SME
Electronics · Problem type
Shockley’s diode equation
Use this for “A diode with IS = 10⁻¹⁴ A and n = 1 has 0.65 V across it. Find the current.”
Diagram
At V = 0.65 V the diode carries 0.832 mA.
Ten times the current takes 59.5 mV more.
This is an example. Type your own values to start a new problem.
I (mA)
=
IS (A)
×
(exp
V (V)
×
VT (mV)
−
1)
ΔV (mV)
=
n
×
VT (mV)
×
ln
10
Formulas
I
=
IS
×
(
e
V
÷ (
n
× VT)
−
1)
0.0008325
=
1 ×
10
-14
×
(
e
0.65 ÷ (1 × 0.02585)
−
1)
ΔV
=
n
×
VT
×
ln(10)
0.05952 = 1 × 0.02585 × ln(10)
Formulas are worked in V, A. The step-by-step shows the conversions.
Assumptions
•
The thermal voltage is VT = kT ÷ q = 25.85 mV at 300 K.
•
The ideality factor n is 1 for an ideal junction and up to 2 for a real one.
•
Ten times the current takes ΔV = nVT ln 10 more, whatever the current.
Step-by-step
WE KNOW
IS = 1 × 10⁻¹⁴ A, n = 1, VT = 25.85 mV, V = 0.65 V
FIND
diode current (I), voltage for ten times the current (ΔV)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to V, A).
VT = 25.85 mV = 0.02585 V (1 V = 1000 mV)
Step 2 · Find diode current (I)
Use I = IS × (eV ÷ (n × VT) − 1)
Shockley’s equation: the current grows by e for every nVT across the diode.
I
=
10
-14
×
(
e
0.65 ÷ (1 × 0.02585)
−
1)
I
=
10
-14
×
(
e
25.14507
−
1)
I
=
10
-14
×
(8.3246
×
10
10
)
I
=
0.0008325
A
Step 3 · Find voltage for ten times the current (ΔV)
Use ΔV = n × VT × ln(10)
ΔV
=
nVT
ln(
I₂
I₁
),
with
the
current
ratio
10.
ΔV = 1 × 0.02585 × ln(10)
ΔV = 0.02585 × 2.303
ΔV = 0.05952 V
Step 4 · Convert the answers
Change each answer back to the unit you chose.
I = 0.0008325 A = 0.8325 mA (1 A = 1000 mA)
ΔV = 0.05952 V = 59.5218 mV (1 V = 1000 mV)
Check
Put the numbers back in. Both sides should match.
0.0008325
=
10
-14
×
(
e
0.65 ÷ (1 × 0.02585)
−
1)
✓
0.05952 = 1 × 0.02585 × ln(10) ✓
Topic check
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