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Field Effect Transistors Circuit Analysis HP PA8000 Fairchild Clipper C100 Fujitsu 68903 EE314 FET Circuit Analysis 1.MOS Small Signal Equivalent 2.Transconductance 3.Common-Source Amplifiers 4.Source Follower 5.Logic gates Chapter 12: Field Effect Transistors Small-Signal Equivalent Circuit for FETs Output signal from an amplifier using FET can be effectively modulated by small changes of input signal current. In this way it is possible to make small changes from the Q point. Symbols: The total quantities: iD(t), vGS(t) The dc point values: IDQ, VGSQ The signal id(t), vgs(t) vGS (t ) VGSQ v gs (t ) i D (t ) I DQ id (t ) Characteristic Small-Signal Equivalent Circuit -Transconductance Schematic Analysis… (a little bit of math) iD K vGS Vt 0 2 I DQ id (t ) K VGSQ v gs (t ) Vt 0 2 2 I DQ id (t ) K VGSQ Vt 0 2K VGSQ Vt 0 v gs t Kv gs (t ) 2 We know that Also we assume that I DQ K VGSQ Vt 0 2 v gs (t ) VGSQ Vt 0 * Small-Signal Equivalent Circuit -Transconductance Schematic 2 I DQ id (t ) K VGSQ Vt 0 2K VGSQ Vt 0 v gs t Kv gs (t ) 2 We know that Also we assume that Drain current generated by signal I DQ K VGSQ Vt 0 2 v gs (t ) VGSQ Vt 0 * id (t ) 2 K VGSQ Vt 0 v gs (t ) Small-Signal Equivalent Circuit -Transconductance We define the transconductance as gm id (t ) vgs (t ) or id (t ) g m v gs (t ) so g m 2 K VGSQ Vt 0 Small-Signal Equivalent Circuit -Transconductance iD K vGS Vt 0 2 so vGS Vt 0 I DQ K Thus the transconductance g m 2 K VGSQ Vt 0 2 KI DQ Small-Signal Equivalent Circuit -Transconductance Exercise The transistor has KP=50mA/V2, Vto=2V, L=10mm, and W=400mm W KP K 1mA / V 2 L 2 g m 2 K VGSQ Vt 0 2(4 2) 4mS Small-Signal Equivalent Circuit Also we assume that ig (t ) 0 g m 2 KI DQ W KP K L 2 Better performance is obtained with higher values of gm. Please notice that gm is proportional to the square root of the Q point drain current. Simply, we can increase gm by choosing a higher value of IDQ. More Complex Equivalent Circuits For more accurate analyses of FET transistor we have to add more components to an equivalent circuit. Small capacitance: for high response FET amplifiers Drain resistor: account for the effect of vDS on the drain current Correction for id id (t ) g m v gs (t ) vds / rd Please read section: Transconductance and … pp.591 Example 12.3 Drain Resistance Calculation 1 i D rd vDS so at vGS=4V 1 iD 10.7 10 mA 0.7 mS 0.175mS 10 6V rd vDS 4 rd 5.7k Common-Source Amplifier Schematic G D S Equivalent circuit The dc supply voltage acts as a short circuit for the ac current. Common-Source Amplifier C1, C2 -coupling C Schematic Cs -bypass C ac signal G D S Equivalent circuit The dc supply voltage acts as a short circuit for the ac current. Common-Source Amplifier: Gain, Rin and Rout Equivalent circuit (once more) 1 R 1 / rd 1 / RD 1 / RL ' L Voltage gain v0 g m v gs RL' vin v gs v0 Av g m RL' vin Input resistance Rin vin RG R1 R2 iin From bias point analysis Common-Source Amplifier: Gain, Rin and Rout To find out the Rout we have to: disconnect the load, replace the signal source by short circuit – Thevenin equivalent resistance No source connected to the input Example 12.4 if vgs=0 then gmvgs=0 Output resistance Rout 1 1 / R D 1 / rd Source Follower Small-Signal Equivalent Circuit –Source Follower Notice that small signal IDS goes up. Why? Small-Signal Equivalent Circuit –Source Follower RL' 1 1 / rd 1 / RS 1 / RL Input resistance Voltage gain v0 g mvgs RL' vin vgs vo vgs 1 gm RL' v0 g m RL' Av 1 ' vin 1 g m RL vin Rin RG R1 R2 iin Since the output voltage is almost equal to the input – hence the name source follower Logic gates – COMS Inverter Logic truth table Vin V out 0 1 1 0 Switch level equivalent circuits Logic gates – COMS NAND gate Logic truth table A B V out 0 0 1 0 1 1 1 0 1 1 1 0 Logic gates – COMS NOR gate Exercise Draw switch level circuits for different inputs and derive the truth table for this gate Logic gates – COMS NOR gate Logic truth table A B V out 0 0 1 0 1 0 1 0 0 1 1 0