Download Example: Single-Supply

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Example: Single-Supply
DC Bias
Consider this small-signal amplifier:
15 V
R1
15 V
RC
vO (t )
  100

vi (t )
+
-
R2
RE

Say we decide that the DC collector current should be
IC  5mA .
Let’s find the resistor values for R1, R2, RC and RE that properly
bias this amplifier!
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Step 1: Write the DC Circuit Schematic
After all, we are designing the DC bias!
15 V
15 V
RC
R1
IC
+
VCE
-
R2
RE
IE
Step 2: Enforce the design goals for VE and VC
Recall that our DC bias “rule-of –thumb” was to divide the VCC
voltage into “thirds” so that:
VE VCC 3 
and
VC  2VCC 3 
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Since we want IC  5mA , we find that the collector resistor
must be:
RC 
VCC VC

IC
Likewise, the emitter resistor is:
RE 
VE

 VE 
IE IC
Step 3: Choose I1 and find R1 and R2
Recall our “rule-of-thumb” for the current I1 is:
0.1 IC  I1  IC
Let’s pick a value in the middle, i.e.:
I1  0.5 IC  2.5mA
Since we know that that the base voltage is approximately:
VB  0.7 VE  5.7 V
and we know that the base current is:
IB 
IC


5.0
 0.05 mA
100
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we can thus determine resistor R1 :
R1 
15.0  5.7
2.5
 3.72 K
15 V
R1
15.0 VB
I1

I1  2.5 mA
Likewise, since we know that the
current I2 is:
IB  0.05 mA
VB  5.7 V
I 2  I1  I B
 2.5  0.05
R2
 2.5 mA
I2
we can find the second resistor R2 :
R2 
VB 5.7

 2.28 K
I2 2.5
Therefore, our completed amplifier design is:
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15 V
15 V
1K
3.7 K
vO (t )
  100

vi (t )
+
-
5.0 mA
2.3K
1K
