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```ELEC 351L
Electronics II Laboratory
Spring 2004
Lab #8: Input and Output Resistances of BJT Diff Amps
Introduction
In order to employ an amplifier as part of an electronic system, it is important to know the gain
of the amplifier. It is almost always equally important to know its input and output resistances as
well, because these two parameters dictate how much voltage will be developed across the input
of the amplifier and across its load. Many methods can be used to measure the input and output
resistances of a wide variety of amplifier circuits; however, diff amps can present a measurement
challenge. Some of the special properties of diff amps (for example, the requirement to maintain
zero input bias voltages) preclude the use of more straightforward techniques. In this lab
experiment you will investigate some of the special considerations involved in measuring the
input and output resistances of a diff amp.
Theoretical Background
Figure 1 shows the basic diff amp circuit that was investigated in the previous lab exercise. Two
resistors (RS1 and RS2) have been added to represent possibly non-negligible source resistances,
that is, the Thévenin equivalent resistances of one or both signal sources. If a single floating
VCC
RC
RC
vOUT1
vOUT2
RS1
RS2
vs1
IREF
R1
Q1
Q2
vin1
vs2
vin2
Io
QA
QB
VEE
Figure 1. Differential amplifier with resistor pull-up loads. Resistors RS1 and RS2
represent the output resistances of the two signal sources.
1
(i.e., differential) source were driving the circuit, RS1 and RS2 would be combined into a single
source resistance RS. In Figure 1, the equivalent source voltages are represented by node
voltages vs1 and vs2, and the actual input voltages of the diff amp are represented by vin1 and vin2.
The single-ended input resistances rin1 and rin2 can be determined with the aid of the small-signal
model of the diff amp shown in Figure 2. Implicit in Figure 2 is the assumption that Q1 and Q2
are matched, as are the pull-up resistors (RC). The BJT output resistances rce have been included
to allow the derivation of more accurate expressions for the output resistances. Each input has
its own single-ended input resistance (rin-se1 and rin-se2), but because of the symmetry of the
circuit, they are equal in value. The amplifier also has a differential input resistance rin-diff, but
we will not consider it here because it is especially difficult to measure.
vout1
vout2
vin1
vin2
rce
rce
RC
r
RC
oib1
r
oib2
ib1
ib2
in
rn
Figure 2. Small-signal model of the diff amp. The triangles represent ground
connections. They are used instead of the traditional symbol in order to save space.
The single-ended input resistance of the first port is defined as
rin se1 
vin1
,
ib1
where it is assumed that vin2 = 0. (All independent sources must be deactivated in order to find a
Thévenin equivalent resistance.) An expression for rin-se1 can be found easily by applying KVL
to obtain
vin1  ib1r  ib 2 r .
Recall that, when vin2 = 0, the right-hand side of the circuit acts in the small-signal sense like a
very low resistance (r / o) in parallel with rn. Thus, almost none of ib1 flows through rn, and
ib2 ≈ −ib1. Consequently,
vin1  ib1r  ib1r  2ib1r ,
2
and
rin se1 
2ib1r
 2r .
ib1
Likewise, rin-se2 = 2r.
The single-ended output resistance rout-se1 is found by deactivating all of the input voltages and
applying a test source vt to the first output port, as shown in Figure 3. The output resistance is
then defined as
v
rout se1  t .
it
i1
it
rce
r
rce
+
−
RC
vt
oib1
RC
ve
ib1
r
oib2
ib2
rn
Figure 3. Small-signal model of the diff amp with test source used to find the singleended output resistance of port #1.
Applying KCL to the upper-left node of Figure 3 (i.e., to the positive side of vt) yields
it 
vt
 i1 ,
RC
where
i1 
vt  v e
  o ib1 ,
rce
and where ve is the small-signal voltage across rn. The right-hand side of the circuit (Q2) acts as
an equivalent small-signal resistance of r / o in parallel with rn, which is also in parallel with r
on the left-hand side. Consequently, the approximate equivalent resistance req measured from
the ve node to ground is given by
3
r
req  r rn
Since i1 also flows through req,
i1 
o

r
o
.
ve ve  o

.
req
r
Equating this to the other expression for i1 given above yields
i1 
v e  o vt  v e
v  ve
v

  o ib1  t
 o e ,
r
rce
rce
r
where we have used the relationship
ib1  
ve
.
r
Solving for ve,
v e  o vt  v e
v

 o e
r
rce
r
 ve 

2ve  o vt ve


r
rce rce
 2
v
1
  o  ve  t
rce 
rce
 r
vt
vt
r
1

 vt
,
2  o rce
1
rce 2  o
2  o rce

1
r
rce
r
which implies that ve is a tiny fraction of vt. Thus, we can assume that the voltage across rce is
approximately equal to vt. Now we can express it totally in terms of vt as
it 
v  v
vt
v
v
v
v
 
r 
 ,
 t   o ib1  t  t   o  e   t  t  o  vt
RC rce
RC rce
 r  RC rce r  2 o rce 
which reduces to
it 
vt
v
v
v
v
 t  t  t  t .
RC rce 2rce RC 2rce
The single-ended output resistance is therefore
rout se1 
vt
1

 RC 2rce  .
1
1
it

RC 2rce
4
If RC is large enough, the existence of the resistance rce could reduce the output resistance by a
significant amount. The presence of rce also leads to gain values that are lower than expected.
The differential output resistance rout-diff can be found using the small-signal model shown in
Figure 4. By symmetry, i1 = i2 and ib1 = ib2. The latter is only possible if ve = 0 (i.e., the ve node
becomes a virtual ground). This in turn implies that ic1 = −ic2. Applying KCL yields
it  ic1  i1 ,
and applying KVL yields
vt  ic1 RC  ic 2 RC  2ic1 RC and vt  i1rce  i2 rce  2i1rce .
The latter is true because oib1 = oib2 = 0. Thus,
vt
v
 t .
2 RC 2rce
it 
The differential output resistance is therefore
rout diff 
vt

it
i1
1
1
1

2 RC 2rce
 2RC rce  .
it
i2
+ −
rce
r
ic1
vt
RC
oib1
ve
ib1
ic2
rce
RC
r
oib2
ib2
rn
Figure 4. Small-signal model of the diff amp with test source used to find the differential
output resistance.
5
Experimental Procedure

Reconstruct the diff amp circuit you investigated in Lab #7 (Figure 1), using the CA3046
transistor array and the same bias conditions and power supply voltages. Remember to
include suitable bypass capacitors across the power supply leads.

Given the known circuit quantities and estimates based on the data sheet, calculate expected
(analytical) values for the single-ended input and output resistances rin-se1 and rout-se1. Also
determine the expected differential output resistance rout-diff.

Assume that the bias conditions of the amplifier will not be affected significantly as long as
the node voltages at the bases of Q1 and Q2 in Figure 1 are no higher than 10 mVDC. Use
reasonable approximations of unknown quantities (or even better, consult the data sheet) to
determine how much resistance could be connected between the base of each BJT and
ground without raising the DC bias voltage at either base to more than 10 mV. The data
sheet for the CA3046 is available at the ELEC 351 Lab web site.

Apply a sinusoidal input voltage at a frequency of 10 kHz to input vin1 of the diff amp. Use
the voltage divider circuit shown in Figure 5 to limit the input voltage. Devise a method to
measure the single-ended input and output resistances rin-se1 and rout-se1. Compare the
measured values to the analytical values calculated previously. Offer a plausible explanation
for any discrepancy.
RS
RA
vin1
vs
+
−
50 
RB
Figure 5. Voltage divider for reducing the output voltage from a function
generator. Devices vs and RS together represent the function generator. Resistor
values RA and RB must be chosen to reduce the voltage by an appropriate amount
while simultaneously creating an appropriate source resistance.

Measure the differential-mode output resistance rout-diff. You will need to devise a method to
ensure that the bias conditions of the diff amp are preserved while you make your
measurement. Compare the measured values to the analytical values calculated previously.
Offer a plausible explanation for any discrepancy.
6
```
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