ESE319 Introduction to Microelectronics Bode Plot Review High Frequency BJT Model


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1 Bode Plot Review High Frequency BJT Model 1
2 Logarithmic Frequency Response Plots (Bode Plots) Generic form of frequency response rational polynomial, where we substitute jω for s: H s=k sm a m 1 s m 1 a 1 sa 0 s n b n 1 s n 1 b 1 sb 0 HH(jω) j can represent an impedance, an admittance, or a gain transfer function. If we are lucky, we can factor each of the polynomials as a product of first degree terms: H s=k sz 1sz 2 sz m s p 1 s p 2 s p n 2
3 Logarithmic Frequency Response Plots (Bode Plots) Determination of a frequency response requires evaluating the complex number expression: H j =K j z 1 j z 2 j z m j p 1 j p 2 j p n Bode's approach was to simplify the calculations, using polar representation of the factors: H j =K z j 1 e j 1 j 1 e j 2 j 1 e j m 1 z 2.. z m z 1 z 2 z m p 1 p 2.. p n j p 1 1 e j 1 j p 2 1 e j 2 j p n 1 e j n where j 1 = z [ j 1 j k z k z k 1]= 2 11 z k k =tan 1 z k 3
4 Bode Plot cont. z k j z k 1 1 j 1 = z [ j 1 j k z k z k 1]= 2 1 z k =z k j z k 1 =2 z k j z k 1 z k k =tan 1 z k z k tan 1 z k 0 o =z k tan 1 z k =45 o zk tan 1 z k 90 o 4
5 Bode Plots cont. Ge j =K j z 1 e j 1 j z 2 e j 2 j z m e j m j p 1 e j 1 j p2 e j 2 j pn e j n We can separate magnitude and phase angle calculations: G = H j =K = 1 2 m 1 2 n m 1 n 1 z i p i j z 1 1 j z 2 1 j z m 1 j p 1 1 j p 2 1 j p n 1 where k =tan 1 z k k =tan 1 p k 5
6 G= H j =K dc j Bode Plots K dc =K z 1 1 j z 2 1 j z m 1 j p 1 1 j p 2 1 j p n 1 Where we define the dc gain, the value of the magnitude of H at ω =0= 0, as: m 1 n 1 z i p i 6
7 Logarithmic Frequency Response Plots (Bode Plots) G= H j =K dc j z 1 1 j z 2 1 j z m 1 j p 1 1 j p 2 1 j p n 1 Another simplification converts the magnitude computation from multiplication to addition by working with its logarithm (in decibels): m G = db 1 20 log 10 j n 1 z i 1 20 log 10 j p i 1 7
8 Logarithmic Frequency Response Plots (Bode Plots) m G = db 1 20 log 10 j n 1 z i 1 20 log 10 j p i 1 Let's calculate the frequency response for a simple transfer function and make some observations: H j =10 j 1 1 j
9 Simple Example Working with logs: H j =10 j j = H db =20 log log 10 j log 10 j 10 1 Note: 1. if the coefficient of j, i.e. 0.1, then j 1 1 and log 10 j 1 =0. a a a 2. If the coefficient of j, i.e. 10, then j 1 and log 10 j 1 =log. a a a 10 a a 3. When = a, j 1 =2 and log 10 j 1 =0.15. a a 9
10 Simple Example H db =20 log 10 K dc 20 log 10 j log j 10 1 Applying the approximation on the previous slide: 1 H db =20 log 10 K dc 110 H db =20 log 10 K dc 20 log H db =20log 10 K dc 20log log
11 Scilab Simulation Kdc=10; //Example Bode Plot KdB= 20*log10(Kdc); omegaz=1; fz=omegaz/(2*%pi); omegap=10; fp=omegap/(2*%pi); f=0.01:0.01:100; magnum=sqrt((f/fz)^2 + 1); magden=sqrt((f/fp)^2 + 1); FreqResp=KdB+20*(log10(magnum)log10(magden)); plot(f,freqresp) term1=kdb*sign(f); //Create constant array of length len(f) term2=max(0,20*log10(f/fz)); //Zero for f < fz; term3=min(0,20*log10(f/fp)); //Zero for f < fp; BodePlot=term1+term2+term3; plot(f,bodeplot); err=bodeplotfreqresp; plot(f,err); 11
12 Scilab Results Bode plot Actual freq. response Error 12
13 Observations The / ωω ratio changes by 20 db for each order of magnitude x a change in frequency (20log 10 (10) = 20). Our rule of 10 scheme for using either 1, or / ωω for x a magnitude estimation is quite accurate. This why the Bode plot is called an asymptotic plot. We can plot a system transfer function, then position straight line segments of ±x 20 db/decade on the Bode plot. The intersection of the lines occurs at the break frequencies. 13
14 Bode Plot Used to Estimate Zeros & Poles 14
15 Bode Plot Superposition Directly from the Bode Plot! H j = j 20 1 j j Hz 100 Hz 500 Hz 15
16 Gain of 10 Amplifier Nonideal Transistor Gain starts dropping at about 1MHz. Why! Because of internal transistor capacitances that we have ignored in our models. 16
17 Sketch of Typical Voltage Gain Response for a CE Amplifier A v db Low Frequency Band Due to external blocking and bypass capacitors Midband ALL capacitances are neglected 20 log 10 A v db 3 db High Frequency Band Due to BJT parasitic capacitors C π and C µ f L BW = f H f L f H f H GBP= A v BW f Hz (log scale) 17
18 High Frequency Smallsignal Model r x C C Two capacitors and a resistor added. A base to emitter capacitor, C π A base to collector capacitor, C µ A resistor, r x, representing the base terminal resistance (r x << r π ) C = C 0 1 V CB V 0c m C =C de C je 0 1 V BE V 0 e 18 m
19 High Frequency Smallsignal Model The internal capacitors on the transistor have a strong effect on circuit high frequency performance! They attenuate base signals, decreasing v be since their reactance approaches zero (short circuit) as frequency increases. As we will see later C µ is the principal cause of this gain loss at high frequencies. At the base C µ looks like a capacitor of value k C µ connected between base and emitter, where k > 1 and may be >> 1. This phenomenon is called the Miller Effect. 19
20 Frequencydependent beta h fe shortcircuit current The relationship i c = βi b does not apply at high frequencies f > f H! Using the relationship i c = f(v π ) find the new relationship between i b and i c. For i b (using phasor notation (I x & V x ) for frequency domain node B': = I 1 b sc r sc V where r x 0 (ignore r x ) 20
21 Frequencydependent h fe or beta The ratio of the two equations: = I 1 b sc r sc node C: I c = g m sc V (ignore r o ) Leads to a new relationship between the I b and I c : h fe = I c I b = g m sc 1 s C r s C 21
22 Frequency Response of h fe h fe = g m sc 1 s C r s C multiply N&D by r π h fe = g j C r m 1 j C C r factor N to isolate g m 1 j C g g m r h fe = m 1 j C C r g m = I C V T For small ω s r = V T I C = low : low C g m and: low C C r Note: low C C r = low C C g m low C g m We have: h fe =g m r = 22
23 Frequency Response of h fe cont. 1 j C g g m r h fe = m = 1 j 1 j C C r z 1 j g mr 1 j = 1 f f z j f f C C r =C C g m C g m => f z f h fe db 20log 10 f f z f Hence, the lower break frequency or 3dB frequency is f β f = 1 2C C r = g m 2C C the upper: f z = 1 2C / g m = g m 2C where f z 10 f 23
24 Frequency Response of h fe cont. Using Bode plot concepts, for the range where: h fe =g m r = For the range where: f f f z s.t. 1 j f / f z 1 We consider the frequencydependent numerator term to be 1 and focus on the response of the denominator: h fe = g m r f 1 j f = f 1 j f 24
25 Frequency Response of h fe cont. Neglecting numerator term: h fe = g m r f 1 j f = f 1 j f And for f / f >>1 (but < f / f z ): h fe f = f f f Unity gain bandwidth: h fe =1 f f =1 f T = f f T = T 2 = f BJT unitygain frequency or GBP 25
26 Frequency Response of h fe cont. =100 r =2500 C =12 pf C =2 pf g m = S = 1 = C C r = rps f = 2 = Hz=4.55 MHz f T = f =455 MHz z = g m = C 2 Hz= rps f z = z 2 = Hz=3180 MHz 26
27 Scilab f T Plot //ft Bode Plot Beta=100; KdB= 20*log10(Beta); fz=3180; fp=4.55; f= 1:1:10000; term1=kdb*sign(f); //Constant array of len(f) term2=max(0,20*log10(f/fz)); //Zero for f < fz; term3=min(0,20*log10(f/fp)); //Zero for f < fp; BodePlot=term1+term2+term3; plot(f,bodeplot); 27
28 h fe Bode Plot (db) f T 28
29 Multisim Simulation vpi I c I b ms vpi Insert 1 ohm resistors we want to measure a current ratio. h fe = I c = g m s C I b 1 sc r C 29
30 Simulation Results Low frequency h fe Unity Gain frequency about 440 MHz. 30
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