ESE319 Introduction to Microelectronics Bode Plot Review High Frequency BJT Model 2008 Kenneth R. Laker, update 12Oct09 KRL 1 ESE319 Introduction to Microelectronics Logarithmic Frequency Response Plots (Bode Plots) Generic form of frequency response rational polynomial, where we substitute jω for s: m m−1 s am−1 s ⋯a1 sa0 H s=K n n−1 s bn−1 s ⋯b1 sb0 HH(j jω) 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: sz sz ⋯sz H s=K 1 2 m s p1s p2⋯s pn 2008 Kenneth R. Laker, update 12Oct09 KRL 2 ESE319 Introduction to Microelectronics Logarithmic Frequency Response Plots (Bode Plots) Determination of a frequency response requires evaluating the complex number expression: j z j z ⋯ j z H j =K 1 2 m j p1 j p2⋯ j pn Bode's approach was to simplify the calculations, using polar representation of the factors: j j j ∣ j 1∣e 1∣ j 1∣e 2 ⋯∣ j 1∣e m z ∗z ∗..∗z z z z H j=K 1 2 m 1 2 m p ∗p ∗..∗p j j j 1 2 n ∣ j 1∣e 1∣ j 1∣e 2 ⋯∣ j 1∣e n p1 p2 pn 2 1 where =tan− ∣ j 1∣= [ j 1− j 1]= 11 k z z z z z k k k k k 2008 Kenneth R. Laker, update 12Oct09 KRL 3 ESE319 Introduction to Microelectronics Bode Plot cont. ≪ zk ⇒∣j 1∣≈1 zk 2 ∣ j 1∣= [ j 1− j 1]= 1 =z k ⇒∣ j 1∣=2 zk zk zk zk z k ≫ zk ⇒∣ j 1∣≈ zk zk −1 o ≪ zk ⇒ tan 0 zk −1 −1 o k=tan =z k ⇒ tan =45 z k zk −1 o ≫ zk ⇒ tan 90 zk 2008 Kenneth R. Laker, update 12Oct09 KRL 4 ESE319 Introduction to Microelectronics Bode Plots cont. j j j ∣ j z ∣e 1∣ j z ∣e 2 ⋯∣ j z ∣e m Ge j =K 1 2 m j 1 j 2 j n ∣ j p1∣e ∣ j p2∣e ⋯∣ j pn∣e We can separate magnitude and phase angle calculations: m z ∣ j 1∣∣ j 1∣⋯∣ j 1∣ ∏ i z z z G =∣H j ∣=K 1 1 2 m n j 1 j 1 j 1 ∏ pi ∣ ∣∣ ∣⋯∣ ∣ 1 p1 p2 pn = 12⋯m − 12⋯n −1 −1 where k=tan k=tan zk pk 2008 Kenneth R. Laker, update 12Oct09 KRL 5 ESE319 Introduction to Microelectronics Bode Plots ∣ j 1∣∣ j 1∣⋯∣ j 1∣ z z z G=∣H j ∣=K 1 2 m dc ∣ j 1∣∣j 1∣⋯∣j 1∣ p1 p2 pn Where we define the dc gain, the value of the magnitude of H at ω == 00, as: m ∏ zi 1 K dc=K n ∏ pi 1 2008 Kenneth R. Laker, update 12Oct09 KRL 6 ESE319 Introduction to Microelectronics Logarithmic Frequency Response Plots (Bode Plots) ∣ j 1∣∣j 1∣⋯∣ j 1∣ z z z G=∣H j ∣=K 1 2 m dc ∣ j 1∣∣ j 1∣⋯∣j 1∣ p1 p2 pn Another “simplification” converts the magnitude computation from multiplication to addition by working with its logarithm (in decibels): m n GdB=∑ 20 log10∣ j 1∣−∑ 20 log10∣ j 1∣ 1 z i 1 pi 2008 Kenneth R. Laker, update 12Oct09 KRL 7 ESE319 Introduction to Microelectronics Logarithmic Frequency Response Plots (Bode Plots) m n GdB=∑ 20 log10∣ j 1∣−∑ 20 log10∣ j 1∣ 1 z i 1 pi Let's calculate the frequency response for a simple transfer function and make some observations: j 1 1 H j=10 j 1 10 2008 Kenneth R. Laker, update 12Oct09 KRL 8 ESE319 Introduction to Microelectronics Simple Example 2 j 1 1 1 1 H j =10 =10 2 j 1 10 1 Working with logs: 10 ∣H ∣=20log ∣10∣20log ∣ j 1∣−20log ∣ j 1∣ dB 10 10 1 10 10 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. , then and . ≥10 ∣ j 1∣≈ log10∣ j 1∣=log10 a a a a a 3. When , and . =a ∣j 1∣=2 log10∣ j 1∣=0.15 a a 2008 Kenneth R. Laker, update 12Oct09 KRL 9 ESE319 Introduction to Microelectronics Simple Example ∣H ∣=20log ∣K ∣20log ∣ j 1∣−20log∣ j 1∣ dB 10 dc 10 1 10 Applying the approximation on the previous slide: 1⇒∣H dB∣=20log10 K dc 110⇒∣H ∣=20log K 20 log dB 10 dc 10 1 10⇒∣H ∣=20log K 20log −20log dB 10 dc 10 1 10 10 2008 Kenneth R. Laker, update 12Oct09 KRL 10 ESE319 Introduction to Microelectronics 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); 2008 Kenneth R. Laker, update 12Oct09 KRL 11 ESE319 Introduction to Microelectronics Scilab Results Bode plot Actual freq. response Error 2008 Kenneth R. Laker, update 12Oct09 KRL 12 ESE319 Introduction to Microelectronics Observations The ω/ωxa ratio changes by 20 dB for each order of magnitude change in frequency (20log10(10) = 20). Our “rule of 10” scheme for using either 1, or ω/ωxa for 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. 2008 Kenneth R. Laker, update 12Oct09 KRL 13 ESE319 Introduction to Microelectronics Bode Plot Used to Estimate Zeros & Poles 2008 Kenneth R. Laker, update 12Oct09 KRL 14 ESE319 Introduction to Microelectronics Bode Plot Superposition Directly from the Bode Plot! j 1 20 H j = j j 1 1 100 500 Hz 20 Hz 500 100 Hz 2008 Kenneth R. Laker, update 12Oct09 KRL 15 ESE319 Introduction to Microelectronics 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. 2008 Kenneth R. Laker, update 12Oct09 KRL 16 ESE319 Introduction to Microelectronics Sketch of Typical Voltage Gain Response for a CE Amplifier ∣Av∣dB Low Midband High Frequency Frequency Band ALL capacitances are neglected Band Due to exter- nal blocking 3dB Due to BJT parasitic and bypass capacitors C and C capacitors π µ 20 log10∣Av∣dB f Hz f L f H (log scale) BW f f f = H − L≈ H GBP=∣Av∣BW 2008 Kenneth R. Laker, update 12Oct09 KRL 17 ESE319 Introduction to Microelectronics High Frequency Small-signal Model C r x C C 0 C = V m Two capacitors and a resistor added. 1 CB A base to emitter capacitor, V 0c Cπ A base to collector capacitor, C C µ C =C je 0 A resistor, r , representing the base de V m x 1 BE terminal resistance (r << r ) V 0 e x π 2008 Kenneth R. Laker, update 12Oct09 KRL 18 ESE319 Introduction to Microelectronics 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 vbe since their reactance approaches zero (short circuit) as frequency increases. As we will see later is the principal cause of this gain loss at Cµ high frequencies. At the base looks like a capacitor of value Cµ connected between base and emitter, where and may k Cµ k > 1 be >> 1. This phenomenon is called the Miller Effect. 2008 Kenneth R. Laker, update 12Oct09 KRL 19 ESE319 Introduction to Microelectronics Frequency-dependent “beta” hfe short-circuit current The relationship ic = βib does not apply at high frequencies f > fH! Using the relationship – ic = f(Vπ ) – find the new relationship between i and i . For i (using phasor notation (I & V ) for b c b x x frequency domain analysis): 1 @ node B': I = sC sC V where r x≈0 (ignore r ) b r x 2008 Kenneth R. Laker, update 12Oct09 KRL 20 ESE319 Introduction to Microelectronics Frequency-dependent hfe or “beta” The ratio of the two equations: 1 I = sC sC V @ node C: I c=g m−sC V (ignore ro) b r Leads to a new relationship between the Ib and Ic: I c g m−sC h fe= = I b 1 s C s C r 2008 Kenneth R. Laker, update 12Oct09 KRL 21 ESE319 Introduction to Microelectronics Frequency Response of hfe g m−sC h = I C V T fe g = r = 1 m V s C s C T I C r multiply N&D by r π C 1 For small ωs= low : low ≪1 g m− j C r g m 10 h fe= 1 j C C r and: 1 low C C r≪1 factor N to isolate gm 10 C C Note: C C r = C C ≫ 1− j g m r low low g low g g m m h = m fe 1 j C C r We have: h fe=g m r = 2008 Kenneth R.