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Unpublished work 2006 by John R Brews 1
PNP current mirror
Schematic
OUT
+
- {V_CC}20.089mA
PARAMETERS:
V_CC = 10V
R_E = 418.85833
R_R = 495.59372
V_SAT = 0.55V
5.7928V
5.0000V
-+
+-
E1
GAIN = 1
M
0
5.0000V
Sweep
+
-
AC
V_ac
1V
0
E
+{R_R}
10.089mA
.model Q_pVAF PNP (Bf={B_F} Is={I_S} Vaf={V_AF}Nf={N_F} Rb={r_X})
5.7928V+
{R_E}
10.044mA
Q_pVAF
Q_Ref
-44.455uA
-10.000mA
DOT-MODEL:
B_F = 224.9477
I_S = .6506fA
V_AF = 115.7V
N_F = 1.0089535
r_X = 10
+
- {V_A}
V_DC
E
+
{R_E}
10.044mA
INPUT SIGNAL
AMPLITUDE = 1V
V_A = 5
FREQUENCY = 1kHz
0
EM
-
+
Transient
Analysis
{AMPLITUDE}
{FREQUENCY}
V_SIN
792.80mV
10.000V
0
5.0000V
MQ_pVAF
Q_Out-44.455uA
-10.000mA5.0000V
Figure 1
Circuit for pnp current mirror using simple device with dot-model statement shown
Figure 1 shows a schematic for a pnp current mirror.1 The purpose of the mirror is to
emulate an ideal current source, that is, to provide the same DC current through the
output node regardless of the voltage applied, DC or transient.
How does it work?
The basic idea behind the circuit is that the left side draws a current through the
reference transistor setting up a corresponding emitter-base voltage. Because the circuit is
1 Also shown in Figure 1 is an evaluator circuit using PSPICE VCVS Part E just to display the value of VEM
on the schematic for easy comparison with the spreadsheet constructed in this chapter.
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Unpublished work 2006 by John R Brews 2
symmetric (assuming the output and reference transistors are alike) the same VEB appears
at the output transistor, so the same current flows there (it is mirrored), almost
independent of the voltage VA because VA hardly affects VEB.
Unfortunately, the mirror is not entirely successful, having these limitations:
1. It provides a nearly constant DC current only over a limited range of voltages. This
limitation arises at voltages VA > VM (VM = mid-base voltage), where the output
transistor QOut leaves active mode and goes into saturation.
2. Even in the range of voltages VA < VM where QOut is active, the current is not strictly
constant, but varies somewhat with VA. That is, the circuit resembles a Norton source
with a finite Norton resistance instead of an ideal current source. This limitation is
due to the finite output resistance of transistor QOut.
The above limitations are illustrated in Figure 2.
Figure 2
Circuit output behavior for Figure 1; at the compliance voltage VCV where output resistance beginsa rapid drop to low values, the output transistor is in saturation by VBC = VSAT
The top panel in Figure 2 shows the current-voltage I-Vbehavior of the mirror. Itdelivers a DC current of 10 mA for voltages below approximately 5.55V. The lower
panel shows the resistance of the mirror, determined as the inverse of the derivative of the
current by voltage. It shows that this resistance is high (934 k; at VA = 1 V), but not
quite constant, and drops suddenly just above VA = VM = 5 V. The drop-off voltage is
called the compliance voltage VCV of the mirror, and the voltage range where nearly
VSAT
VCV
Active Mode
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Unpublished work 2006 by John R Brews 3
constant DC current is delivered is the compliance range of the mirror. If we choose the
bias at 3 dB roll-off of resistance as the verge of the drop, we find the compliance voltage
is VCV = 5.346 V from Figure 2. At this bias, the output transistor is in saturation by an
amount VBC = VSAT = 0.346 V. At the point where the DC currenthas just begun to
drop (VA = 5.55 V) the output resistance of the mirror already is at a very low value of
only 9.34 k;, showing that the Norton resistance of the mirror is much more sensitive to
saturation of the output transistor than is the DC current itself.
Because the limitations of the mirror depend on the limitations of the transistor, we
need a transistor model that includes the Early voltage of the device. Otherwise, the
mirror would still have a voltage limitation, but would be an ideal current source as long
as QOut was active. Hence, we have the dot-model statement in Figure 1, discussed in
detail shortly.
Design goal
We want to design the circuit of Figure 1 to meet specifications on DC current level
IC at VA = VM (both transistors with VBC = 0 V), on compliance voltage VCV (taken as a
specification on VM because VM is unambiguous and differs from VCV by only the small
voltage VSAT discussed later2), and specifications on output resistance RN (Norton
resistance) of the mirror. The variables at our disposal are the leg resistor value RE andthe reference resistor value RR, so unless we are lucky only two of the three specifications
can be satisfied, and a trade-off will be necessary. For this purpose we will set up a
spreadsheet incorporating the hand analysis below.
AC and DC beta-values
For the dot-model statement of Figure 1, the small-signal ACF-value, which will be
called FAC, is the same as the DC F-value, which is called FDC. However, that is not so formore complex models, so we include this difference in the equations here. EQ. 1 defines
DC F:
2 The value of VSAT is expected to be somewhere around 0.5 V, but its value is unknown without a
simulation. It varies with the type of transistor and with the current and bias conditions.
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Unpublished work 2006 by John R Brews 4
EQ. 1
BI
CIDC!F ,
while AC F is defined by:
EQ. 2
CdI
DCd
DC
CI
DC
CdI
DCd
DC
CI
DC
DCCId
CdI
BdI
CdIAC F
F
F
F
FF
FF
!
!!!
12
1
1
/.
According to EQ. 2, FAC is different fromFDC ifFDC depends on IC. For the dot-model
statement of Figure 1 FDC does not depend on IC, but for real transistors it does. So, for
real transistors, FDC and FAC are the same only at the maximum in the FDC vs. IC curve.
An example is shown in Figure 3 below.
I _E
10nA 1. 0uA 100uA 10mA 1. 0A
I ( C) / I ( B) D( I ( C) ) / D( I ( B) )
0
100
200 AC Bet a
DC Be t a
( 1 0 . 0 0 mA, 2 2 4. 9 4 77 )
Figure 3
Comparison of AC and DC F-values as a function of emitter current IE for the Q2N2907A pnp-transistor with VBC = 0 V
Figure 3 shows the current dependence of the two Fs for the Q2N2907A transistorusing the PSPICE dot model statement for this transistor. It can be seen that the two Fs
agree at the maximum inFDC near an emitter current of IE = 10 mA. In addition,FAC >
FDC when FDC has positive derivative, as predicted by EQ. 2.
Figure 3 is generated using the circuit of Figure 4 with a DC SWEEP analysis to
sweep IE. Zero-bias DC voltage sources are inserted in the base and collector leads and
named B and C to indicate that the currents I(B) and I(C) going through them are the base
and collector currents.
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Unpublished work 2006 by John R Brews 5
+
-
C
DC = 0V
{I_E}
Q2N2907A
I
B
0
I
PARAMETERS:
I_E = 10m
C
+
-
B
DC = 0V
0
Figure 4
Test circuit for generating AC and DC F-plots of Figure 3
The Q_pVAF dot-model parameters
To allow later comparison with the Q2N2907A, the dot model statement of
Figure 1 is introduced, namely
.model Q_pVAF PNP (Bf={B_F} Is={I_S} Vaf={V_AF} Nf={N_F} Rb={r_X})
which can be compared with the dot-model statement for the Q2N2907A:
.model Q2N2907A PNP (Is=650.6E-18 Xti=3 Eg=1.11 Vaf=115.7 Bf=231.7 Ne=1.829+ Ise=54.81f Ikf=1.079 Xtb=1.5 Br=3.563 Nc=2 Isc=0 Ikr=0 Rc=.715+ Cjc=14.76p Mjc=.5383 Vjc=.75 Fc=.5 Cje=19.82p Mje=.3357 Vje=.75+ Tr=111.3n Tf=603.7p Itf=.65 Vtf=5 Xtf=1.7 Rb=10)* National pid=63 case=TO18* 88-09-09 bam creation
provided with PSPICE.
To improve agreement with more realistic dot-model statements like that for the
Q2N2907A, the Q_pVAF dot-model statement includes parameters for Early voltage Vaf,
non-ideal diode-law Nf, and internal series base resistance Rb. This dot-model statement
corresponds in active mode to the I-Vrelation
EQ. 3
-
!
AFV
BCiV
THV
EBiVSICI 11exp
L,
where VEBi, VBCi are the intrinsicemitter-base and base-collector voltages, differing from
the circuit orexternalvalues because of the internal base resistance rX, as discussed
shortly. The parameterL is the ideality factoror, as referred to in the PSPICE
documentation, the forwardcurrent emission coefficient, also discussed shortly.
Notice the intrinsic emitter-base voltage according to EQ. 3 is given by EQ. 4:
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Unpublished work 2006 by John R Brews 6
EQ. 4
!
AFV
BCiVSI
CInTHVEBiV
1
1NL
Lets take a closer look at the effects of these parameters.
Early voltage: parameter VAF
The Early voltage enters the current I-Vrelation as shown in EQ. 3. Somewhat less
obvious is the Early voltage influence on the transistorFs. The DC base current of the
transistor, IB, does not depend upon the base-collector voltage, so EQ. 5 gives the DC F
EQ. 5
!!
!
!!AFVBCV
BCVDCBI
AFV
BCVBCVCI
BICI
DC 10
1)0(
FF .
That is, the DC F increases with VBC because of the Early effect. The AC F does the same
thing if the current dependence ofFDC is contained inFDC(VBC=0). That is,
EQ. 6
!!!
AFV
BCVBCVACAC 10FF .
Base resistance: parameter Rb and the intrinsic base resistance rX
The introduction of the base resistance rX introduces some complications into ourequations because the device behavior is governed by the internalVEBi of the transistor,
which differs from the externalVEB of the circuit by the voltage drop across rX. See
Figure 5.
Figure 5
Internal and external voltages related to rX: VEBi < VEB because of drop across rX, while VBCi > VBC
Xr EBiV
EBV
CI
F/CI
BCV BCiV
F
XrCI
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Unpublished work 2006 by John R Brews 7
According to Figure 5, the internal and external emitter-base voltages are related as:
EQ. 7
XrDC
CIEBVEBiV
F! .
Combining EQ. 7 with EQ. 4 we find VEB is related to the current by
EQ. 8
XrDC
CI
AFV
BCiVSI
CInTHVEBVF
L
!
1
1N .
In addition to its effect on VEBi, the voltage drop across rX causes a non-zero
internalVBCi of the transistor, even though the external circuit VBC = 0 V. (Consider
Figure 5 for the case where VBC = 0 V.)
A non-zero VBCi means the Early effect comes into play, affecting the current andthe beta values of the transistor, so the transistor betas increase according to
EQ. 9
!!
AFV
BCiVVBCVBCiV 1)0()( FF .
The value of VBCi is given by Ohms law as EQ. 10:
EQ. 10
XrBCiV
CIBCiV )(F
! .
Solving the quadratic found by substituting EQ. 9 into EQ. 10, we find VBCi as shown inEQ. 11 next:
EQ. 11
-
!! 1
2/1
)0(
41
2 AFVBCVXrCIAFV
BCiVF
.
Parameter rX is set using dot-model parameter Rb, namely, rX = Rb.
Current dependence of small-signal parameters; parameterL
You may recall the current and voltage dependence of the transistor small-signalparameters for a simple bipolar exhibiting Early effect. In particular,
EQ. 12
)()0( BCVCIBCVAFV
BCVCI
AFVOr
!
!! ,
CI
THVAC
mg
ACrFF
T }! .
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Unpublished work 2006 by John R Brews 8
In EQ. 12, rO = output resistance, rT = base input resistance, gm = small-signal
transconductance, VAF = Early voltage and VTH = thermal voltage (kBT/q 25.864 mV @
27r C). EQ. 12 for rO does not agree with most textbooks, but it does agree with PSPICE
and transistor physics.
In real transistors the ideal diode law is not satisfied. To help match this reality, the
Q_pVAF dot-model statement includes parameter Nf.When this parameter is used, the
transconductance and base resistance are given by the relations:
EQ. 13
THV
CImg
L
1! ,
CI
THVAC
mg
ACrF
LF
T !! ,
where parameterL is the before-mentioned ideality factor also known as the forward
current emission coefficient.
ParameterLis specified by dot-model parameter Nf, namely L = Nf.
SettingL
How can Lbe found? Lets assume we want the value ofL that makes our
Q_pVAF-model fit the Q2N2907A. The same approach works for other models. Figure 5
puts the Q2N2907A and Q_pVAF in identical circuits. These circuits set the external VBC
= 0 V, which is the case for the mirror at the design point. Then we find the externalgm-
values by running a DC sweep of IE and taking derivatives, as shown in Figure 7. We setthe Nfvalue to make the two gm-values the same at the current level of interest, namely
10 mA in this case. We set
EQ. 14
)_(
)29072(
pVAFQmg
ANQmgfN ! = 0.3784089/0.3750509 =1.0089535.
The externalgm-values are related to the internalgm-values by EQ. 15 (using EQ. 7 for
VEBi):
EQ. 15
!
x
x
x
x!
x
x! mig
DC
Xrmig
EBV
EBiV
EBiV
CI
EBV
CIextmgF
1)( ,
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Unpublished work 2006 by John R Brews 9
where the internal transconductance gmi = IC/VEBi. EQ. 15 is interesting mainly because
it shows setting the external gm-values equal also makes the internal gm-values equal.3, 4
Q_pVAF
0
PARAMETERS:
I_E = 10mA
0
DOT-MODEL:
B_F = 224.9477
I_S = .6506fA
V_AF = 115.7V
N_F = 1
r_X = 10
{I_E} {I_E}
E_2907A
Q2N2907A
+
-
C_Nf
DC = 0
0
E_VAF 0
+
-
C_2907A
DC = 0
Figure 6
Test circuit for finding value of dot-model parameter Nf; notice that Nf= 1 in this test
I _E
0A 5mA 10mA 15mA 20mA
D( I ( C_Nf ) ) / D( V( E _VAF ) ) D( I ( C_2 907 A) ) / D( V ( E_ 2907 A) )
0
0. 5
1. 0
( Q_pVAF, 10. 0000m, 375. 0509m)
( Q2N2907A, 10. 0000m, 378. 4089m)
Figure 7
Comparison of external gm values when Nf= 1; we want to increase Nfto make these two values thesame
Finding Nf this way makes the simple transistor model Q_pVAF resemble closely
the more realistic model Q2N2907A, and in particular makes sure that the small-signalparameters gm and rT at the design point are the same.
Hand analysis
Q-point analysis
The design is done for the case VA = VM because that makes analysis simpler. If
instead we choose VA < VM, the two transistors have different VBC values and that affects
the currents andF-values because of the Early effect. For VA = VM, applying KVL to the
output side of the circuit of Figure 1, we find a relation for RE given by EQ. 16 below:
EQ. 16
3 Technically there are two values for gmi for each gm, and we want the value very close to gm.4 It is the internalgm-value that is given by EQ. 13 and is listed in the PROBE output file.
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Unpublished work 2006 by John R Brews 10
DCCIMVEBVCCV
ERF/11
! .
In EQ. 16 the various symbols are: IC = output (collector) current of QOut, VEB = emitter-
base voltage of QOut, VM = base voltage of both transistors, FDC = DC beta of QOut.
Applying KVL to the left side of the mirror we find RR is given by5EQ. 17
DCCIMV
RRF/21
!
Small-signal analysis
Next we ask just how much the current varies for voltages below the base voltage.
That is, what is the slope of the I-Vcurve for VA < VM. The easiest way to find out is to
bias the mirror at some value of VA below VM and superpose a small-signal AC voltage
Vac. Then the small-signal current Iac that flows is
EQ. 18
acVNR
acVAV
CIacI
1|
x
x! ,
where RN is the Norton resistance of the mirror, and indicates the rate of variation of the
current with applied voltage. The small-signal circuit corresponding to this approach is
shown in Figure 8 below.6 Test current Ix is applied and RN = Vx/Ix.
In Figure 8 the parasitic base resistance rX is included to allow a closer comparison
with the Q2N2907A later on. This resistance is included in Figure 1 by specifying the
dot-model parameter Rb, set in the dot-model PARAMETERbox to Rb = rX.
+
0
+
+
+
0
++
0
+
Figure 8
5 Use KCL at the base of QOut to derive the factor (1+2/FDC).6 The replacement of transistor QRefby the resistor rREF is explained in the Appendix.
xV
ER
ER
REFr
Tr
Or
RR
bIxI
bIACxI F
bI
xI
bIACF
Xr
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Unpublished work 2006 by John R Brews 11
Small-signal circuit corresponding to Figure 1
An easy way to solve circuits like this is to determine all the currents and then use
KVL. Taking KVL on the left side of the circuit noting that (rREF + RE) is in parallel with
RRand following Ib through rT and RE we find EQ. 19:
EQ. 19
? A 0// ! ERbIxIXrrRRERREFrbI T ,
which determines Ib in terms of Ix as
EQ. 20
RRERREFrXrrERER
xIbI //!
T.
Then KVL through the right side of the circuit provides
EQ. 21
ERbIxIOrbIACxIxV ! F ,
Collecting terms in Ix and Ib and substituting for Ib from EQ. 20 we find RN as EQ. 22next:
EQ. 22
RRERREFrXrrERrRRERREFr
ERRRERREFrXrrER
ERACOr
xI
xVNR //
//
//1
!!
T
T
TF
.
Notice that for large RE, the leading term in RN approaches (FAC+1) rO, while for
small RE it approaches rO. So RN is a large resistance, and increases with RE. The most
ideal current source from the viewpoint of voltage-independent current occurs at large
RE. Resistance RN has a complex dependence on the specifications for current value and
compliance voltage, and an easy way to see the connections is through graphs generated
using a spreadsheet. (For example, see Figure 19 and Figure 21 later on.)
Transient analysis
When a large-signal sinusoidal AC voltage of amplitude Vac is applied to the mirror
output with DC voltage VA applied, the instantaneous applied voltage is
EQ. 23
)2sin()( tfacVAVtA TY ! .
To avoid driving the output transistor into saturation, where its low resistance will cause
a large AC current spike, the DC bias must be chosen below the compliance voltage VCV
by at least the AC signal amplitude Vac, that is, we require
EQ. 24
acVCVVAV e .
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Unpublished work 2006 by John R Brews 12
Spreadsheet
The hand analysis is put into the spreadsheet as shown in Figure 9. The diode-
connected reference transistor resistance is denoted by r_REF, following the analysis in
the appendix. To avoid round-off error, a series expansion is used for V_BCi at smallvalues (an IF STATEMENT represents (1+x)1 by a series for argumentsx < 2 105).7
The numerical values corresponding to Figure 9 are shown in Figure 10.
In Figure 10 the values for the transistor parameters are selected to represent the
QN2907A pnp bipolar transistor parameters included with PSPICE. The values forFDC0
andFAC0 are determined as shown in Figure 3 for the specified current level of
IC = 10 mA at VBC = 0V.
The Norton resistance is found using the AC beta from EQ. 9 and the small-signalcircuit of Figure 8.
Figure 9
Input worksheet for current mirror design project
7 Syntax of the IF STATEMENT is described in Chapter 3, or in EXCEL help. Click on HELP and type if
function in the SEARCH BOX.
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Unpublished work 2006 by John R Brews 13
Figure 10
Numerical values for the design in Figure 1
Verification of spreadsheet
Q-point verification
When the spreadsheet values for RE and RRare pasted into PSPICE, the Q-point
results are seen in Figure 1. They agree with the specifications of IC = 10 mA and
VM = 5 V. In addition, we can look at the small-signal results. The PROBE output file is
shown in Figure 11 below. Parameters rO, rT and gm agree with the spreadsheet.
Figure 11
PROBE output file for case of Figure 1; VBC of QOut is not quite zero, indicating some inaccuracy
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Unpublished work 2006 by John R Brews 14
Small-signal verification
The Norton resistance is checked using a small-signal ACSWEEP analysis, as shown
in Figure 12. The discrepancy with the spreadsheet is about 2/100 %.
F r equ enc y
1. 0Hz 1. 0KHz 1. 0MHz 1. 0GHz 1. 0THz
1 / I ( V _ a c )
0
1 . 0 M
2 . 0 M
( 1 . 000000000, 888 . 21755144K)
Figure 12
Determination of Norton resistance using small-signal ACSWEEP with a 1 V AC input signal
To check that the discrepancy between PSPICE and the spreadsheet is not some
algebraic problem in our small-signal analysis, we can check the analysis of Figure 8
using PSPICE.We set up the PSPICE circuit shown in Figure 13 below:
0
0
+
{r_PI}
PARAMETERS:
r_REF = 2.5982005
R_E = 418.85833
R_R = 495.59372
r_X = 10r_O = 11570.04445
r_PI = 587.01615
+ {R_E}
0
F1
GAIN = 224.9485643
+{R_E}
+
{r_REF}
+
{R_R}
888.43KV
+
{r_O}
+
{r_X}
I_DC
1A1.0000A
Figure 13
Small-signal circuit corresponding to Figure 8 to check analysis for Norton resistance RN
The circuit of Figure 13 contains no capacitances so a DC analysis is sufficient. The
circuit is linear, so the ratio of the voltage across the test source to the current in the test
source does not depend on the value of the current, which we take as 1 A to make
calculation easy. Then the resistance looking into the circuit is equal numerically to the
voltage at the input.Running the BIAS POINT analysis the results shown in Figure 13 indicate the Norton
resistance is RN = 888.43 k;, compared to 888.43 k; from the spreadsheet. Therefore,
the analysis of the circuit is accurate and the discrepancy in Figure 12 comes from
another source. It does not appear large enough to have practical importance.
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Unpublished work 2006 by John R Brews 15
Transient behavior
The spreadsheet has not been extended to treat behavior where the output transistor
is saturated. However, we can make transient analyses to compare with the behavior seen
in Figure 2, providing a check on the concepts behind the mirror design. We set up theDC bias using the parameter V_SAT, as shown in Figure 14. When V_SAT = 0V, the DC
bias is set below VB by the AC amplitude, so EQ. 24 is satisfied and the output transistor
always is active. As V_SAT is increased, the output transistor goes further and further
into saturation, and the mirror resistance falls rapidly.
INPUT SIGNAL
AMPLITUDE = 1V
V_A = {5-AMPLITUDE+V_SAT}
FREQUENCY = 1kHz
DOT-MODEL:
B_F = 224.9477
I_S = .6506fA
V_AF = 115.7V
N_F = 1.0089535r_X = 10
PARAMETERS:
V_CC = 10V
R_E = 418.85833
R_R = 495.59372
V_SAT = 0.346V
Figure 14
Introduction of V_SAT to describe how far into saturation the output transistor is driven at the topof the AC signal
Figure 15
Output current for various values of V_SAT
Figure 15 shows that only slight peaking of output current occurs at V_SAT =
0.346V, the point in Figure 2 where the mirror resistance begins to drop, but becomes
evident as V_SAT is increased to 0.55 V, the point in Figure 2 where the mirror
resistance has dropped substantially. For larger V_SAT, the peak increases very rapidly.
Detection of this peak in AC current is one way to determine the compliance voltage ofthe mirror.
Comparison with a more realistic transistor model
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Unpublished work 2006 by John R Brews 16
We replace the simple model of Figure 1 with one of the transistor models provided
with PSPICE, namely the Q2N2907A, as shown in Figure 16.8 The Q-point agrees fairly
closely with the spreadsheet because the dot-model parameters in Figure 1 were chosen
to agree with this transistor. However, the value of VEB =VEM is not the same as with the
simpler model Q_pVAF, VEM = 786 mV compared to 793 mV in Figure 1. The model
Q2N2907A is much more complex than the model Q_pVAF, and parameterL only
approximates its behavior. An investigation of just what leads to the discrepancy could be
a big job. It wont be done here.
0
Q_Out
Q2N2907A
-44.486uA
-10.007mA
+
- {V_CC}20.104mA
0
PARAMETERS:
V_CC = 10VR_E = 418.85833
R_R = 495.59372
V_SAT = 0.346V
5.0037V+
- {V_A}
V_DC
E
0
EM
Sweep
+
-
AC V_AC
1V
5.7897V 786.03mV
+{R_E}
10.052mA
-
+
Transient
Analysis
{AMPLITUDE}
{FREQUENCY}
V_SIN
-+
+
-
E1
GAIN = 1
10.000V
E
Q_Ref
Q2N2907A
-44.488uA
-10.007mA
M
0
+ {R_E}
10.052mA
M
5.7897V
5.0000V
+{R_R}
10.096mA
INPUT SIGNAL
AMPLITUDE = 1V
V_A = 5V
FREQUENCY = 1kHz
Figure 16
The pnp mirror with the dot-model statement for the Q2N2907A provided with PS PICE
The PROBE output file is shown in Figure 17 below. Comparison with Figure 11 shows
the small-signal parameters of the simple model agree, as we intended when setting L.
8 The Q2N2907A is pasted on the schematic by selecting the schematic, opening menu PLACE/PART and
typing Q2N2907A into the PART window. The EVAL library must be enabled.
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Unpublished work 2006 by John R Brews 17
Figure 17
PROBE output file for the mirror using Q2N2907A transistors
The small-signal Norton resistance is found in Figure 18. Unlike Figure 12,
Figure 18 shows frequency roll-off due to the parasitic capacitances of the Q2N2907A.
The ideal transistor model Q_pVAF in Figure 1 doesnt include any capacitances.
F r equency
1 . 0Hz 10KHz 100MHz 1. 0THz
1 / I ( V_AC)
0
0. 5M
1. 0M
( 7 . 7 46 K, 6 28 . 4 8K)( 1 Hz , 8 87 . 7 56 7K)
Figure 18
AC Norton resistance vs. frequency for Figure 16The value RN = 887.8 k; in Figure 18 is close to the spreadsheet prediction of RN =
888.4 k;.
Using the spreadsheet as a design tool
The example of Figure 1 satisfies a specification of 10 mA output current for
voltages below VA = 5 V. However, other specifications involving the output resistance
of the mirror could arise. To see what compromises are necessary, trade-off charts areeasily set up as shown in Figure 19 below.
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Unpublished work 2006 by John R Brews 18
496
4190
1000
2000
3000
4000
5000
6000
0 2 4 6 8 10 12 14
Output Current I_C (mA)
Leg
R_
E,
RefR_
R
(;
)
R_E
R_R
DESIGN
DESIGN
V_M (V) = 5
8.8843E+05
0.E+00
2.E+06
4.E+06
6.E+06
8.E+06
1.E+07
0 2 4 6 8 10 12 14
Output Current I_C (mA)
Norton
R_
N
(;
)
R_N
Design
V_M (V) = 5
8.8843E+05
9.2, 11570.0
1.E+03
1.E+04
1.E+05
1.E+06
1.E+07
0 2 4 6 8 10
Midbase Voltage V_M (V)
Norton
R_
N
(;
)
R_N
DESIGN
r_O
I_C (mA) = 10
419
496
0
200
400
600
800
1000
0 2 4 6 8 10
Midbase Voltage V_M (V)
Leg
R_
E,
RefR_
R
(;
)
R_E
R_R
DESIGN
DESIGN
I_C (mA) = 10
Figure 19
Trade-off charts based on the spreadsheet of Figure 9
As usual, we first try to understand the trends shown in the charts to gain some
understanding of the circuit behavior. Lets begin by thinking about the downward trends
in RN as either the mid-base voltage VM increases or the output current increases.
Trend of RN with mid-base voltage VM when IC = constant
For VM to increase, the voltage drop across RE must be reduced because VM is
mainly determined by VCC and the drop across RE. At a fixed output current the only way
to reduce this drop is by reduction of RE. EQ. 22 shows that reduction of RE reduces RN
because the contribution to RN from the term
RRERREFrXrrERER
ACOr // TF
drops as RE goes down. For RE = 0 ;, RN = rO, which is as low as it gets.
Trend of RN with output current IC when VM = constant
For IC to increase, the current on the left side of the mirror must increase. One way
this can happen is to reduce RR, because the current basically is determined by VM/RR.
However, such an increase in current will naturally tend to increase the drop across RE,which cannot happen if VM is maintained. Therefore, RE must drop as IC increases, and as
already observed, this causes RN to drop.
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Unpublished work 2006 by John R Brews 19
Trend of RR and RE with VM when IC = constant
It already is argued that RE drops as VM increases. Also, as VM increases the current
tends to increase because it is controlled by VM/RR. However, an increase in current is not
allowed if IC is held fixed, so RRmust increase to maintain the current. Thus, RRand REhave opposite trends, as shown in the upper right panel of Figure 19.
Trend of RR and RE with IC when VM = constant
From the above arguments RRand RE both must drop as IC increases to maintain
VM.
This discussion clarifies how RRand RE affect the mirror properties.
Example designAs a different design problem, lets consider a case where we want to specify VM
and RN but dont care much about the current IC, except it should be as large as possible.
To be specific, lets request that VM = 2 V and RN 1 M; for VA VM = 2 V.We
require the amplifier be built with Q2N2907 transistors.
Answer: The design is found using GOAL SEEK.We set VM = 2 V in the input
worksheet, and ask GOAL SEEKto set the output resistance at RN = 1 M; by varying
I_C_mA. See Figure 20.
Figure 20
Using GOAL SEEKto find correct IC for RN = 1 M;
The resulting design curves are shown in Figure 21.
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Unpublished work 2006 by John R Brews 20
155
560
0
2000
4000
6000
8000
0 2 4 6 8 10 12 14
Output Current I_C (mA)
Leg
R_
E,
RefR_
R
(;
)
R_E
R_R
DESIGN
DESIGN
V_M (V) = 5
1.0000E+06
0.E+00
5.E+06
1.E+07
2.E+07
0 2 4 6 8 10 12 14
Output Current I_C (mA)
Norton
R_
N
(;
)
R_N
Design
V_M (V) = 5
1.0000E+06
9.2, 9043.5
1.E+03
1.E+04
1.E+05
1.E+06
1.E+07
0 2 4 6 8 10
Midbase Voltage V_M (V)
Norton
R_
N
(;
)
R_N
DESIGN
r_O
I_C (mA) = 12.79
560
1550
200
400
600
800
0 2 4 6 8 10
Midbase Voltage V_M (V)
Leg
R_
E,
RefR_
R
(;
)
R_E
R_R
DESIGN
DESIGN
I_C (mA) = 12.79
Figure 21
Design curves with design point for the requested specs of VM = 2 V and RM = 1 M;
The spreadsheet suggests RE = 560 ; and RR= 155 ;, and that the output current
will be IC = 12.8 mA. This current is larger than the IC = 10 mA used to calibrate our
Q_pVAF model parameters, so our values forFs and L may be off a bit, requiring some
recalibration.We check out the design using PSPICE.
Figure 22
PROBE output file for design
Figure 22 shows the PROBE output file. At this current levelFDC and FAC still are
close to the values used before. The value of rT = 458 ; compares with a spreadsheet
value of rT = 459 ;, so the design shouldnt be too far off.
Figure 23 shows the Q-point at the design condition VA = 2 V = VM. Indeed VM is
very close to the designed-for VM = 2 V.
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Unpublished work 2006 by John R Brews 21
Q_Ref
Q2N2907A
-56.941uA
-12.803mA
0
0
E
-
+
Transient
Analysis
{AMPLITUDE}
{FREQUENCY}
V_SIN
M
0
Q_Out
Q2N2907A
-56.940uA
-12.803mA
INPUT SIGNAL
AMPLITUDE = 1V
V_A = 2V
FREQUENCY = 1kHz
E
0
EM
10.000V
M
2.7941V
2.0000V+
- {V_A}
V_DC
2.0015V
+ {R_E}
12.860mA
-+
+
-E1
GAIN = 1
+
- {V_CC}25.720mA
2.7941V
PARAMETERS:
V_CC = 10V
R_E = 560.33533R_R = 154.94888
V_SAT = 0.346V
Sweep
+
-
AC
V_AC1V
792.59mV
+{R_E}
12.860mA
+{R_R}
12.917mA
Figu
re 23
Q-point check on design when VA = VM = specified VM = 2 V
Finally, the real comparison of mirror characteristics is shown in Figure 24.
Figure 24
Mirror properties for example design
Figure 24 shows the mirror RN > 1 M; for VA 1.5 V, while the spec calls for
more room with RN > 1 M; for VA 2 V.We have a couple of choices to improve the
design. The first is simply to use the trade-off curves to pick better values for RE and RR.
The second is to recalibrate the model to see if that makes the spreadsheet design
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Unpublished work 2006 by John R Brews 22
adequate. Because the calibration doesnt seem too far off, the first choice looks more
promising.What has to be done to bring the design within specs?
The problem with the design is that VM is too small. So we could try the spreadsheet
again with a larger VM. That tends to lower the RN, so IC has to be adjusted. Iteration is
needed, so we set VM = 2.5 V and use GOAL SEEKto find the design with RN = 1 M;.
For this design the resistance plot is shown in Figure 25. The spreadsheet values are RE =
550 ; and RR= 204.5 ; with IC = 12.12 mA. This design meets the spec with RN 1
M; for VA < 2.3 V, but we are forced to a lower current.
V_A
0V 2V 4V 6V 8V 10V- 1 / D ( I ( V_D C ) )
0
0 . 5 M
1 . 0 M
( 2 . 275, 1 . 0 00M)( 2 . 8 99 , 7 20 . 5 2K)
( 3 . 0 83 , 1 0. 8 0K)
1 . 0 00 , 1 . 0 08M)
Figure 25Mirror resistance for iterated design
Comments
The current mirror, a circuit element ubiquitous in analog design, differs from an
ideal current source strictly because of the limitations of the bipolar transistors used to
build it. For that reason, a somewhat complex model for the transistor is needed tocapture the mirror behavior. Nonetheless, a simple spreadsheet is useful to design a
current mirror to meet specifications on current level, compliance voltage range and
output Norton resistance.
The spreadsheet enables easy exploration of design trade-offs. For example, the
designs presented indicate a trend to lower currents as compliance range is increased at a
fixed Norton resistance. Qualitative design decisions are based upon recognition of these
trade-offs, which can be discovered as well as made quantitative using a spreadsheet to
explore hand analysis. The spreadsheet is also an interface between hand analysis and
PSPICE to aid verification of the assumptions of hand analysis, and to alert us to wrinkles
that must be added to our thinking if we want reasonable designs.
Appendix: the diode-connected transistor
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Unpublished work 2006 by John R Brews 23
The small-signal equivalent circuit for QRef, which has the collector shorted to the
base (diode connection), is shown here to be a simple resistor, rREF. This simple resistor
replaces QRef in the small-signal circuit of Figure 8. The small-signal circuit for QRef is
shown in Figure 26.
++
+
Figure 26Small-signal circuit for reference transistor QRef
Because the base and collector are short-circuited, base and collector are the same
node. The voltage between collector and base is VT, so the currents in the various
branches are as shown in Figure 26. Consequently the total current flowing between
collector and emitter is EQ. 25 below:
EQ. 25
T
Frr
V1
r
VI
X
ceAC
O
cec
! .
That is, the transistor behaves like a resistor of value rREF given by EQ. 26:
EQ. 26
!
1
rr//rr
AC
XOREF
F
T
Rewrite this resistance in terms of the current as shown in EQ. 27:
EQ. 27
}
!
T
T F
F
L
F
r
r1
11
V
I
V
I
rr
1
r
1
r
1
XAC
AC
THF
C
AF
C
X
AC
OREF
.
Because VAF >> VTH (for example, for the Q2N2907A in Figure 1 at 27rC, VAF =
115.7 V and VTH 26 mV), the last term dominates the sum, and to a very good
approximation we can take the small-signal equivalent circuit of QRef to be a single
resistor of value rREF given by EQ. 28:
EQ. 28
Xr Tr Or
bACIF
E
C
+
ecV
X
ecb
rr
VI
!
T
cI
O
ec
r
V
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Unpublished work 2006 by John R Brews 24
mXTHF
CREF
g
1
r
r1
1
V
Ir }
}
T
L
Exercises
1. Go through the chapter using an n-channel mirror with a simple model Q_nVAF
using dot-model statement:
.model Q_nVAF NPN (Bf={B_F} Is={I_S} Vaf={V_AF} Nf={N_F} Rb={r_X}),
and instead of fitting the pnp Q2N2907A make the Q_nVAF match the
PSPICE-provided npn Q2N2222 transistor.
2. Use the spreadsheet to make trade-off plots using IC as the dependent variable and RN
and VM as independent x-axes. Describe how you obtain these charts. Discuss the
origins of trends in your charts.
3. Make a pnp mirror design using Q2N2907A transistors similar to the one in the
chapter but for specifications of RN 1 M; for VA 8 V.
Answer: Using the spreadsheet calibrated for IC = 10 mA, we find a design using GOAL
SEEKfor RE =306.5 ; and RR= 1.983 k; with IC = 4 mA.We check the F-values as
shown in Figure 27 and the ideality coefficient as shown in Figure 28 and update the
spreadsheet.
I _E
100uA 1. 0mA 10mA
I ( C) / I ( B) D( I ( I _C) ) / D( I ( I _B) )
2 00 . 0
2 12 . 5
2 25 . 0
2 37 . 5( AC Be t a , 4 . 0 00m, 224. 95 )
( DC Be t a , 4 . 0 00m, 223. 86 )
Figure 27
Check ofF-values at initial guessed IC 4 mA
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Unpublished work 2006 by John R Brews 25
I _E
0A 5mA 10mA
D( I ( C_Nf ) ) / D( V ( E _VAF ) ) D( I ( C _2 907A) ) / D( V( E _2907A) )
0
200m
400m
( 4. 00000m, 152. 3348m)
( 4. 00000m, 152. 9590m)
Figure 28
External gm-values for use in finding new L = 1.0040976
We then use GOAL SEEKagain to design to specification. See Figure 20. The new
design is RE = 305.7 ;, RR= 1971.8 ;, IC = 4.02 mA. These resistance values are put
into PSPICE to check the design. The PROBE output file is shown in Figure 29, and the Q-
point in Figure 30.
Figure 29
PROBE output file for design
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Unpublished work 2006 by John R Brews 26
8.0000V
E
+{R_E}
4.0405mA
0
+{R_R}
4.0585mA
+
- {V_CC}8.0810mA
-
+
Transient
Analysis
{AMPLITUDE}
{FREQUENCY}
V_SIN
M
INPUT SIGNAL
AMPLITUDE = 1V
V_A = 8V
FREQUENCY = 1kHz
EM-+
+
-E1
GAIN = 1
+
- {V_A}
V_DC
E
8.0026V
Q_Ref
Q2N2907A
-17.967uA
-4.0225mA
0
10.000V
0
Sweep
+
-
AC
V_AC1V
PARAMETERS:
V_CC = 10V
R_E = 305.73953R_R = 1971.82545
V_SAT = 0.346V
Q_Out
Q2N2907A
-17.967uA
-4.0225mA
8.7647V
+ {R_E}
4.0405mA
8.7647V
0
762.05mV
M
Figure30
Q-point for VA = 8 V
Figure 29 shows rT = 1.46 k;, compare to the spreadsheet value 1.45 k;, rO =
28.8 k; compared to 28.8 k;, good agreement. Figure 31 shows the mirror
characteristics.
Figure 31
Mirror characteristics for designThe mirror satisfies RN 1 M; for VA 8.2 V, exceeding the 8V specification. The
mirror provides only 4 mA of current however, compared to 12 mA for the design
presented in the chapter.
Software elements presented
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CAPTURE andPSPICE
Finding AC and DC betas (Figure 3)
Introduction to the pnp dot-model statement:
.model Q_pVAF PNP (Bf={B_F} Is={I_S} Vaf={V_AF} Nf={N_F} Rb={r_X})
Introduction to the pnp dot-model statement for the Q2N2907A transistor.
Dealing with base resistance rX
Finding the ideality factorL to match a more realistic bipolar model (Figure 7)
Another example implementing a small-signal circuit (Figure 13)
EXCEL
Another example of implementing a hand analysis in the spreadsheet, verifying it and
using it for design