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EE 4345 – Semiconductor Electronics Design Project Transistor Current Sources and Active Loads Anuj Shah Himanshu Doshi Jayaprakash Chintamaneni Nareen Katta Nikhil Patel Preeti Yadav Current Sources Current sources made by using active devices have come to be widely used in analog integrated circuits as Biasing elements and as load devices for amplifier stages The use of current sources in biasing can result in superior insensitivity of circuit performance to power supply variations and to temperature A simple 2-transistor current source Simple current source Since Q1 and Q2 have the same base-emitter voltage. Their collector currents are equal, Ic1=Ic2 Summing currents at the collector of Q1 gives, I ref I c1 2 I ref I c1 Ic2 2 1 I c1 f 0 I c 2 I ref Vcc Vbe (on) R f Thus, For identical devices Q1 and Q2, the output and the reference Currents are equal Collector characteristics for an npn transistor Norton equivalent representation Of a transistor current source Thevenin equivalent Representation of a Transistor current source Simple current source with current gain The collector current Ic2 is different form the reference Current by a factor [ 1 + ( 2/ßf )] The emitter current of the transistor Q3 is given by, I c1 f Ic2 f 2I c 2 f Summing currents at the collector of Q1, I ref 2I c 2 I c1 0 f ( f 1) Since Ic1 and Ic2 are equal, I ref Io Ic2 2 1 2 f f Thus, the reference and the output current differ only by a factor of 1/ßf2 Widlar current source Widlar current source Assuming that the transistors are identical and that Vce2 is high enough to allow Q2 to operate in the active region,The collector current of Q2 is given by, Vbe2 I c 2 I s exp VT Vbe1 I C1 VT ln IS Vbe2 Ic2 VT ln IS Voltage equation around the base-emitter loop gives, Vbe1 Vbe2 R2 I e 2 Solving for R2 gives I C1 VT R2 ln Ic2 IC 2 Neglecting the base currents, I c1 I ref Vcc Vbe1 R1 Design a widlar current source , given Vcc=15volts. Assume identical transistors.Also, design a 10uA current mirror and compare total resistance required in the two circuits. Let Iref =1mA 15v 0.6v 14.4k 1mA I c1 R2 VT ln I c2 1mA 10A 12k 0.025 ln Design of current mirror, Vcc Vbe R I ref 15v 0.5v 1.45M 10A Cascode current source with bipolar transistors Performing a full small-signal analysis on this circuit, including the finite small-signal resistance of Q1 and Q4 We obtain, Ro =ß0 r0/2 Bipolar Wilson current source Conventional Differential Amplifier Adm I c Rc g m Rc VT Resistors are used as load elements Large Voltage Gain – requires large power-supply voltage and large values of resistors Example: and A voltage gain of 500 would require I c Rc 13V I c 100A, Rc would have to be 130K Current Sources as Active Load Common-Emitter Amplifier with Active Load Emitter-Coupled Pair with Active Load Input Offset Voltage of the Emitter-Coupled Pair with Active Load Common-Mode Rejection Ratio of the Emitter-Coupled Pair with Active Load Common-Source Amplifier with Active Load Fig.Common-emitter amplifier with active load I c1 I c 2 Vce 2 Vcc Vce1 I c1 I c 2 [( Vcc Vce1 ), VBE 2 ] Fig.npn collector characteristic with pnp Load line superimposed Initially at Vi=0 at point 1 As Vi Increases at point 2 at point 3 at point 4 Transistors NPN PNP Off Saturates On On On On Saturated Saturated Large – Signal Model I c1 Ic2 I ref V 1 ce1 VAN Vbe 2 1 I s 2 exp VT V I s1 exp i VT VBE3 I c3 I s 2 exp VT Vce1 V0 Vce 2 Vcc V0 Vce 2 VAP VBE ( on) 1 VAP Fig.dc transfer characteristics Of common-emitter Amplifier with active load V0 [Vcc VBE ( on) ]( Where, VA( eff ) VAN ) VA( eff ) [1 VAN VAP I s1 exp I ref Vi VT ] VANVAP & [V V CC CE ( sat ) ] V0 [VCE ( sat ) ] ( ) VAN VAP VAN V0 [VCC VBE (on) ]( ) VAN VAP Fig.Small-signal equivalent Circuit for common-emitter Amplifier with active load g m1 g m1 Av g m1 (ro1 || ro 2 ) 1 1 npn g m1 pnp g m 2 ro1 r02 R0=(ronop+ropon) Typical values for voltage gain are 1000 to 2000 range. Fig.Emitter-coupled pair with Active load Drawback: The quiescent value of the common-mode output voltage is very sensitive to the value of the emitter-biasing current source and the active-load current sources. Fig.Common-mode half-circuit for emitter-coupled pair with active load Output voltage Voc is very sensitive to the voltage at the base of Q6, which is influence by Iref2. Example: If Iref1 and Iref2 are different by 4% , the output voltage Voc will change by about 2V. The same change result from a 1mV mismatch between Q6 And Q7. This circuit eliminates the common-mode problems but provides a single output with much better rejection of common-mode Input signals. Large signal behavior model Fig.Active-load stage with output connected to a voltage source Load resistance is zero. Vbias is adjusted in order to keep Q2 and Q4 in forward Active region. Ic2 Vbe2 Vce 2 I s (exp )(1 ) VT VAN Vbe3 Vce3 I c 3 I s (exp )(1 ) VT VAP Vbe4 Vce 4 I c 4 I s (exp )(1 ) VT VAP Vce1 Vcc Vbe3 Vbe1 Vcc Vce 3 VBE ( on) Vid Vbe1 Vbe2 Vid 2VA( eff ) tanh 2VT Vo Vcc VBE (on) V V V 1 AN AP tanh id 2V V V AN AP T where VANVAP VA(eff ) VAN VAP Vid Vo Vcc VBE (on) 2VA( eff ) tanh 2VT Vcc Vo VBE ( on) VCE ( sat) Avd VA( eff ) VT 1 VT VANVAP 1 VAN VAP VT VT VAN VAP 1 Avd g m (ronpnllropnp ) npn pnp V T EarlyFacto r ,Where VA Early factor is on the order of 2 * 10-4 Device Mismatch Effects: Presence of Component mismatches within the Amplifier itself and drifts of component values with Temperature produce differential voltages at the output that are indistinguishable from the signal being amplified. For Transistor Differential Amplifiers the effect of mismatches is represented by two Quantities,the input offset voltage and the input offset current. Circuit Containing Mismatches Equivalent Circuit with identically matched devices Input Offset Voltage is given by the expression VOS where VT I c1 I s 2 VT I c 2 I s1 kT 26mV q at 300K Is represents the Saturation Current 2 qni Dn I sn An N AWBn VCB where Wb(VCB) is the base width of the function VCB NA is the acceptor density in the base and A is the emitter Area Input Offset Voltage of the Emitter Coupled Pair with Active Load For a Resistive Load the Input Offset Voltage arises mainly due to the mismatches in Is in the Input Transistors and from mismatches in the collector load Resistors. For an Active Load the input offset voltage results from mismatches in the input Transistors and load devices and from the base current of the load devices. Emitter Coupled Pair with Active Load Input Offset Voltage of the Emitter Coupled Pair with Active Load Vce3 Vce 4 Vce1 Vce 2 and The Collector Current Ic4 is related to Ic3 by Is4 I c 4 I c 3 I s3 The Collector Current Q2 is equal to (-Ic4) and thus Is4 I c 2 I c3 I s3 The Current Ic1 equals (-Ic3),plus the base currents in the pnp transistors 2 I c1 I c 3 1 f The input offset voltage is given by I I 2 Voc VT ln s 3 s 2 1 I s 4 I s1 f In a Worst Case, the Offset Voltage is 0.2VT These Circuits have significantly higher offset Voltage than the Resistively loaded case The offset Voltage can be reduced by inserting resistors in the Emitters of Q3 and Q4. Actively Loaded Emitter Coupled Pair for improved Offset Common-Mode Rejection Ratio of the Emitter Coupled Pair with Active Load CMRR is defined as the Magnitude of the ratio of differential to Common Mode Gain. Provides conversion from a differential signal to a signal that is referenced to ground.This type of Conversion is required in all differential input,single-ended output amplifiers. Simplest differential to single-ended converter is the resistively Loaded emitter coupled pair. Differential to Single ended Conversion Using resistively loaded Emitter coupled Pair Differential to Single ended Conversion Using actively loaded Emitter coupled Pair The output is given by Vo Adm 2Vic Vid 2 CMRR 1 CMRR 1 2 g m REE 1 Transistor Q3 operates at twice the current of Q1 and Q2 CMRR g m 3 ro 3 1 3 CMRR of the resistively loaded stage is the inverse of the of the current source Transistor when a simple current source is used as a biasing element. For an active-loaded case In case of Resistively loaded case ,changes in the common Mode input will cause changes in the bias current IEE because of the finite output resistance of the biasing current source.This will cause a change in Ic2 and an identical change in Ic1. Because of the active load ,the change in Ic1 will produce a change in the currents flowing in the pnp load transistors ,which produces a change in the collecter current of Q2.So the output does not change at all in response to common Mode Inputs. Analytically CMRR Adm Vos Acm Vic 1 Common-Source Amplifier with Active Load Common Source Amplifier using an NMOS driver and PMOS active load. When M1 and M2 are forward active,the small signal gain is g m1 Av g o1 g o 2 Gain for NMOS depletion-load transistor g m1 Av g mb 2 Common-Source Amplifier with p channel transistor Current source load Transfer Characteristic I-V Characteristic of n-channel depletion mode load transistor Common Source Amplifier with depletion mode transistor load and dc transfer characteristic Common Source Amplifier with Enhancement-Mode load Common-Source Amplifier with enhancement-mode load and transfer characteristic Av W / L1 g m1 W / L2 gm2 Source Coupled Pair with Active load Av g m1 go2 go4 Widely used in CMOS Circuit Design References: Analysis and Design of Analog Integrated Circuits, 3rd Edition by Paul R.Gray and Robert G.Meyer. Analog integrated circuit design,1st edition by David Johns and Ken Martin Electronics,2nd Edition, by Allan R.Hambley