section 6 5 The Common Source Amp with Active Loads

1 5/4/2011 section 6_5 The Common Source Amp with Active Loads 1/2. The Common Source Amp with Active Loads Reading Assignment: pp. 582-587. Amplifiers are frequently made as integrated circuits ( , op-amps). Although both BJTs and MOSFET integrated circuit amplifiers are implemented as ICs, we find that MOSFETs amplifiers are almost exclusively implemented as integrated circuits ( , rarely are MOSFET amps made of discrete . components). Making integrated circuit amplifiers has many positives, but a few negatives: Positives: Jim Stiles The Univ.

2 Of Kansas Dept. of EECS. 5/4/2011 section 6_5 The Common Source Amp with Active Loads 2/2. The amplifier circuit can be quite complex, yet still small and inexpensive. Thus, current sources are no big deal . Negatives: We cannot make large capacitors ( , COUS), so that DC. blocking capacitors are not possible this makes bias solutions more complex, particularly for multi-stage amplifiers. Additionally, it if difficult to make resistors in integrated circuits. Instead, we use resistors constructed from transistors so-called Active Loads .

3 HO: Enhancement Loads HO: The Common Source Amp with an Active Load The sensitivity problem of the previous circuit can be solved using a current Source as a load . HO: The Common Source Amp with a Current Source Jim Stiles The Univ. of Kansas Dept. of EECS. 5/4/2011 Enhancement Loads 1/7. Enhancement Loads Resistors take up far too much space on integrated circuit substrates. Therefore, we need to make a resistor out of a transistor! Q: How can we do that!? After all, a resistor is a two terminal device, whereas a transistor is a three terminal device.

4 A: We can make a two terminal device from a MOSFET by connecting the gate and the drain! + +. v v - - i i Enhancement Load Resistor Load Q: How does this enhancement load resemble a resistor? A: Consider the i-v curve for a resistor: 5/4/2011 Enhancement Loads 2/7. i v i =. R. v Now consider the same curve for an enhancement load. Since the gate is tied to the drain, we find vG = vD , and thus vGS = vDS . As a result, we find that vDS > vGS Vt always. Therefore, we find that if vGS > Vt , the MOSFET will be in saturation (iD = K (vGS Vt )2 ), whereas if vGS < Vt , the MOSFET.

5 Is in cutoff (iD = 0 ). Since for enhancement load i = iD and v = vGS , we can describe the enhancement load as: 0 for v < Vt . i = .. ( ). 2. K v Vt for v > Vt 5/4/2011 Enhancement Loads 3/7. Plotting this equation: i i = K (v Vt )2. v Vt So, resistors and enhancement Loads are far from exactly the same, but: 1) They both have i = 0 when v = 0 . 2) They both have increasing current i with increasing voltage v. i Resistor Enhancement Load v 5/4/2011 Enhancement Loads 4/7. Therefore, we can build a Common Source amplifier with either a resistor, or in the case of an integrated circuit, an enhancement load.

6 VDD. VDD. vO vO. vI vI. For the enhancement load amplifier, the load line is replaced with a load curve (v = VDD vDS )! iD. ID ,VDS. vDS. VDD Vt VDD. And the transfer function of this circuit is: 5/4/2011 Enhancement Loads 5/7. vO. Q in saturation dvO. 1. dv I. vI. Vt Q: What is the small signal behavior of an enhancement load? A: The enhancement load is made of a MOSFET device, and we understand the small-signal behavior for a MOSFET! Step 1 - DC Analysis If V > Vt , then I = K (V Vt ). 2. +. or: V. I. V = +Vt - K I.

7 5/4/2011 Enhancement Loads 6/7. Step 2 Determine gm and ro gm = 2K (VGS Vt ) = 2K (V Vt ). 1 1 1. ro = = =. ID I K (V Vt ). 2. Step 3 Determine the small-signal circuit Inserting the MOSFET small-signal model, we get: i = id G D. +. v =vgs ro - gm v gs S. Redrawing this circuit, we get: i G. D. +. v ro - gm v S. 5/4/2011 Enhancement Loads 7/7. Or, simplifying further, we have the small-signal equivalent circuit for an enhancement load: It is imperative that you understand that the circuit to my right is the small-signal equivalent circuit for an enhancement load.

8 Please replace all enhancement Loads with this small- signal model whenever you are attempting to find the small-signal circuit of any MOSFET amplifier. i +. v ro gm v - Enhancement Load Small-Signal Model 5/4/2011 The Common Source Amp with Enhancement Load 1/9. The Common Source Amp with Enhancement Load VDD. Consider this NMOS amplifier using an enhancement load. * Note no resistors or Q2. capacitors are present! vO(t). * This is a Common Source Q1. amplifier. +. vi(t) _. * ID stability could be a problem VG. Q: What is the small-signal open-circuit voltage gain, input resistance, and output resistance of this amplifier?

9 A: The values that we will determine when we follow precisely the same steps as before!! 5/4/2011 The Common Source Amp with Enhancement Load 2/9. Step 1 DC Analysis Note that: The DC circuit of this amplifier is: ID 1 = ID 2 ID. VDD. and that: ID1. VGS 1 =VG 0 =VG. Q2. and also that: VO. Q1 VDS 2 = VGS 2. ID2 and finally that: VG. VDS 1 =VDD VDS 2. Let's of course ASSUME that both Q1 and Q2 are in saturation. Therefore we ENFORCE: ID 1 = K1 (VGS 1 Vt 1 ). 2. = K1 (VG Vt 1 ). 2. Note that there are no unknowns in the previous equation.

10 The drain current is explicitly determined from K1 , VG , and Vt 1 ! Continuing with the ANALYSIS, we can find the drain current through the enhancement load (ID2), since it is equal to the current through Q1: 5/4/2011 The Common Source Amp with Enhancement Load 3/9. ID 2 = ID 1 = K1 (VG Vt 1 ). 2. Yet we also know that VGS2 must be related to this drain current as: ID 2 = K2 (VGS 2 Vt 2 ). 2. and therefore combining the above equations: ID 1 = I D 2. K1 (VG Vt 1 ) = K2 (VGS 2 Vt 2 ). 2 2. Note this last equation has only one unknown (VGS 2 ) !