SME
Electronics · Problem type
Zener regulator
Use this for “A 5.1 V zener regulates 12 V for a 20 mA load, keeping 5 mA in the zener. Find R.”
Diagram
Vₛ − VZ
R
=
12 V − 5.1 V
276
Ω
Vₛ − VZ
R
=
IL
+
IZ
PZ
=
VZ(Vₛ − VZ)
R
PZ
=
5.1
V
×
12 V − 5.1 V
276
Ω
PZ = 127.5 mW
The zener holds VZ across the load; R drops the rest, and its current splits between the zener and the load.
PZ is with no load.
This is an example. Type your own values to start a new problem.
R (Ω)
=
Vₛ (V)
−
VZ (V)
IL (mA)
+
IZ (mA)
PZ (mW)
=
VZ (V)
×
Vₛ (V)
−
VZ (V)
R (Ω)
Formulas
R
=
Vₛ − VZ
IL + IZ
0.276
=
12 − 5.1
20 + 5
PZ
=
VZ
×
Vₛ − VZ
R
127.5
=
5.1
×
12 − 5.1
0.276
Formulas are worked in kΩ. The step-by-step shows the conversions.
Assumptions
•
The zener holds VZ across the load while at least IZ flows through it.
•
R carries the load current and the zener current together.
•
PZ is the worst case: with the load removed, the zener takes all of R’s current.
•
V ÷ kΩ gives mA, and V × mA gives mW.
Step-by-step
WE KNOW
Vₛ = 12 V, VZ = 5.1 V, IL = 20 mA, IZ = 5 mA
FIND
series resistance (R), zener power with no load (PZ)
Step 1 · Find series resistance (R)
Use R = (Vₛ − VZ) ÷ (IL + IZ)
R
drops
what
the
zener
doesn’t
and
carries
both
currents:
V
mA
gives
kΩ.
R
=
12 − 5.1
20 + 5
R
=
6.9
25
R
=
0.276
kΩ
Step 2 · Find zener power with no load (PZ)
Use PZ = VZ × (Vₛ − VZ) ÷ R
With
no
load
the
zener
carries
all
of
R
’s
current,
Vₛ − VZ
R
:
V
×
mA
gives
mW.
PZ
=
5.1
×
12 − 5.1
0.276
PZ
=
5.1
×
6.9
0.276
PZ
=
35.19
0.276
PZ = 127.5 mW
Step 3 · Convert the answers
Change each answer back to the unit you chose.
R = 0.276 kΩ = 276 Ω (1 kΩ = 1000 Ω)
Check
Put the numbers back in. Both sides should match.
0.276
=
12 − 5.1
20 + 5
✓
127.5
=
5.1
×
12 − 5.1
0.276
✓
Topic check
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