Download Single Supply OpAmp Example

J.P.O’Rourke
Using the Linear Transfer Function to solve single supply op-amp designs.
The vast majority of projects over the past few years stress the need for portability and devices
which run off a single supply. From the design reviews it appeared that many designs were
attempting to power dual supply op-amps from a single supply. For the record, dual supply opamps are not designed to operate properly off of a single battery supplies. All though DC to DC
converters offer a means of obtaining dual supplies from a single battery. A more economical
and less sophisticated solution would be to use devices specifically design to run off a single
battery.
This handout will specifically address analytic solutions to non-inverting and inverting designs
using the linear transfer function model of the op-amp. Using two examples, a solution for each
will be obtained and then tested out using Multisim to verify both designs.
Let’s get started by noting that the linear op-amp transfer function will be limited to the equation
of a straight line. The equation has four possible solutions depending on the sign of the slope(m)
and the y intercept(b).
Y  mX  b
(1)
Let’s select one of those solutions of the transfer function for the inverting mode of operation.
Vo  mVi  b
(2)
Next we will define a set of two data points on this straight line that fall within the operation
range of the selected op-amp. So it is very important that you have a curve noting the linear
operating range of the op-amp of choice. The following Transfer Curve was obtained from data I
measured in the laboratory using a LMC6484 OP-Amp with a 9.0 volt supply.
These devices work very well provided appropriate limitations are observed in the transfer
function. The LMC6484 is a newer op-amp which was designed for rail-to-rail operation and as
the graph indicates with textbook performance that justifies the use of ideal assumptions.
Select a set to two data points for Vo (output voltage) and Vi (input voltage) with limitations
noted above. These points will define a set of simultaneous equations that can be solved to
determine m and b that will satisfy the given data.
Let’s assume an inverting OP-Amp configuration with an input sensor whose output ranges from
0.01 volts to -0.01 volts and needs to interface with an A/D converter which has an input range
of 0 volts to 8.0 volts(the OP-Amps output range). The power supply and voltage reference will
be 9.0 volts and the load 10K. This time we will be using the LMC6484 in the inverting mode.
To do this we will use the following linear transfer function for this mode of operation.
VO  mvi  b
(3)
The following circuit configuration will be used to implement this transfer function.
XSC1
XFG1
R2
Ext Trig
+
_
390kΩ
0
+
0
5
11
R1
2
2
4
4
0
V1
9V
0
0
_
LMC6484AIN
1
R410kΩ
12Ω
+
3
1
R3
_
U1A
1.0kΩ
3
B
A
R5
10kΩ
0
Using the above circuit, we can derive the output voltage as sum of the inverting and noninverting input voltages. The inverting contribution would be,
R 
VOinv    2 Vi
 R1 
(4)
The non-inverting contribution would be,
 R4   R2 
VOnon  Vref 
   1
 R4  R3   R1 
(5)
The final output then would be,
R 
VO    2 Vi  Vr
 R1 
(6)
 R4   R2 
R 
VO   2 Vi  Vref 
   1
 R1 
 R4  R3   R1 
(7)
Comparing the above equation to the straight line transfer function equation (24), yields,
m
R2
R1
and
 R4   R2 
b  Vref 
   1
R

R
3   R1

 4
(8a), (8b)
Next, we will write down the input/output requirements in a more logical manner making it
easier to insert them into the transfer function equation.
VO  0volts @Vi  0.01volts
(9)
VO  8.0volts @Vi  0.01volts
Vref  9.0volts
(10)
(11)
Inserting these data points into the transfer function equation (3), you get
These are the two simultaneous equations you need to solve.
0  0.01m  b
8.0  0.01m  b
(12)
.
(13)
Solving equation (12) for b,
b  0.01m
(14)
Substituting equation (14) into equation (13) above,
8.0  0.01m  0.01m  0.02m
(15)
Yielding,
m  400
(16)
Substituting m =400 into equation (14) and solving for b,
b  0.01m  (0.01)( 400)  4
(17)
Substituting m= 400 and b=4 into equation (4) yields the solution to the straight line plot on
the LMC-6484 's input/output graph .
VO  400Vi  4
(18)
Now substituting the results of equations (16) and (17) into equation (8a) and (8b). The results
for equation (8a),
400 
R2
R1
(19)
Lets chose a value for R1 since it allows us to define the OP-Amps input resistance. However,
remember that the feedback resistor R2 is equal to the voltage gain (here called "m") times R1 .
So for accuracy here it should not exceed a few megohms.
For this circuit R1 = 1K then,
𝑅2 = (𝑚)(𝑅1 ) = (400)(1𝐾) = 400𝐾
(The closest standard value is 390K ohms)
(20)
Now the results for equation (8b),
 R4   R2 
4  Vref 
   1
 R4  R3   R1 
(21)
 R4   400K 
4  (9) 
 1

R

R
1
K


3
 4
 R4 
4  (9)(401) 

 R4  R3 
(22)
Since there are two unknowns a value must be chosen for one of the resistor values, here
R3 is chosen to equal 10K.
𝑅3 = 10𝐾
(23)
Substituting R3 = 10K into equation (22),
 R4

4  (3609) 

 R4  10 K 
(24)
Solving the above,
R4 = 11.1 ohms
( The closest standard value is 12 ohms)
(25)
Again we see that the circuit simulates as expected with the output being a linear inverted
amplification of the sinewave input.
JOR/2012