SME
Embedded Systems · Problem type
A divider down to 3.3 V
Use this for “A 5 V signal reaches a 3.3 V pin through 2.2 kΩ over 3.3 kΩ. What voltage does the pin see?”
Diagram
Vᵢₙ = V₁ + Vₒᵤₜ = 2 V + 3 V
Vᵢₙ = 5 V
I
=
Vᵢₙ
R₁ + R₂
=
5
V
2.2 kΩ + 3.3 kΩ
I = 0.9091 mA
Drag the source or a resistor up or down to change it.
This is an example. Type your own values to start a new problem.
Vₒᵤₜ (V)
=
Vᵢₙ (V)
×
R₂ (kΩ)
R₁ (kΩ)
+
R₂ (kΩ)
V₁
Voltage across R₁
calculated
I
Divider current
worked out (not typed)
Formulas
Vₒᵤₜ
=
Vᵢₙ
×
R₂
R₁
+
R₂
3
=
5
×
3.3
2.2 + 3.3
V₁
=
Vᵢₙ
−
Vₒᵤₜ
2 = 5 − 3
I
=
Vᵢₙ
R₁
+
R₂
0.9091
=
5
2.2 + 3.3
Assumptions
•
R₁ runs from the signal to the pin and R₂ from the pin to ground, so they are in series.
•
The input pin draws almost no current, so the same current I flows through both resistors.
•
The pin’s voltage must stay at or below the chip’s supply, here 3.3 V.
Step-by-step
WE KNOW
Vᵢₙ = 5 V, R₁ = 2.2 kΩ, R₂ = 3.3 kΩ
FIND
pin voltage (Vₒᵤₜ), voltage across R₁ (V₁), divider current (I)
Step 1 · Find pin voltage (Vₒᵤₜ)
Use Vₒᵤₜ = Vᵢₙ × R₂ ÷ (R₁ + R₂)
R₂
takes
its
share
of
the
signal:
its
part
of
the
total
resistance.
Vₒᵤₜ
=
5
×
3.3
2.2 + 3.3
Vₒᵤₜ
=
5
×
3.3
5.5
Vₒᵤₜ
=
16.5
5.5
Vₒᵤₜ = 3 V
Step 2 · Find voltage across R₁ (V₁)
Use V₁ = Vᵢₙ − Vₒᵤₜ
Kirchhoff’s
voltage
law:
R₁
drops
what
R₂
leaves
of
the
signal.
V₁
=
5
−
3
V₁
=
2
V
Step 3 · Find divider current (I)
Use I = Vᵢₙ ÷ (R₁ + R₂)
Ohm’s law for the two resistors in series: the signal over their total.
I
=
5
2.2 + 3.3
I
=
5
5.5
I
=
0.9091
mA
Check
Put the numbers back in. Both sides should match.
3
=
5
×
3.3
2.2 + 3.3
✓
2 = 5 − 3 ✓
0.9091
=
5
2.2 + 3.3
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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