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Diode Applications
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Shokley Equation where VT is ~25.8mV at room temp T= 300K or 27C
•The forward bias current is closely approximated by where VT is ~25.8mV at room temp T= 300K or 27C k = Boltzman’s constant = 1.38 x joules/kelvin T = absolute temperature q = electron charge = x 10-19coulombs η = constant dependent on structure, between 1 and 2 (if not given assume n = 1) IS= reverse saturation current –Notice there is a strong dependence on temperature – 2
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We can approximate the diode equation for i>> IS
In reverse bias (when v << 0 by at least VT), then In breakdown, reverse current increases rapidly… can be approximated by a vertical line
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Dynamic resistance
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Load Line Analysis The load of a circuit determines the point or the region of operation of a diode (or device) The method: A line is drawn on the characteristic of the device. The intersection point gives the point of operation (called Q-point)
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Load line •Look at the simple diode circuit below. We can write two equations:
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Diode in DC Series Circuit: Forward Bias
VD KVL: E – VD – VR = 0 E = VD + IDR ID = IR= E/R → VD = 0V
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Diode in DC Series Circuit:
Reverse Bias E = VD + IDR →ID =0A • VD = E • VR = IDR = 0V VD 8
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Example1 For the series diode configuration below, employing the diode characteristics of figure below, determine VDQ, IDQ and VR. In general, approximate model of diode is used in applications because of non- ideal real life conditions (tolerance, temperature effect, etc) never allow an ideal case to be applied 9 9
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Solution Step1: Find the maximum ID. VD = 0V→ ID = IR= E/R= 10mA
Step 2: Find the maximum VD. ID =0A → E = VD + IDR=10V Step 3: Plot the load line Step 4 : Find the intersection between the load line and the characteristic curve. This is the Q-point From curve : VDQ=0.78 V and IDQ=9.25mA Step 5: Checking : VR=IRR=IDQR=9.25 Or VR =E-VD =9.22V 10 10
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Example Repeat the analysis of previous example using the practical diode model VDQ=0.7 V and IDQ=9.25mA 11
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Example Repeat the analysis of previous example using the ideal diode model VDQ=0 V and IDQ=10 mA
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Repeat the analysis of Example 1 with R = 2 k
VD = 0V→ ID = IR= E/R= 5mA ID =0A → E = VD + IDR=10V VDQ=0.7 V IDQ=4.6 mA 13
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Example Repeat the analysis of previous example using the practical diode model VDQ=0.7 V IDQ=4.6 mA 14
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PWLDM (Piecewise Linear Diode Model)
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SERIES DIODE CONFIGURATIONS WITH DC INPUTS
For the series diode configuration below, determine VD, VR and ID, given that E=8V and R=2.3 k VD Solution: KVL VR = E – VD = = 7.3V ID = IR = VR/R = 7.3/2.2k = 3.32m A Si 8V 2.2 k 16
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Example Determine ID, VD2 and Vo for the circuit.
Remember, the combination of short circuit in series with an open circuit always results in an open circuit and ID=0A.
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Example Determine I, V1, V2 and Vo
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Example : Determine Vo and ID for the series cct below:
Vo=E-VD1-VD2= =11V ID=IR=11/5.6=1.96 mA 19
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Parallel Configurations
Solve this circuit like any Parallel circuit, knowing VD = 0.7V (or up to 0.7V) in forward bias and as an open in reverse bias. VD1 = VD2 = Vo =0 .7V VR = 9.3V Diodes in parallel are used to limit current: IR = E – VD = 10V -0 .7V = 28mA R k ID1 = ID2 = 28mA/2 = 14mA
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Example Determine the resistance R for the network when I=200mA. Si Si
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Example Determine the currents I1, I2, and ID2 for the network
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Diode as logic gate OR Gate
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AND Gate
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