, the upper node 1 now has a negative voltage on it while the lower node 2 has a positive voltage on it; this causes diodes D4 and D3 to conduct and steer the signal out to the + and — terminals.

So, no matter what happens at the Vin input terminals the output voltage is always positive going—very important.

Now that we have a fully rectified signal, can we use this for DC applications? The answer is no, the signal is not AC, but it’s varying DC, still no good to us. It has to be stable at some voltage (preferably 5.0V for TTL logic), so we need to "regulate" it more and filter it. One possibility that comes to mind is to place a filter capacitor on the output Vout; let’s take a quick look at that since we are going to see more of this design trick shortly.

The right half of Figure 3.42 shows the full wave rectifier circuit with an output capacitor on it, if we choose the output capacitor such that

- R
_{load}*C >> 1/f_{ripple}

Where R_{load} is across the output, then the discharge time of the
RC network will be much larger than the charging frequency, so the capacitor
will never discharge. Where f_{ripple} is the ripple frequency
calculated at 2 times the AC line frequency, for USA 60Hz AC this would be
120Hz. So, the question is what size capacitor are we talking about here? Well,
let’s say that we have a 100 ohm load resistance and a 120Hz
f_{ripple}; we must choose C such that

- C >> 1/(f
_{ripple}*100 Ω)

Let’s assume "much greater" is 100x; then C would be

- C = 100 / (120Hz*100 Ω) = 8333μF

This is typical for this kind of application. I would go ahead with a
10,000μF or maybe even a 20,000μF to be safe. About as big as a D size
battery. Next, let’s just see how much ripple voltage we would cause by
draining the power with a load. We start with I_{load }= C*dv/dt:

- I
_{load}= C*dv/dt

rearranging a bit

- dv = I
_{load }*dt/C

and dt is just the small change in time per charging cycle; this is just the period (T) of the charging cycle, which is T = 1/f. Therefore we can write

- dv = I
_{load}/ F*C

And at 120Hz with a filter capacitor of 20,000μF, and a load current of say 10mA, we are going to have a voltage ripple of

- dv = 10mA / (120Hz * 20,000μF) = 4.1mV—not bad!

Although this seems reasonable, it’s still not stable enough for digital design’ we need the voltage to stay very stable over large current drains. For example, a decent sized microcontroller can pull a couple hundred milliamperes; along with ancillary hardware, the current drain of a basic embedded system (the XGS for example) can be in the range of 500–1000mA. Let’s plug 500mA to see the results:

- dv = 500mA / (120Hz * 20,000μF) = 208mV = .208 V

This is starting to get dangerous. The point being that we need something more advanced that can actively adjust the voltage and keep it steady under any current drain; this is what a voltage regulator does, which we will get to in a moment. But, for now, realize this analysis and design is the first stage in your power system: rectification, filtering, and then finally regulation.