|Applications of Diodes - Electronics & Communication tutorials|
Applications of Diodes
Now we will briefly consider a couple of applications for diodes. We’ll cover these in much more detail later.
Conversely, when vI < 0 , the ideal diode is "off" and there is no current through R.
Therefore, v0 = 0. We have "rectified" the input voltage with this circuit. This process of rectification will work for any type of input signal, whether it is periodic (as shown above) or not.
Digital logic gate. Diodes and resistors together can be used to make rudimentary logic gates. For example:
Assuming the voltages are 0 V for "low" signals and 5 V for the "high" signals, then the circuit shown above is a two-input OR gate:
o If vA = 5 V and vB = 0 or 5 V, then vY = 5 V.
o If vB = 5 V and vA = 0 or 5 V, then vY = 5 V.
for ideal diodes. This is an OR function Y=A+B.
Why do we have the resistor in this circuit? It’s a "clamp down" resistor and forces the voltage vY = 0 when vA = vB = 0. (What if there was no R? Wouldn’t the voltage be zero in this case as well?)
One huge complication of diodes (or any nonlinear circuit element) in an electrical circuit is the use of superposition in the analysis is disallowed! (The exception to this is if the analysis has been linearized for only small amplitude signals. We’ll see this linearization throughout the course.)
Consequently, nonlinear circuit analysis is usually much more complicated than linear circuit analysis. One often needs numerical analysis for solution, such as that provided by circuit simulation software.
Example N1.1 Determine the voltage V and current I in the circuit below.
This is a nonlinear circuit, so a completely different analysis procedure is required than what you’ve used in the past for linear circuits.
One process you can use to solve this problem is to try (i.e. guess) different on/off combinations for the diodes D1 and D2 until you achieve a physically plausible and self consistent solution.
For example, if D1 is "on" and D2 is "on," then V = 1 V and 3 V.
It is simply impossible to have two different voltages simultaneously at a node. Voltages must be single valued at all nodes. We conclude that D1 and D2 cannot both be "on" simultaneously.
Next, we try D1 "off" and D2 "on." This leads to V = 3 V and I = 8 mA.
This result is physically realistic and self consistent since with V = 3 V then the voltage drop across D1 requires that it be "off," which is what we have assumed.