Sep 17, 2012 #1 E ECE101 Newbie level 5 Joined Nov 25, 2011 Messages 8 Helped 0 Reputation 0 Reaction score 0 Trophy points 1,281 Activity points 1,342 Hi, For an nMOS Invertor with Depletion load, I'm trying to calculate the switching voltage and current at this point. Both devices will be operating in the Linear region but how do I calculate this? Thresholds are 1V and -3V. (W/L)pd / (W/L)pu = 4
Hi, For an nMOS Invertor with Depletion load, I'm trying to calculate the switching voltage and current at this point. Both devices will be operating in the Linear region but how do I calculate this? Thresholds are 1V and -3V. (W/L)pd / (W/L)pu = 4
Sep 17, 2012 #2 erikl Super Moderator Staff member Joined Sep 9, 2008 Messages 8,108 Helped 2,695 Reputation 5,370 Reaction score 2,308 Trophy points 1,393 Location Germany Activity points 44,123 Neglecting the Vds dependency (λ=0): I2 = Kn*( W/L)*(|Vgs2| - Vth2)2 I1 = Kn*(4W/L)*(Vgs1 - Vth1)2 I2 = I1 4(Vgs1 - Vth1)2 = (|Vgs2| - Vth2)2 2(Vgs1 - Vth1) = (|Vgs2| - Vth2) Vgs1 = Vi 2*(Vi - 1V) = 0 + 3V Vi - 1V = 3/2 V = 1.5V Vi = 2.5V = Switching Voltage I1 = I2 = Kn*(W/L)*9 = current at this point Nice homework!
Neglecting the Vds dependency (λ=0): I2 = Kn*( W/L)*(|Vgs2| - Vth2)2 I1 = Kn*(4W/L)*(Vgs1 - Vth1)2 I2 = I1 4(Vgs1 - Vth1)2 = (|Vgs2| - Vth2)2 2(Vgs1 - Vth1) = (|Vgs2| - Vth2) Vgs1 = Vi 2*(Vi - 1V) = 0 + 3V Vi - 1V = 3/2 V = 1.5V Vi = 2.5V = Switching Voltage I1 = I2 = Kn*(W/L)*9 = current at this point Nice homework!
Sep 18, 2012 #3 E ECE101 Newbie level 5 Joined Nov 25, 2011 Messages 8 Helped 0 Reputation 0 Reaction score 0 Trophy points 1,281 Activity points 1,342 Thanks for the answer erikl! "I2 = Kn*( W/L)*(|Vgs2| - Vth2)2" Can you explain how this is derived? Why can we neglect the Vds dependency? Apologies for the stupid questions! This is the answer we have been given, however I'm struggling to understand how he gets his answer of 2.31V I understand why both are operating in the linear region, "equating the currents" is the problem... Last edited: Sep 19, 2012
Thanks for the answer erikl! "I2 = Kn*( W/L)*(|Vgs2| - Vth2)2" Can you explain how this is derived? Why can we neglect the Vds dependency? Apologies for the stupid questions! This is the answer we have been given, however I'm struggling to understand how he gets his answer of 2.31V I understand why both are operating in the linear region, "equating the currents" is the problem...
Sep 18, 2012 #4 erikl Super Moderator Staff member Joined Sep 9, 2008 Messages 8,108 Helped 2,695 Reputation 5,370 Reaction score 2,308 Trophy points 1,393 Location Germany Activity points 44,123 ECE101 said: "I2 = Kn*( W/L)*(|Vgs2| - Vth2)2" Can you explain how this is derived? Click to expand... This is standard for the medium to strong inversion region. Any book on MOS will tell you this. ECE101 said: Why can we neglect the Vds dependency? Click to expand... At least in first approximation. ECE101 said: "equating the currents" is the problem... Click to expand... That's what I did.
ECE101 said: "I2 = Kn*( W/L)*(|Vgs2| - Vth2)2" Can you explain how this is derived? Click to expand... This is standard for the medium to strong inversion region. Any book on MOS will tell you this. ECE101 said: Why can we neglect the Vds dependency? Click to expand... At least in first approximation. ECE101 said: "equating the currents" is the problem... Click to expand... That's what I did.