STABILISER CIRCUITS
EXAMPLE: VOLTAGE STABILISER
Using a 400 mW/4.7 V breakdown diode and a resistor, design a simple, stabilised voltage supply to
deliver a constant 10 mA current from a 12 V power supply. Allow a 10 mA zener current.
Solution
• The constant output voltage is set by the breakdown voltage of the diode
i.e Vo = Vz = 4.7 V
• Total current needed through the series resistor is
I = IL + IZ = 10 mA + 10 mA = 20 mA
• The corresponding voltage across Rs is
VR = Vi - Vo = 12 V - 4.7 V = 7.3 V
• Series resistor needed to produce this is
Rs = (Vi - Vo)/I = 7.3 V/20 mA = 365 Ω
• A SV of 330 Ω will make sure that the minimum load current is always delivered
• Now check that power rating of diode is not exceeded.
• Max current through diode occurs if the load fails and all current flows through diode
i.e Iz(max) = (Vi - Vo)/ Rs = 7.3 V/330 Ω = 22 mA
Power dissipated by diode is
Pmax = Vz Iz(max) = 4.7 V x 22 mA = 104 mW
This is within limits.
EXAMPLE: VOLTAGE STABILISER
If the previous supply actually fluctuates by ±0.5 V, how much does the output vary by if the
breakdown diode has a dynamic resistance of 40 Ω at 10 mA?
Solution
• The variation in output voltage is given by
∆Vz = [rz/(rz+Rs)]∆Vi
= [40 Ω / (40 Ω + 330 Ω)] x ±0.5 V = ±54 mV
SR = 54 mV/500 mV = 0.11
EXAMPLE: VOLTAGE STABILISER
Design a circuit which will stabilise the output of a power supply which fluctuates in the range 15-20
V and needs to always deliver 50 mA to the load. You have available a 5.1 V, 1.3 W breakdown
diode with a dynamic resistance rz = 7 Ω at 10 mA_
Solution
• Max permissible current through the diode is limited by its’ power dissipation
Iz,max = Pz/Vz
Rs
I
IL = 50 mA
= 1.3 W/5.1 V
= 255 mA
Iz
• Total current which must always
Vi
Vz
flow through Rs
Vo
5.1
V
I = Iz + IL
15 - 20 V
= 10 mA + 50 mA
= 60 mA
• Calc the value of Rs to allow this current to flow
→ the worst case is when Vi is at its' min. value of 15 V, (since this is the hardest case to achieve
the desired current), thus
Rs = [Vi(min)-Vz]/I = (15 V - 5.1 V)/60 mA = 165 Ω
→ Choose 150 Ω SV to ensure current always delivers at least 60 mA
Imin = (Vi-Vz)/Rs = (15 V - 5.1 V)/150 Ω = 66 mA
→ load draws 50 mA, diode takes the excess 16 mA
• Output regulation
→ if Vi increases to 20 V, then I increases to 99 mA (for 150 Ω)
I'min = (Vi-Vz)/Rs = (20 V - 5.1 V)/150 Ω = 99 mA
∆Iz = 99 mA - 66 mA = 33 mA
∆Vo = rz ∆Iz = 7 Ω x 33 mA = 231 mV
SR = ∆Vo/∆Vi = 231 mV/5 V ≈ 4.6%
SR = [rz/(rz+Rs] = 7 Ω/(7 Ω + 150 Ω) = 0.045 (4.5%)
• A 28% change in i/p (5 V/17.5 V) voltage produces a 4.5% o/p change
• Check power dissipation of diode (when load fails and all current flows through diode)
Pz = Vz Imax = 5.1 V x 99 mA = 505 mW
This diode is within limits.