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# LEARN ELECTRONICS - DIODES

### INTRODUCTION

The basic function of a diode is to conduct very well in one direction and very poorly in the other. When optimized to turn AC into DC, it is called a rectifier diode which by the way, is the oldest and most widely used type. You will find these most commonly used in power supplies. In addition, several other types have been created including Light Emitting Diodes (LED’s), Small signal diodes, schottky diodes, zener diodes, tunneling diodes, and avalanche diodes.

### SCHEMATIC SYMBOL Diode Schematic Symbol

The image shows the schematic symbol of a diode as well as the p and n sections. With the p section being positively doped silicon and the n section negatively doped silicon. The p side is called the anode and the n side the cathode. The symbol itself looks like an arrow pointing from the p to the n section. This denotes that conventional current flows quite easily through the diode when flowing in the direction of the arrow. Conversely, electron flow goes in the opposite direction to the arrow. The point where the p and the n meet creates a depletion zone which gives rise to the barrier voltage which will be described shortly.

### KNEE VOLTAGE Diode Voltage/Current Graph

The image shows the graph of a silicon diode with voltage along the horizontal axis and current on the vertical axis. From this it can be seen that the diode doesn’t conduct until the applied voltage reaches the barrier voltage. Which is why the current is 0 below this value. As the voltage crosses 0.7 V, electrons flow rapidly and the current rises sharply. The point at which the current rises quickly is called the knee voltage and is typically 0.7 V for a silicon diode and 0.3 V for a germanium diode. This barrier voltage arises because some of the electrons from the n material spill across the pn junction making a small area of electrically neutral material. This area requires 0.7 V in order to push through the barrier and start conducting.

### NON LINEAR DEVICE

A resistor is an example of a linear device. That is, as the voltage across it increases, the current through it increases proportionally. For example, if the voltage doubles, the current doubles.

Whereas, a diode is a non linear device. Below 0.7 V the diode conducts very little, and above 0.7V it conducts readily. Non linear devices will never have a straight line on a graph and this can be seen from the VI graph above.

### BULK RESISTANCE

All semiconductors have some residual resistance in their materials. Once the voltage on a diode has surpassed the barrier voltage, the only thing that impedes the current flow is this small resistance. The total of which is simply the sum of the p resistance and n resistance and depends on the doping level of the silicon. Typically this total is less than 1 ohm. Generally this characteristic is not on the datasheet but it can be calculated from other parameters namely the forward voltage at a specified forward current. For example, in the datasheet below, the Maximum Forward Voltage Drop (Vf) is stated to be 0.93 Volts typically to a maximum of 1.1 Volts at 1 Amp. So the typical bulk resistance is:

Rbulk = (0.93 - 0.7) / 1

= 0.23 Ohms

### MAXIMUM DC FORWARD CURRENT

Most manufactures specify a maximum forward current. This value is the maximum current that can be safely passed continuously without fear of damage to the diode. This is labeled Io in the datasheet for the 1N4000 rectifier diode datasheet below. 1N4000 series datasheet

This is why there needs to be some series resistance in the circuit of a diode. Whether it’s a simple resistor, or the load that the diode sees, it is necessary in order to prevent excess current flow and damage to the diode. This is because without a series resistor, the only thing limiting the current is the bulk resistance of the diode and as mentioned above, is typically less than 1 ohm and will result in excessive current and likely destroying the diode.

### MAXIMUM POWER DISSIPATION

Some manufactures specify a maximum power dissipation. This is the max power that a diode can dissipate without damage and is equal to the current times the voltage. As mentioned above, it is easy to exceed this value without a series resistance. For example, if you have a diode with a max power of 1 watt and a bulk resistance of 0.23 ohms connected to a 5 Volt / 1 Amp source, the current through the diode is:

I = V / R

= 5 / 0.23

= 21.7 Amps!

Since this value exceeds the power supply rating we will calculate the power using 1 Amp.

P = V x I

= 5 x 1

= 5 Watts

This power dissipation is 5 times the rating of the diode and will certainly cause it to fail and may even release the magic blue smoke.

### THE REVERSE REGION

When a diode is reversed biased, a very small amount of current flows up until the breakdown voltage when the current rises rapidly for small increases in voltage. In the case of a rectifier diode, it’s imperative that the applied reverse voltage is less than the breakdown voltage. It’s important to note that regardless of the diode circuit, the diode voltage will follow the same current plot as the graph above. However the voltage and current values will vary depending on the specific diode characteristics.

### FIRST APPROXIMATION

There are 3 basic approximations when working with diodes and analyzing their operation. The first approximation treats the diode as a perfect unidirectional switch. That is, when the diode is forward biased, the switch is closed and full current flows. When the diode is reverse biased, the switch is open and no current flows. The VI graph shows that current rises to maximum instantly when the voltage is above 0 and drops to 0 when the voltage is below 0.

When trying to analyze an existing circuit, using the first approximation is often good enough. For example, if the output of a power supply is should be 50 V and you measure 20 V, it’s obvious that something is wrong. However there are times when this approximation gives terrible answers and more accuracy is needed.

An example showing all approximations on the same circuit will be shown below.

### SECOND APPROXIMATION

The second approximation takes into account the barrier voltage of 0.7 V for silicon or 0.3 V for germanium.

If the source voltage is much greater than the barrier voltage the error produced by using the first approximation is small. For example, if the source is 70 V, the error is 1%. But if the source voltage is close to the barrier voltage, it needs to be taken into account. For example, if the source is 7 V, the error is 10% which can have a major effect on the operation of the circuit.

Think of the equivalent circuit as a switch in series with a voltage source of 0.7 V. If the source voltage is less than 0.7 V, no current flows and if it is above 0.7 V, full current flows as shown in the graph.

### THIRD APPROXIMATION

The third approximation also included the bulk resistance of the diode. This diode graph includes the effect of this resistance.

As the forward current increases, the forward voltage also increases slightly because we have to add the voltage drop across this resistance (I x R) and the equivalent circuit is a switch in series with a voltage source for the barrier potential in series with a resistor.

When the source voltage is greater than the barrier potential of 0.7 V, the diode conducts and the voltage across the diode is:

Vd = 0.7 + Id x Rb

For germanium, use 0.3 V instead of 0.7 V.

Using a current of 1 Amp and a bulk resistance or 0.23 Ohms, the voltage across a silicon diode becomes:

Vd = 0.7 + 1 x 0.23

=0.723 Volts

A difference of 0.023 V from the 2nd approximation.

Since the bulk resistance is usually very low, the effect on the circuit is negligible and not worth the effort calculating unless working on a very high precision instrument where that small voltage can affect its operation.

Bulk resistance is often not specified on datasheets but can be calculated from the information given. For example, if the diode has a forward voltage of 0.9 V at 1 A then the resistance is:

Rd = (0.9 – 0.7) / (1 – 0) = 0.2 / 1

= 0.2 Ohms

Replace 0.7 V with 0.3 V for germanium diodes.

The 0.7 and 0 in the equation are due to the fact that no current flows at the knee voltage.

### WHICH APPROXIMATION TO USE

When analyzing a circuit, it’s important to know which approximation to use to save time. If analyzing a power supply with large voltages, the first approximation is adequate as the error from the second and third approximation is small enough to ignore.

If analyzing a circuit with lower source voltages, i.e. 15 V or less, the effect of the barrier voltage should be included as the error at 15 V is about 5% and at 10 V is 7% and will affect the expected voltage values.

The third approximation is only needed for very precise situations. For example if the diode is in a circuit with a source voltage of 1V, diode resistance of 0.23 ohms, and with a varying load current of 10 mA to 50 mA, the load voltage after the diode can vary from 0.2885 V to 0.2977 V, a variation of 3%. If your circuit must maintain 1% error, the resistance becomes important.

However, for most circuit analysis, the second approximation gives accurate results and is easy to implement and calculate.

### APPROXIMATION EXAMPLE

Now we will analyze a simple diode circuit using all 3 approximations in 2 different situations, one with a large source voltage, and the second with a small voltage source. The circuit will use a resistor of 1000 ohms with a silicon diode having a bulk resistance of 0.2 Ohms. The first situation will use a Vs of 50 Volts and the second with 5 Volts. We will calculate the voltage across the resistor and diode for each approximation.

### Vs = 50 Volts

FIRST APPROXIMATION

In the 1st approximation we assume the diode as a switch. Which means the diode voltage is 0. So the current in the circuit is:

I = V / R

= 50 / 1000

= 50 mA

The voltage across the diode is 0 so the full source voltage is across the resistor being 50 Volts.

SECOND APPROXIMATION

Now we take the barrier voltage into consideration. Following the same process, the voltage across the resistor now is:

Vr = 50 - 0.7

= 49.3

And the current in the circuit is:

I = 49.3 / 1000

= 49.30 mA

THIRD APPROXIMATION

The third and final calculation is to include the bulk resistance of 0.2 Ohms. Following above, the total resistance in the circuit is:

Rt = 1000 + 0.2

= 1000.2 Ohms

The voltage across the resistance total is the same as above being:

Vr = 50 - 0.7

= 49.3 Volts

The current through the circuit is now:

I = 49.3 / 1000.2

= 49.29 mA

Including the bulk resistance, the diode voltage is now:

Vd = 0.7 + (.04929 x 0.2)

= 0.7099 Volts

ERROR

The error for the 1st approximation resistor voltage is:

Vr1err% = (50 / 49.29) - 1 x 100

= 1.44 %

And the 2nd approximation error is:

Vr2err% = (49.3 / 49.29) - 1 x 100

= 0.02 %

Since many component tolerances are in the 5% range, the worst error of 1.44% is well below other components and it serves no purpose to use a more accurate approximation for most analysis. You'll notice that the 2nd approximation is almost 0 error and shows that including the bulk resistance is completely unnecessary for large voltage sources.

Next let's look at a voltage source of 5 Volts.

### Vs = 5 Volts

FIRST APPROXIMATION

Resistor voltage is:

Vr = 5 Volts

Resistor current is:

I = 5 / 1000

= 5 mA

SECOND APPROXIMATION

Resistor voltage is:

Vr = 5 - 0.7

= 4.3 Volts

Resistor current is:

I = 4.3 / 1000

= 4.3 mA

THIRD APPROXIMATION

Total resistance is:

Rt = 1000 + 0.2

= 1000.2 Ohms

Resistor current is:

I = 4.3 / 1000.2

= 4.299 mA

Resistor voltage is:

Vr = .004299 x 1000

= 4.299 Volts

Finally, diode voltage is:

Vd = 0.7 + (.004299 x 0.2)

= 0.701 Volts

ERROR

1st approximation error is:

Vr1err% = (5 / 4.299) - 1 x 100

= 16.3 %

2nd approximation error is:

Vr2err% = (4.3 / 4.299) - 1 x 100

= 0.02 %

From this you can see that there is an error of 16.3% if using the 1st approximation, so it is wise in this instance to use the 2nd approximation which is only 0.02% away from the 3rd approximation. Meaning that even in this case, taking the bulk resistance into account is unnecessary.

### CONCLUSION

Hopefully this overview of diodes and how they work helps to demystify them and allows you to better use them in your own creations. As a general rule of thumb, the 2nd approximation will almost always serve you well as should be the go to when analyzing most circuits.

In a future article, I will describe Zener diodes and how to go about designing with them including the current limiting resistor and power calculations needed.