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Measurements & Instrumentation
Signal conditioning and data acquisition
Use this for “A 12-bit ADC spans 0 to 10 V. What is one LSB, the code for 3.3 V and the quantization error?”
Chart
12
bits:
4096
codes,
0
to
4095;
1
LSB
=
10
V
4096
=
2.441
mV
3.3
V
2.441
mV
=
1,352,
rounded
down:
D
=
1351;
back,
D
×
LSB
=
3.298
V
Zoomed to codes 1344 to 1359 of 0 to 4095
This is an example. Type your own values to start a new problem.
Q (mV)
=
FSR (V)
2
D
=
⌊
Vᵢₙ (V)
Q (mV)
⌋
eq
Quantization error
calculated
Formulas
Q
=
FSR
2
n
0.002441
=
10
2
12
D
=
⌊
Vᵢₙ
Q
⌋
1,351
=
⌊
3.3
0.002441
⌋
eq
=
Q
2
0.001221
=
0.002441
2
Formulas are worked in V. The step-by-step shows the conversions.
Assumptions
•
A unipolar converter: the codes 0 to 2ⁿ − 1 cover 0 up to the full-scale range.
•
It truncates: the code counts the whole steps of Q below the input.
•
Reading a code as the middle of its step, the error is at most ±Q ÷ 2 (half an LSB).
•
The input stays inside the range; at FSR and above the code sticks at the top.
Step-by-step
WE KNOW
n = 12, FSR = 10 V, Vᵢₙ = 3.3 V
FIND
resolution (1 LSB) (Q), output code (D), quantization error (eq)
Step 1 · Find resolution (1 LSB): Q
Use Q = FSR ÷ 2ⁿ
n
bits
make
2ⁿ
codes,
so
the
range
is
cut
into
2ⁿ
equal
steps.
Q
=
10
2
12
Q
=
10
4,096
Q
=
0.002441
V
Step 2 · Find output code (D)
Use D = ⌊Vᵢₙ ÷ Q⌋
Count the whole steps below the input; the converter drops the part of a step left over.
D
=
⌊
3.3
0.002441
⌋
D
=
⌊3.3
×
4,096
10
⌋
D
=
⌊1,351.68⌋
D
=
1,351
Step 3 · Find quantization error (eq)
Use eq = Q ÷ 2
Every input in a step reads as its middle, which is never more than half a step away.
eq
=
0.002441
2
eq = 0.001221 V
Step 4 · Convert the answers
Change each answer back to the unit you chose.
Q = 0.002441 V = 2.4414 mV (1 V = 1000 mV)
eq = 0.001221 V = 1.2207 mV (1 V = 1000 mV)
Check
Put the numbers back in. Both sides should match.
0.002441
=
10
2
12
✓
1,351
=
⌊
3.3
0.002441
⌋
✓
0.001221
=
0.002441
2
✓
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