Subject Name MICROELECTRONIC CIRCUITS Subject Code 10 EC

Subject Name : MICROELECTRONIC CIRCUITS Subject Code: 10 EC 63 Prepared By: Sreepriya Kurup, Arshiya Sultana Department: ECE Date: 3/4/2015 28/02/2015 1

UNIT 2 SINGLE STAGE INTEGRATED CIRCUIT AMPLIFIERS 3/4/2015 2

Topic Details Ø MOSFET Scaling ØComparison between MOSFET and BJT Ø IC Biasing – MOSFET Current sources ØMOS Current Steering Circuits ØBJT current mirror ØBJT current source ØBJT current steering circuits ØHigh Frequency Response Of Amplifiers ØMiller’s Theorem 3/4/2015 3

MOSFET Scaling Ø In 1965 , G. E Moore predicted that the number of transistors in integrated circuits would double after every two years. This prediction has come TRUE. ØTo assemble a large number of transistors in given silicon area we have to reduce the size of the transistor. The process of reducing vertical and horizontal dimensions of MOSFETs by some scaling factor S , which is greater than 1, is called scaling. Ø Two types of schemes commonly used for MOSFET scaling are constant voltage scaling and constant field scaling. 3/4/2015 4

Constant Field Scaling Ø The MOSFET dimensions as well as supply voltages are scaled by the same scaling factor S, greater than 1. Since, the scaling of supply and terminal voltage maintains the same electric field as that of original device such scaling is termed constant-field scaling. Ø Such scaling is also called full scaling, as the geometric dimensions and supply voltages are scaled simultaneously. Constant Voltage Scaling Ø The geometrical dimensions of the MOSFET are only scaled by the scaling factor S while the supply and terminal voltage are kept constant. Hence such scaling is termed as constant voltage scaling. ØSuch scaling is also called partial scaling, as the scaling is applied only to geometric dimensions and not supply voltages. 3/4/2015 5

The impact of constant-field scaling and constant voltage scaling on the physical parameters of the MOSFET is shown in the table below: summarized in the. Before table below: Parameters Scaling After Scaling(constant Scaling(Constant Field) Voltage) Channel Width W W/S Channel Length L L/S WL WL/S 2 Oxide Thickness Tox / S Oxide Capacitance Cox S Cox Threshold Voltage Vt Vt Supply Voltage VDD / S VDD Gate Voltage Drain Voltage VGS / S VGS VDS / S VDS NA Area Doping Density 3/4/2015 Vt S 2 NA 6

COMPARISON OF THE MOSFET AND THE BJT 3/4/2015 7

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IC BIASING - MOSFET CURRENT SOURCES Biasing in integrated circuit design is done using constant current sources. BASIC MOSFET CURRENT SOURCE Figure shows the circuit of a simple MOS constant-current source. The transistor Q 1 has the drain shorted to its gate, thereby forcing it to operate in the saturation mode. Neglecting channel length modulation, we get ID 1 = ½ kn’ (W/L)1 (VGS – Vtn)2 ……………. . (1) 3/4/2015 11

The drain current of Q 1 is supplied by VDD through resistor R. Since the gate currents are zero, ID 1 = IREF = VDD – VGS …………. (2) R Where IREF is the reference current. i. e the current through the resistor R Equations (1) and (2) can be used to determine the value of R For transistor Q 2, VGS is the same as that of transistor Q 1 Therefore transistor Q 2 is also in saturation. From figure, neglecting the channel length modulation we get, ID 2 = I 0 = ½ kn’ (W/L)2 (VGS – Vtn)2 ……………. (3) Where I 0 is the output current of the current source 3/4/2015 12

From equation (1) and (3) we get the relation between the reference current and the output current from the current source. I 0 = (W/L)2 IREF (W/L)1 ………………. . (4) Where (W/L)1 and (W/L)2 are the aspect ratios of transistor 1 & 2 respectively. For identical transistors, I 0 = IREF i. e the circuit replicates or mirrors the reference current at the output terminal. Hence the name current mirror. The current gain or the current transfer ratio is give by equation (4).

Effect of V 0 on I 0 For finding the effect of V 0 on I 0, we assume that the transistor Q 2 is in saturation. For transistor Q 2 to be in saturation, the circuit to which the drain voltage Q 2 is to Be connected must establish a drain voltage V 0 , that satisfies the relationship V 0 ≥ VGS – Vt or V 0 ≥ Vov Where, Vov is the overdrive voltage 3/4/2015 14

Figure shows the output characteristic of the current mirror circuit show in figure above , considering the effect due to channel modulation also. The current mirror has a finite output resistance, R 0 = ∆ V 0 = r 02 = VA 2 ∆ I 0 Where VA 2 is the early voltage of Q 2 When finite output resistance is also considered, we can express the output current I 0 as I 0 = IREF (W/L)2 1 + V 0 -VGS (W/L)1 VA 2

MOS Current Steering Circuits Ø On an IC chip with a number of amplifier stages, a constant dc current called a reference current, is generated at one location and is then replicated at various other locations for biasing the various amplifier stages through a process known as current steering. ØAdvantages 1. The effort expended on generating a predictable and stable reference current, usually utilizing a precision resistor external to the chip, need not be repeated for every amplifier stage. 2. Furthermore, the bias currents of the various stages track each other in case of changes in power-supply voltage or in temperature. 3/4/2015 16

Figure shows a simple current steering circuit. Transistors Q 1, Q 2 and Q 3 form a two output current mirror. So, I 2 = IREF (W/L)2 (W/L)1 and , I 3 = IREF (W/L)3 (W/L)1 For transistors Q 2 and Q 3 to be in saturation, Or, 3/4/2015 VD 2, VD 3 ≥ -Vss + VGS 1 – Vtn VD 2, VD 3 ≥ -Vss + VOV 1 17

From the figure, I 3 = I 4 I 5 = I 4 (W/L)5 (W/L)4 To keep Q 5 in saturation, VD 5 ≤ VDD - |VOV 5| NOTE: From figure its seen that , transistor Q 2 pulls its current I 2 from a load (not shown in figure) , so transistor Q 2 acts as current sink. But, transistor Q 5 pushes current into a load (not shown in figure) , so transistor Q 5 acts as current source.

BJT Current Mirror ØThe basic BJT current mirror is shown in Figure. It works in the similar manner to that of the MOS mirror. ØThe two important differences: 1. The nonzero base current of the BJT (i. e finite β) causes an error in the current transfer ratio of the bipolar mirror. 2. The current transfer ratio is determined by the relative areas of the emitter-base junctions of Q 1, and Q 2.

The current transfer ratio is given by I 0 = Is 2 = Area of EBJ Of Q 2 IREF Is 1 Area of EBJ of Q 1 where Is 1 and Is 2 are the scale currents of transistor Q 1 and Q 2 respectively. Effect of finite β on the current transfer ratio Transistor Q 1 and Q 2 are matched and have the same VBE , so their collector currents will be equal. From the figure, node equation at the collector of Q 1 gives IREF = Ic + 2 Ic / β = Ic 1 +

Since I 0 = Ic , the current transfer ratio is In general, if Is 2 = m Is 1, then the actual current transfer ratio is given by Finite output resistance of BJT current mirror The current mirror has a finite output resistance, R 0 = ∆ V 0 = r 02 = VA 2 ∆ I 0 Taking both finite β and finite R 0 into account , we get the current transfer ratio as

BJT CURRENT SOURCE The basic BJT current mirror can be used to implement a simple current source. From the figure , and The output resistance of this current source is r 0 of Q 2. R 0 = r 02 = VA I 0 = VA IREF

BJT CURRENT STEERING CIRCUITS The diode-connected transistor Q 1, resistor R, and the diode-connected transistor Q 2, generate the required IREF. From the figure it can be seen that , to generate a dc current twice the value of IREF, two transistors Q 5 and Q 6 each of which is matched to Q 1 are connected in parallel. So, I 3 = 2 IREF Similarly , to generate a dc current three times the value of IREF, three transistors Q 7, Q 8 and Q 9, Each of which is matched to Q 2 are connected in parallel. So, I 4 = 3 IREF

HIGH FREQUENCY RESPONSE OF AMPLIFIERS ØThe frequency response of these direct-coupled or DC amplifiers is shown in figure. It is clear that the gain remains constant at its midband value AM down to zero frequency (DC). Compared to the capacitively coupled amplifiers that use bypass capacitors, direct-coupled IC amplifiers do not suffer gain reduction at low frequencies. ØHowever, gain falls off at the high-frequency end due to the internal capacitances of the transistor. These capacitances represent the charge storage phenomena that take place inside the transistors and are included in the highfrequency device models.

The High Frequency Gain Function The amplifier gain considering the effect of internal capacitances can be expressed as A(s) = AM FH(s) where AM is the midband gain which is the low frequency or dc gain found by neglecting the effect of transistor’s internal capacitances. where, ωp 1, ωp 2, …, ωpn are positive numbers indicating the frequencies of n real poles. ωz 1, ωz 2, …, ωzn are positive numbers indicating the frequencies of n real zeros.

The 3 d. B frequency , f. H ØThe region of the high-frequency band which is close to the midband, is the region of interest for an amplifier designer. The designer needs to estimate—and if need be modify—the value of the upper 3 -d. B frequency f. H or ωH = 2πf. H. Ø It should be known that in many cases the zeros are either at infinity or such high frequencies as to be of little significance to the determination of ωH. If one of the poles, say ωP 1, is of much lower frequency than any of the other poles, then this pole will have the greatest effect on the value of the amplifier ωH. It means that, this pole referred as dominant pole will dominate the highfrequency response of the amplifier. In those cases the function FH(s) can be approximated by

So, if a dominant pole exists, then the value of ωH can be written as ØIf a dominant pole does not exist, then ωH can be derived as follows: Consider the case of a circuit having two zeros and two poles in the high frequency band. So, Substituting s=jω, and considering the square of magnitude , we get

By definition at ω = ωH , |FH|2 = ½ So, we get By solving for ωH we get , This relationship can be extended to any number of poles and zeros as

Open Circuit Time Constants Another method for finding the value of f. H is open circuit time constant. Let we expand the FH(s) expression given below: We get, The coefficients b and a are related to the frequencies of poles and zeros respectively.

The coefficient b 1 can be written as The value of b 1 is obtained by considering the various capacitances in the high-frequency equivalent circuit one at a time while all other capacitors are set to zero (replacing them with open circuits)and then adding the individual time constants. Hence the name open circuit time constant. i. e If the zeros are not dominant and if one of the poles, say P , is dominant, then we may write, Therefore,

MILLER’S THEOREM Consider that impedance Z is connected between two nodes 1 and 2, as shown in figure. Nodes 1 and 2 are connected to other parts of the circuit also. It is assumed that voltage at node 2 is known and it is V 2 = KV 1 , where K is a gain factor Miller's theorem states that impedance Z can be equivalently replaced by two impedances: Z 1 connected between node 1 and ground and Z 2 connected between node 2 and ground, where