diode circuits analysis

EE-203 Diode Circuits Analysis 1 Problem 1: Plot the load line and find the Q-point for the diode circuit in Figure 1 if V = 5 V and R = 10 kΩ. Use the i-v characteristic in Figure 2. Figure 1 Figure 2 Solution: 4 DD D D DD 4 5 10 | V 0 I 0.500 | I 0 V 5 4.5 Forward biased - V 0.5 I 0.450 10 DD I VmAV V VmA =+ == == === Ω EE-203 Diode Circuits Analysis 2 123 4 1 mA 2 mA Q-point i D v D 5 Problem 2: Find the Q-point for the circuit in Figure 3 using the ideal diode model and constant voltage drop model with V on =0.6V. Figure 3 Solution : Ideal diode model: I D = 1V/10kΩ = 100 µA; (100 µA, 0 V) Constant voltage drop model: I D = (1-0.6)V/10kΩ = 40.0 µA; (40.0 µA, 0.6 V) EE-203 Diode Circuits Analysis 3 Problem 3: Find the Q-point for the diode in Figure 4 using (a) the ideal diode model and (b) the constant voltage drop model with V on = 0.6 V. (c) Discuss the results. Which answer do you feel is most correct? Figure 4 Solution : Using Thévenin equivalent circuits yields and then combining the sources 0.4 V + - + - V I 2.2 k Ω (a) Ideal diode model: The 0.4 V source appears to be forward biasing the diode so we will assume it is "on". Substituting the ideal diode model for the forward region yields 0.4 0.182 2.2 V I mA k == Ω . This current is greater than zero, which is consistent with the diode being "on". Thus the Q-pt is (0 V, +0.182 mA). + - + - + - V I 1 k Ω 1.2 k Ω 2 V1.6 V EE-203 Diode Circuits Analysis 4 Ideal Diode: 0.4 V +- + - V I 2.2 k Ω CVD: 0.4 V + - 0.6 V I 2.2 k Ω +- on V (b) CVD model: The 0.4 V source appears to be forward biasing the diode so we will assume it is "on". Substituting the CVD model with V on = 0.6 V yields 0.4 0.6 90.9 2.2 VV I A k µ − ==− Ω . This current is negative which is not consistent with the assumption that the diode is "on". Thus the diode must be off. The resulting Q-pt is: (0.4 V, 0 mA). 0.4 V + - I=0 2.2 k Ω -+ V (c) The second estimate is more realistic. 0.4 V is not sufficient to forward bias the diode into significant conduction. For example, let us assume that I S = 10 -15 A and assume that the full 0.4 V appears across the diode. Then 15 0.4 10 exp 1 8.89 0.025 D V iA nA V − ⎡⎤ ⎛⎞ =−= ⎜⎟ ⎢⎥ ⎝⎠ ⎣⎦ , a very small current. EE-203 Diode Circuits Analysis 5 Problem 4: (a) Find I and V in the four circuits in Figure 5 using the ideal diode model. (b) Repeat using the constant voltage drop model with V on = 0.7 V. Figure 5 (a) ( ) () () D 55 ( ) Diode is forward biased: = 5+0= 5 | I= 0.500 20 ( ) Diode is reverse biased: =0 | V=7 20 7 | V 10 37 ( ) Diode is forward biased: =3 0=3 | I= 0.500 20 ( ) Diode is reverse aVVmA k bIkIVV cVVmA k d −− −− = Ω −Ω= =− −− −= Ω () D biased: =0 | V= 5 20 5 | V 10 I kI V V−+ Ω =− =− (b) ( ) () () D 54.3 ( ) Diode is forward biased: = 5+0.7= 4.3 | I= 0.465 20 ( ) Diode is reverse biased: =0 | V=7 20 7 | V 10 2.3 7 ( ) Diode is forward biased: =3 0.7=2.3 | I= 0.465 20 ( ) Diode aVVmA k bIkIVV cVVmA k d −− −− = Ω −Ω= =− −− −= Ω () D is reverse biased: =0 | V= 5 20 5 | V 10 I kI V V−+ Ω =− =− EE-203 Diode Circuits Analysis 6 Problem 5: (a) Find I and V in the four circuits in Figure 5 using the ideal diode model if the resistor values are changed to 100 kΩ. (b) Repeat using the constant voltage drop model with V on = 0.6 V. Solution : (a) ( ) () () D 55 ( ) Diode is forward biased: = 5+0= 5 | I= 100 100 ( ) Diode is reverse biased: =0 A | V=7 100 7 | V 10 37 ( ) Diode is forward biased: =3 0=3 | I= 100 100 () Diode is reverse aVVA k bIkIVV cVVA k d µ µ −− −− = Ω −Ω= =− −− −= Ω () D biased: =0 A | V= 5 100 5 | V 10 I kI V V−+ Ω =− =− (b) ( ) () () D 54.4 ( ) Diode is forward biased: = 5+0.6= 4.4 | I= 94.0 100 ( ) Diode is reverse biased: =0 | V=7 100 7 | V 10 2.4 7 ( ) Diode is forward biased: =3 0.6=2.4 | I= 94.0 100 () Diod aVVA k bIkIVV cVVA k d µ µ −− −− = Ω −Ω= =− −− −= Ω () D e is reverse biased: =0 | V= 5 20 5 | V 10 I kI V V−+ Ω =− =− EE-203 Diode Circuits Analysis 7 Problem 6: Find the Q-points for the diodes in the circuits in Figure 6 using the ideal diode model. Figure 6 Solution : Diodes are labeled from left to right () () ()()() 12 3D2 D1 D3 D3 D2 1 123 10 0 ( ) on, D off, D on: I 0 | I 1 37 05 I 1.00 I 1.00 | V 5 10 3000 2 2.5 D : 1.00 mA, 0 V D : 0 mA, 2 V D : 1.00 mA, 0 V D aD mA kk mA mA I V k − == = Ω+ Ω −− += →= =−− =− Ω − () () () ( )()() 12 3D2 D3 D1 D2 1 D3 1 123 ( ) on, D off, D off: I 0 | I 0 10 5 I 0.500 | V 5 10 8000 1.00 81012 V 5 12000 1.00 : 0.500 , 0 : 0 , 1.00 : 0 , 1.00 D D bD mA I V kkk IV DmAVDAVDAV == −− ===−−=− Ω+ Ω+ Ω =− − + =− −− EE-203 Diode Circuits Analysis 8 () () () ()()( ) 123 D1 10K 2 1 10 12K 3 12 10 123 ( ) on, D on, D on 010 02 I 1.25 0 | I 0.200 | 1.05 0 810 25 I 0.583 | 0.783 0 12 : 1.25 , 0 : 1.05 m , 0 : 0.783 m , 0 DD K DKK cD mA mA I I I mA kk mA I I I mA k DmAVD AVD AV −− − ==>==− =+=> ΩΩ −− = = =−= > Ω () () () ()()() 1231 2 D3 D1 3 D2 3 123 ( ) , , : 0, 0 12 5 I 567 0 | V 0 5 10000 0.667 0 30 V 5 12 10000 1.33 0 : 0 , 0.667 : 0 , 1.33 : 567 , 0 DD D D d D off D off D on I I V AIV k IV DA VDA VD AV µ µ == −− ==>=−−+=−< Ω =− − =− < −− Problem 7: Find the Q-points for the diodes in the circuits in Figure 6 using the constant voltage drop model with V on = 0.6 V. Solution: Diodes are labeled from left to right ( ) () () ()()( ) 12 3D2 D1 D3 D3 D2 1 123 10 0.6 0.6 ( ) on, D off, D on: I 0 | I 1.00 37 0.6 5 I 1.00 I 0.760 | V 5 10 0.6 3000 1.40 2.5 D : 1.00 mA, 0.600 V D : 0 mA, 1.40 V D : 0.760 mA, 0.600V D aD mA kk mA mA I V k −−− == = Ω+ Ω −−− += →= =−−− =− Ω − () () () ( )()() 12 3D2 D3 D1 D2 1 D3 1 123 ( ) on, D off, D off: I 0 | I 0 10 0.6 5 I 0.480 | V 5 10 0.6 8000 0.560 81012 V 5 12000 0.760 D : 0.480 mA, 0.600 V D : 0 A, 0.560 V D : 0 A, 0.760 V D D bD mA I V kkk IV == −−− ===−−−=− Ω+ Ω+ Ω =− − + =− −− EE-203 Diode Circuits Analysis 9 () () () () 123 110 2110 12 31210 12 ( ) on, D on, D on 0.6 9.4 0.6 1.4 1.10 0 | 0.200 810 1.4 5 0.900 0 | 0.533 | 0.733 0 12 D : 1.10 mA, 0.600 V D : 0.900 mA, 0.60 DK DD K K D K K cD VV ImAI mA kk V III mA I mAIII mA k −−− −− ==>==− ΩΩ −− =+ = > = = = − = > Ω ()() 3 0 V D : 0.733 mA, 0.600 V () () () ()()() 1231 2 D3 D1 3 D2 3 123 ( ) , , : 0, 0 11.4 5 I 547 0 | V 0 5 10000 0.467 0 30 V 5 11.4 10000 0.933 0 : 0 , 0.467 : 0 , 0.933 : 547 , 0 DD D D d D off D off D on I I V AIV k IV DA VDA VD AV µ µ == −− ==>=−−+=−< Ω =− − =− < −−