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Diodes

Assume you have a pn junciton with a doping of 10 16 donor atoms per cc.

Id

Vd+ -

p n

Id

Vd+ -

Diode pn junction Diode symbol

Diode VI Characteristics:

The current through a diode is exponential with respect to voltage:

I d = I o (e V d / V T 1)

V d = V T ln ( I d / I o + 1)

where

is a constant, approximately 1 for germanium, 2 for silicon

at 300K (the volt equavelent of temperature)V T = 0.026 V

is the reverse saturation current I o

For Silicon, . The VI charactersistic is then: I o 2 10 8 A

VI characteristic of a Silicon diode

NDSU Diodes ECE 321 - JSG

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Problem: Find the current and voltage across a Silicon diode:

Vd

+

-

+5+

-

1000

Solution: Looking left at point AB, the VI characteristics are:

I d = 5V d

200

Looking right, the VI characteristics are that of a diode:

I = I o (e V / V T 1)

This gives two equations for two unknowns.

A graphical solution is to plot these two on the same graph. The intersection is the point (Vd, Id) whichsatisfies both equations:

SciLab Code:

The diode VI characeristics:- - >I o = 2e- 8;

- - >I d = [ 0: 0. 1: 5] ' / 1000;- - >Vd = 0. 052*l og( I d/ I o + 1) ; The r esi st or & 5V suppl y VI char act er si t i cs:

The 5V source and 1k resistor VI characteristic:- - >I 2 = [ 0: 0. 1: 5] ' / 1000;- - >V2 = 5 - 1000*I 2;

- - >I 2 = pl ot ( Vd, I d*1000, V2, I 2*1000)- - >xl abel ( ' Vd' ) ;- - >yl abel ( ' I d (mA) ' )


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Diode VI characteristic (blue) and resistor & 5V supply VI characteristic (red). The intersection is thesolution: (0.7037V, 21.4814mA).

You can also solve this numerically with a f(x)=0 solver.

Guess Id

Solve for Vd using the diode equation

Solve for Vd using the 5V & resistor equation

The error is the difference in Vd

Itterate until the difference is zero

This gives a numerical solution of

Vd = 0.7037V

Id = 21.4814mA

MultiSim

A very useful program is a SPICE simulator, like MultiSim. This lets you draw a circuit on a PC and simulate it's response.

For example, measure the voltage and current through a diode.

First, add the parts in MultiSim. You'll need to search through the menus to find the parts, but it's prettyintuitive.

Click on Show Power Source to select the DC supply. Right click on it to change it to 5V.

The diode icon has diodes, etc.

Right click on a part to rotate it.

Click on the Show Measurement Family to see the volt meter and ammeter.

Once you place the parts, click on one end to add a wire to another part. When you're done, your circuitlooks like the following:


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Click F5 to start and stop the simulation, This then shows you one point on the VI curve: (0.693V,

4.307mA).To change the resistor, right click on it, select replace, and change it to a 10k resistor

Do this over and over and you can get the VI characteristic for the diode.

You can also show this on a plot.

Replace the 5VDC source with an AC source. Right click to change it to 5V.

Add a 1 Ohm resistor in series with the diode. This lets you measure the current through the diode

by measuring the voltage across the 1 Ohm resistor.Add an oscilloscope. Set it up to plot AB (voltage vs. current).


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On the oscilloscope

The X axis is the voltage (off by 1mV due to the 1 Ohm resistor)

The Y axis is the current (1mV = 1mA).

Note that

The diode turns on at about 0.7V.

The voltage across the diode does change with current, but not a lot.

Different types of LEDs turn on at different voltages. A red LED, for example, turns on at about1.7V:

Since the current-voltage relationship is nonlinear, it's somewhat hard to describe this with a singlenumber. Typicaly, LED's are rated as the voltage you get with the current is 20mA. You're trying todescribe a curve with a single point - which isn't all that accurate. The voltage is almost constant,however, so you can assume that the voltage is the same as long as current flows.


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