IC Biasing-Current Sources, Current Mirrors and Current Steering Circuits | EGR 392, Study notes of Engineering

Download IC Biasing-Current Sources, Current Mirrors and Current Steering Circuits | EGR 392 and more Study notes Engineering in PDF only on Docsity! 1 18 6.3 IC Biasing –Current Sources, Current Mirrors and Current-Steering Circuits 6.3.3 BJT Circuits Figure 6.8 The basic BJT current mirror. If Q1 and Q2 are matched, that is, they have the same EBJ area: If the area of the EBJ of Q2 is m times that of Q1, REFO mII = 12 SS mII = tBE VV SC eII /= REFO II = Note: Here, m is the current transfer ratio. In general, it is given by 1 2 1 2 of EBJ of Area of EBJ of Area Q Q I I I I S S REF O == 19 6.3 IC Biasing –Current Sources, Current Mirrors and Current-Steering Circuits 6.3.3 BJT Circuits Figure 6.9 Analysis of the current mirror taking into account the finite β of the BJTs. The effect of finite transistor β on the current transfer ratio ) 2 1(/2 β β +=+= CCCREF IIII CO II = ββ /21 1 )/21( + = + = C C REF O I I I I 1, →∞→ REF O I I β ratioansfer current tr in theoccur error will 2% a ,100 =βIf 2 20 If the current transfer ratio is m, β 1 1 + + = m m I I REF O O A O O O O I V r I V R 22 == ∆ ∆ ≡ )1)( 1 1 ( 2A BEO REFO V VV m m II − + + + = β The finite output-resistance for BJT mirror: Taking both the finite β and the finite RO, the output current of a BJT mirror with a nominal current transfer ratio m: 21
Example Consider a BJT current mirror with a nominal current transfer ratio of unity. Let the transistor have IS = 10 -15A, β = 100, and VA = 100V. For IREF = 1mA, find IO when VO = 5V. Also, find the output resistance. 4 24 Background: A Graphical Interpretation of Poles and Zeros Transfer function: H(s) = 1/(s-1) Transfer function: H(s) = s-1 Ref: http://www.chem.mtu.edu/~tbco/cm416/PolesAndZeros.html 25 6.4: High-frequency response – general considerations 6.4.1 The High-frequency Gain Function )()( sFAsA HM= With the internal transistor capacitances, the general form of the amplifier gain is: AM is the midband gain, equal to low-frequency or DC gain. )/1)...(/1)(/1( )/1)...(/1)(/1( )( 21 21 pnpp znzz H sss sss sF ωωω ωωω +++ +++ = Here are positive numbers representing the frequencies Of the n real poles and are positive, negative, or infinite numbers representing the frequencies of the n transmission zeros. ,,...,, 21 pnpp ωωω ,,...,, 21 znzz ωωω Note: S0, FH(s)1, A(s) = AM 5 26 6.4: High-frequency response – general considerations 6.4.2 Determining the 3-dB Frequency fH 1/1 1 )( P H s sF ω+ ≅ Dominant-pole response: This is the transfer function of a low pass network with .1PH ωω ≅ As a rule of thumb, a dominant pole exists if the lowest-frequency pole is at Least two octaves (a factor of 4) away from the nearest pole or zero. ...) 11 (2...) 11 (/1 2 1 2 1 2 2 2 1 ++−++≅ ZZPP H ωωωω ω If there is no dominant pole existing, the 3-dB frequency ωH can be get by: 27 Example 6.5 The high-frequency response of an amplifier is characterized by the transfer function Determine the 3-dB frequency approximately and exactly. )104/1)(10/1( 10/1 )( 44 5 ×++ − = ss s sFH 6 28 Solution The lowest frequency pole at 104 rad/s is two octaves lower than the second pole and a decade lower than the zero, so it is the dominant pole. srad H / 9800 )10( 2 )104( 1 )10( 1 /1 252424 = − × +=ω 29 Figure 6.13 Normalized high-frequency response of the amplifier in Example 6.5. 9 34 (c) circuit for determining the resistance seen by Cgs; The 3-dB frequency can be determined by the method of open-circuit time constant. =gsR Ω=ΩΩ= kkk RR sigin 8.80100||420 || The open circuit time constant of Cgs is ns RC gsgsgs 8.80108.80101 312 =×××= = − τ The resistance Rgs seen by Cgs is 35 (d) circuit for determining the resistance seen by Cgd. The resistance Rgd seen by Cgd is by setting Cgs = 0 and short-circuit Vsig. A test current IX is applied in the circuit shown below. in gs sig gs x R V R V I −−= L xgs gsmx R VV VgI ′ + += At node G: At node D: RIV xgs ′−= sigin RRR ||=′ L xx xmx R VRI RIgI ′ +′− +′−= )( )( L x L mx R V R R RgI ′ = ′ ′ +′+ )1( 10 36 Ω=′′+′+′== MRRgRR I V R LmL x x gd 16.1 The resistance Rgd seen by Cgd is The open circuit time constant of Cgd is ns RC gdgdgd 11601016.1101 612 =×××= = − τ The upper 3-dB frequency ωH can now be determined from ≅Hω gdgs ττ + 1 skrad /806 10)11608.80( 1 9 = ×+ = − kHzf HH 3.128 2 == π ω