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Berkeley ELENG 105 - Discussion Section 101

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EE 105 --- Spring 2005 --- Discussion Notes (written by Amin) 1 Monday, April 18, 2005 EE 105 Discussion Section 101 Announcements High Voltage Gain Amplifier Design Suppose that it is desired to design an amplifier with large voltage gain using a single stage. Based upon our knowledge of single-stage amplifiers, one potential candidate is the common-source amplifier. Let’s examine the maximum attainable voltage gain from such a stage. Basic common-source amplifier Figure 1 shows the circuit diagram of a PMOS common-source stage. The voltage gain of such a stage can be shown to equal the following, 11 ominoutrgvv⋅−= (1) In order to obtain a feel for how high the expression in (1) can be in value, let’s use appropriate equations for 1mg and 1or . More specifically, substituting 1112DSATSDmVIg⋅= and 1111SDoIr⋅=λ in (1) yields the following 111111212λλ⋅−=⋅⋅⋅−=DSATSDDSATSDinoutVIVIvv (2) Substituting typical values for 1DSATV (200mV) and 1λ (0.05V-1) into (2), yields a voltage gain of 200 in magnitude. In order to increase the voltage gain of a common-source stage, one has two options from (2): 1. Reduce 1DSATV , and/or 2. Reduce 1λ Unfortunately, one can not reduce 1DSATV endlessly because the MOSFET transistor enters the subthreshold region of operation (see EE 130) for DSATV values below approximately 150mV. This is why 100-150mV is often set as the lower bound for DSATV in when it comes to amplifier design in EE105/140/240. Hence, to increase the voltage M1 vout vin VDD Figure 1. PMOS CS with ideal current sourceEE 105 --- Spring 2005 --- Discussion Notes (written by Amin) 2 gain of a common-source stage beyond 200 in magnitude, one can only reduce λ in practice. In order to reduce λ of a transistor, one needs to increase the length. In order to not upset the corresponding DSATV of the resulting transistor, one needs to increase the width by the same proportion. As an example, to increase the voltage gain of a common-source stage from 200 to 40E3, one would need to increase the length (and correspondingly width) by a factor of 200. Increasing the size of a transistor has two disadvantages: (i) eats up silicon area, (ii) degrades the frequency response because the intrinsic capacitances associated with the MOSFET are larger. The latter is of much more concern. Hence, a different topology needs to be employed if voltage gain larger than 200 in magnitude is desired. Note: In practice, one does not have access to ideal current sources. As a result, the ideal current source depicted in Figure 1 would be implemented using a current mirror in practice. Subsequent to this, the voltage gain expression in (1) would need to be modified. The result is the following )//(311 oominoutrrgvv⋅−= (3) where, 3or is the resistance “looking” down into the current mirror-based current source. Cascode amplifier In order to arrive at a topology that yields large voltage gain, one needs to step back and recall the big picture of how voltage gain is set. As mentioned on previous occasions, the voltage gain of an unloaded amplifier is given by the following expression outmvRGgainvoltageunloadedA⋅−== (4) From (4), one has two options to increase the voltage gain of an amplifier: 1. Increase mG , and/or 2. Increase outR Here, we will focus on the latter option, i.e. increasing the output resistance. The resulting amplifier is referred to as a cascode amplifier, a schematic of which is shown in Figure 2. Let’s next examine the voltage gain of such a structure by evaluating, in turn, its mG and outR . Gm of a cascode amplifier Figure 3 shows the circuit from which Gm for the cascode amplifier shall be computed. Note that a test voltage (i.e. tv ) has been applied to the gate and the output has been shorted to ground, in accordance with the rules for calculating Gm. Gm can be calculatedEE 105 --- Spring 2005 --- Discussion Notes (written by Amin) 3 as follows: 1. tv causes a current to be generated in M1 through the 1mg voltage-dependent current source. More specifically, the current generated has a value of tmvg ⋅−1. This current enters node β from the left, and divides between the right and down branches. One can use the current divider formula to see how the current tmvg ⋅−1 divides between the aforementioned branches. 2. Using the current divider formula, one obtains the following expression for the current flowing through the downward branch from node β, Rup Rout βiout )(111tmsgmvgvg−⋅=⋅1or Rright RdownFigure 2. PMOS cascode with ideal current source M2M2 VBIAS M1 vout vin VDD Figure 3. Calculating Gm M2M1Figure 4. Calculating RoutEE 105 --- Spring 2005 --- Discussion Notes (written by Amin) 4 downrightrighttmdownRRRvgi+⋅⋅−= )(1 (5) But, using the resistance formulas )(1)()(1,22222222221mbmombmoombmodownorightggrggrrggrRandrR+=⋅+≈⋅++== (6) Substituting the expressions in (6) into (5) yields )(1)(22111mbmootmdownggrrvgi++⋅⋅−= (7) Assuming that 1>>⋅omrg , (7) can be simplified as follows tmootmmbmootmdownvgrrvgggrrvgi ⋅−=⋅⋅−≈++⋅⋅−=111122111)()(1)( (8) 3. From Figure 3, downoutii −= . Thus, using (8) tmoutvgi ⋅≈1 (9) But, toutmviG = by definition. Hence, (9) yields the following 1mmgG ≈ (10) Note: The Gm of a cascode is approximately equal to the Gm of a basic common-source stage. Rout of a cascode amplifier Figure 4 shows the circuit from which Rout for the cascode amplifier shall be computed. Assuming that the current source is ideal, upoutRR=. Using the resistance formulas from Discussion 8, 212221)(oombmooupoutrrggrrrr⋅⋅+++== (11) Again assuming that 1>>⋅omrg , (11) can be simplified as follows 2122)(oombmupoutrrggRR ⋅⋅+≈= (12)EE 105 --- Spring 2005 --- Discussion Notes (written by Amin) 5 Note: The Rout of a cascode is greater than the Rout of a basic common-source stage by approximately a factor of 222)(ombmrgg⋅+ Voltage gain of a cascode amplifier Using (10) and (12) in (4), yields the following 21221)(oombmminoutrrgggvv⋅⋅+⋅−≈ (13) In order to obtain a feel for how high the expression in (13) can be in value, let’s use appropriate equations for 1mg , 2mg , 1or , and 2or (we will ignore 2mbg for now). Doing so yields, 21212211221141122λλλλ⋅⋅⋅−=⋅⋅⋅⋅⋅⋅⋅−≈DSATDSATSDSDDSATSDDSATSDinoutVVIIVIVIvv (14) Substituting


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Berkeley ELENG 105 - Discussion Section 101

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