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°c Copyright 2009. W. Marshall Leach, Jr., Professor, Georgia Institute of Technology, School of Electrical and Computer Engineering.
The BJT
Notation
The notations used here for voltages and currents correspond to the following conventions: Dc bias values are indicated by an upper case letter with upper case subscripts, e.g. VDS , IC. Instantaneous values of small-signal variables are indicated by a lower-case letter with lower-case subscripts, e.g. vs, ic. Total values are indicated by a lower-case letter with upper-case subscripts, e.g. vBE , iD. Circuit symbols for independent sources are circular and symbols for controlled sources have a diamond shape. Voltage sources have a ± sign within the symbol and current sources have an arrow.
Device Equations
Figure 1 shows the circuit symbols for the npn and pnp BJTs. In the active mode, the collector-base junction is reverse biased and the base-emitter junction is forward biased. For the npn device, the active-mode collector and base currents are given by
iC = IS exp
μ vBE VT
iB =
iC β
where VT is the thermal voltage, IS is the saturation current, and β is the base-to-collector current gain. These are given by
VT =
kT q
= 0.025 V for T = 290 K = 25.86 mV for T = 300 K (2)
IS = IS 0
μ 1 +
vCE VA
β = β 0
μ 1 + vCE VA
where VA is the Early voltage and IS 0 and β 0 , respectively, are the zero bias values of IS and β. Because IS /β = IS 0 /β 0 , it follows that iB is not a function of vCE. The equations apply to the pnp device if the subscripts BE and CE are reversed. The emitter-to-collector current gain α is defined as the ratio iC /iE. To solve for this, we can write
iE = iB + iC =
μ 1 β
iC =
1 + β β iC (5)
It follows that
α = iC iE
β 1 + β
Thus the currents are related by the equations
iC = βiB = αiE (7)
Figure 1: BJT circuit symbols.
Transfer and Output Characteristics
The transfer characteristics are a plot of the collector current iC as a function of the base-to-emitter voltage vBE with the collector-to-emitter voltage vCE held constant. From Eqs. 1 and 3, we can write
iC = IS 0
μ 1 +
vCE VA
exp
μ vBE VT
It follows that iC varies exponentially with vBE. A plot of this variation is given in Fig. 2. It can be seen from the plot that the collector current is essentially zero until the base-to-emitter voltage reaches a threshold value. Above this value, the collector current increases rapidly. The threshold value is typically in the range of 0. 5 to 0. 6 V. For high current transistors, it is usually smaller. The plot shows a single curve. If vCE is increased, the current for a given vBE is larger. However, the displacement between the curves is so small that it can be difficult to distinguish between them. The small-signal transconductance gm defined below is the slope of the transfer characteristics curve evaluated at the quiescent or dc operating point for a device.
Figure 2: BJT transfer characteristics.
The output characteristics are a plot of the collector current iC as a function of the collector-
Figure 4: (a) BJT dc bias circuit. (b) Circuit for Example 1.
Example 1 Figure 4(b) shows a BJT dc bias circuit. It is given that V +^ = 15 V, R 1 = 20 kΩ, R 2 = 10 kΩ, R 3 = R 4 = 3 kΩ, R 5 = R 6 = 2 kΩ. Solve for IC 1 and IC 2. Assume VBE = 0.7 V and β = 100 for each transistor.
Solution. For Q 1 , we have VBB 1 = V +R 2 / (R 1 + R 2 ), RBB 1 = R 1 kR 2 , VEE 1 = −IB 2 R 4 = −IC 2 R 4 /β, VEE 1 = 0, and REE 1 = R 4. For Q 2 , we have VBB 2 = IE 1 R 4 = IC 1 R 4 /α, RBB 2 = R 4 , VEE 2 = 0, REE 2 = R 6. Thus the bias equations are
V +^
R 2
R 1 + R 2
IC 2
β
R 4 = VBE +
IC 1
β
R 1 kR 2 +
IC 1
α
R 4
IC 1
α
R 4 = VBE +
IC 2
β
R 4 +
IC 2
α
R 6
These equations can be solved simultaneously to obtain IC 1 = 1.41 mA and IC 2 = 1.74 mA.
Hybrid-π Model
Let each current and voltage be written as the sum of a dc component and a small-signal ac component as follows: iC = IC + ic iB = IB = ib (12) vBE = VBE + vbe vCE = VCE + vce (13)
If the ac components are sufficiently small, we can write
ic =
∂IC
∂VBE
vbe +
∂IC
∂VCE
vce ib =
∂IB
∂VBE
vbe (14)
where the derivatives are evaluated at the dc bias values. Let us define the transconductance gm, the collector-to-emitter resistance r 0 , and the base-to-emitter resistance rπ as follows:
gm =
∂IC
∂VBE
IS
VT
exp
μ VBE VT
IC
VT
r 0 =
μ ∂IC ∂VCE
IS 0
VA
exp
μ VBE VT
VA + VCE
IC
rπ =
μ ∂IB ∂VBE
IS 0
β 0 VT exp
μ VBE VT
VT
IB
The collector and base currents can thus be written
ic = i^0 c + vce r 0 ib = vπ rπ
where i^0 c = gmvπ vπ = vbe (19) The small-signal circuit which models these equations is given in Fig. 5(a). This is called the hybrid-π model. The resistor rx, which does not appear in the above equations, is called the base spreading resistance. It represents the resistance of the connection to the base region inside the device. Because the base region is very narrow, the connection exhibits a resistance which often cannot be neglected.
Figure 5: (a) Hybrid-π model. (b) T model.
The small-signal base-to-collector ac current gain β is defined as the ratio i^0 c/ib. It is given by
β =
i^0 c ib
gmvπ ib = gmrπ =
IC
VT
VT
IB
IC
IB
Note that ic differs from i^0 c by the current through r 0. Therefore, ic/ib 6 = β unless r 0 = ∞.
T Model
The T model replaces the resistor rπ in series with the base with a resistor re in series with the emitter. This resistor is called the emitter intrinsic resistance. The current i^0 e can be written
i^0 e = ib + i^0 c =
μ 1 β
i^0 c =
1 + β β i^0 c =
i^0 c α
where α is the small-signal emitter-to-collector ac current gain given by
α = β 1 + β
Thus the current i^0 c can be written i^0 c = αi^0 e (23)
Figure 7: (a) Circuit with the i^0 c source replaced by identical series sources. (b) Simplified T model.
Norton Collector Circuit
The Norton equivalent circuit seen looking into the collector can be used to solve for the response of the common-emitter and common-base stages. It consists of a parallel current source ic(sc) and a resistor ric from the collector to signal ground. Fig. 8(a) shows the BJT with Thévenin sources connected to its base and emitter. To solve for the Norton equivalent circuit seen looking into the collector, we use the simplified T model in Fig. 8(b). By superposition of vc, αi^0 e, vtb, and vte, the following equations for ic and i^0 e can be written
ic = vc r 0 + r^0 ekRte
- αi^0 e − vtb r^0 e + Rtekr 0
Rte Rte + r 0
− vte Rte + r e^0 kr 0
r^0 e r e^0 + r 0
i^0 e = vtb r^0 e + Rtekr 0
vte Rte + r^0 ekr 0 − vc r 0 + r e^0 kRte
Rte r e^0 + Rte
These can be solved to obtain
ic =
vtb r^0 e + Rtekr 0
μ α −
Rte Rte + r 0
vte Rte + r e^0 kr 0
μ α +
r^0 e r e^0 + r 0
vc r 0 + r^0 ekRte
μ 1 − αRte r^0 e + Rte
This equation is of the form ic = ic(sc) + vc ric
where ic(sc) and ric are given by ic(sc) = Gmbvtb − Gmevte (34)
ric = r 0 + r e^0 kRte 1 − αRte/ (r e^0 + Rte)
Figure 8: (a) BJT with Thevenin sources connected to the base and the emitter. (b) Simplified T model.
Figure 9: (a) Circuit for calculating ric. (b) Norton collector circuit.
and Gmb and Gme are given by
Gmb =
r^0 e + Rtekr 0
μ α −
Rte Rte + r 0
or
α r e^0 + Rtekr 0
r 0 − Rte/β r 0 + Rte
Gme =
Rte + r e^0 kr 0
μ α +
r^0 e r e^0 + r 0
or
α r^0 e + Rtekr 0
r 0 + r e^0 /α r 0 + Rte
The Norton equivalent circuit seen looking into the collector is shown in Fig. 9. For the case r 0 À Rte and r 0 À r^0 e, we can write
ic(sc) = Gm (vtb − vte) (38)
where Gm =
α r e^0 + Rte
Figure 11: (a) Circuit for calculating ie(sc). (b) Thévenin emitter circuit.
where ve(oc) and rie are given by
ve(oc) = vtb r 0 + (1 − α) Rtc r e^0 + r 0 + (1 − α) Rtc or = v tb
r 0 + Rtc/ (1 + β) r^0 e + r 0 + Rtc/ (1 + β)
rie = r e^0 r 0 + Rtc r^0 e + r 0 + (1 − α) Rtc =^ or r^0 e
r 0 + Rtc r e^0 + r 0 + Rtc/ (1 + β)
The Thévenin equivalent circuit seen looking into the emitter is shown in Fig. 10.
Thévenin Base Circuit
Although the base is not an output terminal, the Thévenin equivalent circuit seen looking into the base is useful in calculating the base current. It consists of a voltage source vb(oc) in series with a resistor rib from the base node to signal ground. Fig. 12(a) shows the BJT symbol with a Thévenin source connected to its emitter. Fig. 12(b) shows the Pi model for calculating the base voltage. By superposition of vte, ib, and βib, treating each branch of βib separately in the superposition, we can write the following equation for vb
vb = vte r 0 + Rtc Rte + r 0 + Rtc
- ib [rx + rπ + Rtek (r 0 + Rtc)]
+βib r 0 Rte Rte + r 0 + Rtc = vte r 0 + Rtc Rte + r 0 + Rtc
rx + rπ + Rtek (r 0 + Rtc)
+β r 0 Rte Rte + r 0 + Rtc
This equation is of the form vb = vb(oc) + ibrib (48)
where vb(oc) and rib are given by
vb(oc) = vte
r 0 + Rtc Rte + r 0 + Rtc
Figure 12: (a) BJT with Thevenin source connected to the emitter. (b) T model for calculating vb(oc).
Figure 13: (a) Circuit for calculating vb. (b) Thévenin base circuit.
rib = rx + rπ + Rtek (r 0 + Rtc)
+β
r 0 Rte Rte + r 0 + Rtc or = rx + (1 + β) re + Rtek (r 0 + Rtc) +β r 0 Rte Rte + r 0 + Rtc
The equivalent circuit which models these equations is shown in Fig. 13.
The r 0 Approximations
The r 0 approximations approximate r 0 as an open circuit in all equations except the one for ric. In this case, the equations for ic(sc), Gm, ric, ve(oc), rie, vb(oc), and rib are
ic(sc) = i^0 c = Gm (vtb − vte) Gm = α r e^0 + Rte
Figure 16: Summary of the small-signal equivalent circuits.
resistance can be calculated by replacing the circuit seen looking into the collector by the Norton equivalent circuit of Fig. 9(b). With the aid of this circuit, we can write
vo = −ic(sc) (rickRtc) = −Gmb (rickRtc) vtb (55)
rout = rickRtc (56)
where Gmb and ric, respectively, are given by Eqs. (36) and (35). The input resistance is given by
rin = Rtb + rib (57)
where rib is given by Eq. (50).
The Common-Collector Amplifier
Figure 17(b) shows the ac signal circuit of a common-collector amplifier. We assume that the bias solution and the small-signal resistances r^0 e and r 0 are known. The output voltage and output resistance can be calculated by replacing the circuit seen looking into the emitter by the Thévenin equivalent circuit of Fig. 10(b). With the aid of this circuit, we can write
vo = ve(oc)
Rte rie + Rte
r 0 + Rtc/ (1 + β) r^0 e + r 0 + Rtc/ (1 + β)
Rte rie + Rte vtb (58)
rout = riekRte (59)
where rie is given by Eq. (46). The input resistance is given by
rin = Rtb + rib (60)
where rib is given by Eq. (50).
Figure 17: (a) Common-emitter amplifier. (b) Common-collector amplifier. (c) Common-base amplifier.
The Common-Base Amplifier
Figure 17(c) shows the ac signal circuit of a common-base amplifier. We assume that the bias solution and the small-signal parameters r^0 e and r 0 are known. The output voltage and output resistance can be calculated by replacing the circuit seen looking into the collector by the Norton equivalent circuit of Fig. 9(b). The input resistance can be calculated by replacing the circuit seen looking into the emitter by the Thévenin equivalent circuit of Fig. 10(b) with ve(oc) = 0. With the aid of these circuits, we can write
vo = −ic(sc) (rickRtc) = Gme (rickRtc) vte (61) rout = rickRtc (62) rin = Rte + rie (63)
where Gme, ric, and rie, respectively, are given by Eqs. (37), (35), and (46).
The CE/CC Amplifier
Figure 18(a) shows the ac signal circuit of a two-stage amplifier consisting of a CE stage followed by a CC stage. Such a circuit is used to obtain a high voltage gain and a low output resistance. The voltage gain can be written
vo vtb 1
ic1(sc) vtb 1
×
vtb 2 ic1(sc)
×
ve2(oc) vtb 2
×
vo ve2(oc)
= Gmb 1 × [− (ric 1 kRC 1 )] × r 0 r e^02 + r 0
×
Rte 2 rie 2 + Rte 2
where r e^02 is calculated with Rtb 2 = ric 1 kRC 1. The input and output resistances are given by
rin = Rtb 1 + rib 1 (65) rout = rie 2 kRte 2 (66)
Although not a part of the solution, the resistance seen looking out of the collector of Q 1 is Rtc 1 = RC 1 krib 2.
Figure 19: (a) Second cascode amplifier. (b) Differential amplifier.
Looking out of the emitter of Q 1 , the Thévenin voltage and resistance are given by
vte 1 = ve2(oc)
RQ
RQ + RE + rie
= vtb 2
r 0 + Rtc/ (1 + β) r^0 e + r 0 + Rtc/ (1 + β)
RQ
RQ + RE + rie
Rte 1 = RE + RQk (RE + rie) (68) The small-signal collector voltage of Q 1 is given by
vo 1 = −ic1(sc) (rickRtc) = − (Gmbvtb 1 − Gmevte 1 ) (rickRtc) = −Gmb (rickRtc) vtb 1
+Gme r 0 + Rtc/ (1 + β) r e^0 + r 0 + Rtc/ (1 + β)
RQ
rie + RE + RQ
vtb 2 (69)
By symmetry, vo 2 is obtained by interchanging the subscripts 1 and 2 in this equation. The small- signal resistance seen looking into either output is
rout = Rtckric (70)
where ric calculated from Eq. (35) with Rte = RE + RQk (RE + rie). Although not labeled on the circuit, the input resistance seen by both vtb 1 and vtb 2 is rin = rib. A second solution of the diff amp can be obtained by replacing vtb 1 and vtb 2 with differential and common-mode components as follows:
vtb 1 = vi(cm) +
vi(d) 2
vtb 2 = vi(cm) −
vi(d) 2
where vi(d) = vtb 1 −vtb 2 and vi(cm) = (vtb 1 + vtb 2 ) / 2. Superposition of vi(d) and vi(cm) can be used to solve for vo 1 and vo 2. With vi(cm) = 0, the effects of vtb 1 = vi(d)/ 2 and vtb 2 = −vi(d)/ 2 are to cause
vq = 0. Thus the vq node can be grounded and the circuit can be divided into two common-emitter stages in which Rte(d) = RE for each transistor. In this case, vo1(d) can be written
vo1(d) =
ic1(sc) vtb1(d)
×
vo1(d) ic1(sc) vtb1(d) = Gm(d) × (−rickRtc)
vi(d) 2
= Gm(d) × (−rickRtc) vtb 1 − vtb 2 2
By symmetry vo2(d) = −vo1(d). With vi(d) = 0, the effects of vtb 1 = vtb 2 = vi(cm) are to cause the emitter currents in Q 1 and Q 2 to change by the same amounts. If RQ is replaced by two parallel resistors of value 2 RQ, it follows by symmetry that the circuit can be separated into two common-emitter stages each with Rte(cm) = RE + 2RQ. In this case, vo1(cm) can be written
vo1(cm) =
ic1(sc) vtb1(cm)
×
vo1(cm) ic1(sc) vtb1(cm) = Gm(cm) (−rickRtc) vi(cm)
= Gm(cm) × (−rickRtc) vtb 1 + vtb 2 2
By symmetry vo2(cm) = vo1(cm). Because Rte is different for the differential and common-mode circuits, Gm and rib are different. However, the total solution vo 1 = vo1(d) + vo1(cm) is the same as that given by Eq. (69), and similarly for vo 2. Note that ric is the same for both solutions and is calculated with Rte = RE + RQk (RE + rie). The small-signal base currents can be written ib 1 = vi(cm)/rib(cm) + vi(d)/rib(d) and ib 2 = vi(cm)/rib(cm) −vi(d)/rib(d). If RQ → ∞, the common-mode gain is very small, approaching 0 as r 0 → ∞. In this case, the differential solutions can be used for the total solutions. If RQ À RE +rie, the common-mode solutions are often approximated by zero.
Small-Signal High-Frequency Models
Figure 20 shows the hybrid-π and T models for the BJT with the base-emitter capacitance cπ and the base-collector capacitance cμ added. The capacitor ccs is the collector-substrate capacitance which in present in monolithic integrated-circuit devices but is omitted in discrete devices. These capacitors model charge storage in the device which affects its high-frequency performance. The capacitors are given by
cπ = cje + τ (^) F IC VT
cμ =
cjc [1 + VCB /φC ]mc^
ccs = cjcs [1 + VCS /φC ]mc^
where IC is the dc collector current, VCB is the dc collector-base voltage, VCS is the dc collector- substrate voltage, cje is the zero-bias junction capacitance of the base-emitter junction, τ (^) F is the forward transit time of the base-emitter junction, cjc is the zero-bias junction capacitance of the base-collector junction, cjcs is the zero-bias collector-substrate capacitance, φC is the built-in potential, and mc is the junction exponential factor. For integrated circuit lateral pnp transistors, ccs is replaced with a capacitor cbs from base to substrate, i.e. from the B node to ground. In these models, the currents are related by i^0 c = gmvπ = βi^0 b = αi^0 e (78)
These relations are the same as those in Eq. (26) with ib replaced with i^0 b.
