SME
Electronics
Diodes
Use this for “A silicon diode and a 1 kΩ resistor are in series with 5 V. Find the current.”
Diagram
I
=
5 − 0.7
1
=
4.3
mA
This is an example. Type your own values to start a new problem.
VR (V)
=
Vₛ (V)
−
VD (V)
I (mA)
=
VR (V)
R (kΩ)
PD (mW)
=
VD (V)
×
I (mA)
Formulas
VR = Vₛ − VD
4.3 = 5 − 0.7
I
=
VR
R
4.3
=
4.3
1
PD
=
VD
×
I
3.01 = 0.7 × 4.3
Assumptions
•
Constant-drop model: the diode takes VD whenever it conducts (0.7 V for silicon, 1.8–3.3 V for LEDs).
•
The diode points with the current, so it is forward biased.
•
V ÷ kΩ gives mA, and V × mA gives mW.
Step-by-step
WE KNOW
Vₛ = 5 V, VD = 0.7 V, R = 1 kΩ
FIND
resistor voltage (VR), current (I), diode power (PD)
Step 1 · Find resistor voltage (VR)
Use VR = Vₛ − VD
Kirchhoff’s voltage law: the diode takes its drop and the resistor takes the rest.
VR = 5 − 0.7
VR = 4.3 V
Step 2 · Find current (I)
Use I = VR ÷ R
Ohm’s
law
for
the
resistor;
the
diode
carries
the
same
current.
V
kΩ
gives
mA.
I
=
4.3
1
I
=
4.3
mA
Step 3 · Find diode power (PD)
Use PD = VD × I
The diode’s drop times its current: V × mA gives mW.
PD = 0.7 × 4.3
PD = 3.01 mW
Check
Put the numbers back in. Both sides should match.
4.3 = 5 − 0.7 ✓
4.3
=
4.3
1
✓
3.01 = 0.7 × 4.3 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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