Bode Plot Review High Frequency BJT Model 1
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
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
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
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
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
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
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 10 1 8
Simple Example Working with logs: H j =10 j 1 1 2 1 1 j =10 10 1 2 10 1 H db =20 log 10 10 20 log 10 j 1 1 20 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
Simple Example H db =20 log 10 K dc 20 log 10 j 1 1 20 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 10 1 10 H db =20log 10 K dc 20log 10 1 20log 10 10 10
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=bodeplot-freqresp; plot(f,err); 11
Scilab Results Bode plot Actual freq. response Error 12
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
Bode Plot Used to Estimate Zeros & Poles 14
Bode Plot Superposition Directly from the Bode Plot! H j = j 20 1 j 100 1 j 500 1 20 Hz 100 Hz 500 Hz 15
Gain of 10 Amplifier Non-ideal Transistor Gain starts dropping at about 1MHz. Why! Because of internal transistor capacitances that we have ignored in our models. 16
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
High Frequency Small-signal 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
High Frequency Small-signal 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
Frequency-dependent beta h fe short-circuit 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 analysis): @ node B': = I 1 b sc r sc V where r x 0 (ignore r x ) 20
Frequency-dependent h fe or beta The ratio of the two equations: = I 1 b sc r sc V @ 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
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 1 1 10 and: low C C r 1 1 10 Note: low C C r = low C C g m low C g m We have: h fe =g m r = 22
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
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 frequency-dependent 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
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 unity-gain frequency or GBP 25
Frequency Response of h fe cont. =100 r =2500 C =12 pf C =2 pf g m =40 10 3 S = 1 = 1012 10 3 C C r 122 2.5 =28.57 106 rps f = 2 = 28.57 6.28 106 Hz=4.55 MHz f T = f =455 MHz z = g m = 40 10 3 10 12 C 2 Hz=20 10 9 rps f z = z 2 =3.18 109 Hz=3180 MHz 26
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
h fe Bode Plot (db) f T 28
Multisim Simulation v-pi I c I b ms v-pi 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
Simulation Results Low frequency h fe Unity Gain frequency about 440 MHz. 30