ESE319 Introduction to Microelectronics High Frequency BJT Model 2008 Kenneth R. Laker, update 12Oct10 KRL 1 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 12Oct10 KRL 2 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 20log10∣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 12Oct10 KRL 3 ESE319 Introduction to Microelectronics High Frequency Small-signal Model C SPICE r x CJC = C µ0 CJE = C je0 C TF = τF RB = r x C 0 C = m Two capacitors and a resistor added. V CB 1 V 0c A base to emitter capacitor, Cπ C A base to collector capacitor, C C =C je0 ≈C 2C µ de V m de je0 A resistor, r , representing the base 1− BE x V 0 e C de=F g m terminal resistance (r << r ) x π = forward-base transit time τF 2008 Kenneth R. Laker, update 12Oct10 KRL 4 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 12Oct10 KRL 5 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 = – find the new relationship – ic f(Vπ ) 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 NOTE: s = σ + jω, in sinusoidal steady-state s = jω. 2008 Kenneth R. Laker, update 12Oct10 KRL 6 ESE319 Introduction to Microelectronics Frequency-dependent hfe or “beta” 1 (ignore r ) I = sC sC V @ node C: I c=g m−sC V o 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 12Oct10 KRL 7 ESE319 Introduction to Microelectronics Frequency Response of hfe g −sC m I V h fe= C T 1 g m= r = V I s C s C T C r multiply N&D by r and set s = j π ω C 1 For small ωs= low : low ≪1 g m− j C r g m 10 h = fe 1 j C C r 1 and: low C C r≪1 factor N to isolate gm 10 C C C C r C C 1 j g r Note: low =low ≫low − m g m g m gm h = fe 1 j C C r We have: h fe=g m r = 2008 Kenneth R. Laker, update 12Oct10 KRL 8 ESE319 Introduction to Microelectronics Frequency Response of hfe cont. C f h dB 1− j 1 j fe 1− j g m r − g z f z m 20log h fe= = g m r = 10 1 j C C r f 1 j 1 j f f f f z C C C r=C C ≫ => f z ≫ f g m g m Hence, the lower break frequency or frequency is – 3dB fβ 1 g m 1 g m f = = the upper: f = = 2C C r 2C C z 2C / g m 2C where f z10 f 2008 Kenneth R. Laker, update 12Oct10 KRL 9 ESE319 Introduction to Microelectronics Frequency Response of hfe cont. Using Bode plot concepts, for the range where: f f h fe=g m r= 1− j f / f ≈1 For the range where: f f f z s.t. ∣ z∣ We consider the frequency-dependent numerator term to be 1 and focus on the response of the denominator: g m r h = = fe f f 1 j 1 j f f 2008 Kenneth R. Laker, update 12Oct10 KRL 10 ESE319 Introduction to Microelectronics Frequency Response of hfe cont. g m r Neglecting numerator term: h = = fe f f 1 j 1 j f f 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 BJT unity-gain fre- f T = = f quency or GBP 2 2008 Kenneth R. Laker, update 12Oct10 KRL 11 ESE319 Introduction to Microelectronics Frequency Response of hfe cont. −3 =100 r=2500 C =12 pF C =2 pF g m=40⋅10 S 12 −3 1 10 ⋅10 6 = = =28.57⋅10 rps 12 2 2.5 C C r ⋅ 28.57 6 f f 455 MHz f = = 10 Hz=4.55 MHz T = = 2 6.28 −3 12 g m 40⋅10 ⋅10 9 z= = Hz=20⋅10 rps C 2 f = z =3.18⋅109 Hz=3180 MHz z 2 2008 Kenneth R. Laker, update 12Oct10 KRL 12 ESE319 Introduction to Microelectronics Scilab fT 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); 2008 Kenneth R. Laker, update 12Oct10 KRL 13 ESE319 Introduction to Microelectronics hfe Bode Plot (dB) f T 2008 Kenneth R. Laker, update 12Oct10 KRL 14 ESE319 Introduction to Microelectronics Multisim Simulation v-pi I c I b v-pi mS Insert 1 ohm resistors – we want to measure a current ratio. I c g m−s C h fe= = I b 1 sC C r 2008 Kenneth R. Laker, update 12Oct10 KRL 15 ESE319 Introduction to Microelectronics Simulation Results Theory: f T = f =455 MHz Low frequency |hfe| Unity Gain frequency about 440 MHz. 2008 Kenneth R. Laker, update 12Oct10 KRL 16.