AN61

A Product Line of
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AN61
Designing with References Extending the operating voltage range
Peter Abiodun A. Bode, Snr. Applications Engineer, Diodes Zetex Ltd
Introduction
There may be times when it is required to shunt-regulate a higher voltage than the 3-terminal
reference is designed for. This may be either as a stand alone shunt regulator or it may be as part
of a series regulator arrangement.
The following circuits offer a number of suggestions on how this might be done.
Figure 1 simply cascades two references together. The output voltage is the sum of the two stages
combined. It is worth noting that the output voltage and its accuracy is affected by both reference
devices as well as the four resistors R1, R2, R4 and R5 meaning the errors are cumulative.
Vin
R3
IL
I R3
IKA
REF2
V X2
VREF2
VOUT = V X 1 + V X 2
Vout
R1 ⎞
R4 ⎞
⎛
⎛
= VREF 1 ⎜1 +
⎟ + VREF 2 ⎜1 +
⎟
R2 ⎠
R5 ⎠
⎝
⎝
R4
Assuming VREF1 = VREF2 = VREF
R5
C1
Then,
R1
R3 =
REF1
V X1
VREF1
R2
R1 R 4 ⎞
⎛
VOUT = VREF ⎜ 2 +
+
⎟
R 2 R5 ⎠
⎝
VIN − VOUT
IR 3
Conditions:
IKA(min) < I R 3 ≤ IKA(max)
VREF ≤ (V X 1,V X 2 ) ≤ VKA(max)
GND
Figure 1 Higher voltage shunt regulator
This is not the case for either Figure 2 or Figure 3. In both cases, a device (another reference or a
zener diode) is used to drop any excess voltage within the circuit whilst a reference, REF1, is
primarily responsible for regulation.
The top device is "invisible" to the output voltage and is there purely as a protective device for the
controlling reference to take up the excess voltage that would otherwise damage the bottom device. The output voltage and its accuracy are entirely determined by the controlling reference, R1
and R2.
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I R3
R3
Vin
IL
R1 ⎞
⎛
VOUT = VREF 1 ⎜1 +
⎟
⎝ R2 ⎠
Vout
IKA
REF2
R4
R4 ⎞
⎛
V X = VREF 2 ⎜1 +
⎟
R5 ⎠
⎝
R5
R3 =
VX
VREF2
C1
R1
VREF1
VIN − VOUT
IR3
Conditions:
REF1
I KA(min) ≤ I R 3 ≤ I KA(max)
R2
(VKA(min) + V X ) ≤ VOUT ≤ (VKA(max) + V X )
GND
Figure 2 Improved higher voltage shunt regulator
R3
Vin
I R3
2.7k
IL
R1 ⎞
⎛
VOUT = VREF ⎜1 +
⎟
R2 ⎠
⎝
Vout
IKA
VZ ( nom ) = VOUT − VKA
VZ
ZD1
68V
VRE F
R3 =
R1
390k
Conditions:
AP431
C1
0.1μF
VIN − VOUT
IR3
R2
10k
I KA(min) ≤ I R 3 ≤ I KA(max)
GND
(VKA(min) + VZ ) ≤ VOUT ≤ (VKA(max) + VZ )
Figure 3 Higher voltage shunt regulator with no limit
Calculated Example 1
Requirement
Supply Voltage: 60V to 75V
Output voltage: 50V
Load current:
5mA
Assume the use of AP431.
Discussion
The required voltage of 50V is higher than what could be handled by a single reference but within
the capability of two references. The AP431 with its VKA(max) rating of 36V is ideal for this problem.
It is assumed therefore that the 2-reference solution in Figure 2 will be used.
Solution
First, determine R1 assuming R2 = 10k.
⎛V
⎞
R1 = R 2⎜⎜ OUT − 1⎟⎟
V
⎝ REF
⎠
⎛ 50
⎞
= 10k ⎜
− 1⎟
⎝ 2.495 ⎠
Or
Issue 1 - September 2008
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= 190.4k
R1 = 191k
to the nearest E192 value
and within 0.32%.
2
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Next determine IR3.
This will be the required load current plus the minimum operating current of the AP431, IKA(min)
which can be as much as 0.5mA.
Hence, IR3 = 5.5mA.
Next, determine R3 and VX, R4 and R5. The optimum thing to do is to ensure that the circuit is
able to supply the required current under worst case conditions and then check that all devices
still work within their design parameters at the opposite extreme.
Worst case conditions are full load current and minimum input voltage.
Hence,
R3 =
=
VIN (min) − VOUT
IR3
10
0.0055
= 1.81k⍀
R3 = 1.8k
to the nearest lower value
in E12.
It is best to ensure VOUT is equidistributed across the two references.
Therefore,
VX = VOUT/2
= 25V
Assuming R5 = 10k
⎛V
⎞
R 6 = R 5⎜⎜ OUT − 1⎟⎟
⎝ VREF
⎠
⎛ 25
⎞
= 10k ⎜
− 1⎟
⎝ 2.495
⎠
= 90.2k
R4 = 90.9k
Or
to the nearest E48 value.
Figure 4 below shows the circuit with all circuit values. The last exercise now is to verify that all
will be well even at the opposite extreme of the worst conditions that were used to calculate these
values.
R3
Vin
I R3
1.8k
IL
Vout
IKA
AP431
VX
VREF2
C1
AP431
VREF1
R4
90.9k
R5
10k
R1
191k
R2
10k
GND
Figure 4 50V shunt regulator using two AP431s
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The worst case conditions are full load current and minimum input voltage. The opposite
extremes will be maximum input and no load. What would happen to all circuit elements under
these conditions?
The output voltage will remain at 50V and equally distributed across the references but all of IR3
will now flow into them. However, IR3 is now IR3(max) which is given by
I R 3(max) =
VIN (max) − VOUT
=
R3
25
1800
= 13.9mA
Since the AP431 can sink up to 200mA this is not a problem. However the power dissipation in
each device will only be 25V x 13.9mA = 347.5mW. Therefore a suitable package to handle this
power will have to be chosen.
The AP431 comes in several package options. These range from the SOT23 handling 400mW up
to SOT89 handling 800mW. A SOT23 device might just be good enough for the above solution
but this is at 25°C and will have to be derated for higher ambient temperatures. A bigger package
may be needed.
Lastly the power rating of R3 needs to be decided.
Thus,
PR 3(max) =
(VIN (max) − VOUT )2
R3
25 2
=
1800
PR3(max) = 347.2mW
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To keep surface temperature rise to a minimum a resistor rated at least 0.5W should be used.
The circuit in Figure 4 was both bench tested and simulated and the graphs below show the results.
70
60
50
V
40
Vin
30
Vout
I(R3)
I(KA)
I(L)
20
10
-0
12
10
mA
8
6
4
2
-0
0
0.2
0. 4
0.6
0.8
1
1.2
1. 4
1.6
1.8
2
Tim e/ mSecs
2.2
2.4
200uSecs/div
Figure 5 Two-reference circuit without a load
70
60
V
50
40
Vin
Vout
I(R3)
I(KA)
I(L)
30
20
10
-0
12
10
mA
8
6
4
2
-0
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
Time/mSecs
2
2.2
2.4
200uSecs/div
Figure 6 Two-reference circuit with a 10k load
Figure 5 shows the test without any load, hence all of the available current goes into the regulator
chain. When loaded with a 10k resistor as shown in Figure 6, the load takes a constant current of
5mA regardless of the input voltage.
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Calculated Example 2
Requirement
Supply Voltage:
Output voltage:
Load current:
115V to 135V
100V
5mA
Assume the use of AP431.
Discussion
The required voltage of 100V is higher than what even two AP431's could handle. Technically,
three or more references could be cascade connected following the same principle in Figure 2. A
better solution might be to replace the dropper reference with a zener diode as shown in Figure 3.
Solution
The design considerations are the same as for Calculated Example 1 except that it is not possible
to have the output voltage equidistributed across the reference and zener diode which needs to
necessarily carry the bulk of the voltage.
Hence, determine R1 assuming R2 = 10k.
⎞
⎛V
R1 = R 2⎜⎜ OUT − 1⎟⎟
⎠
⎝ VREF
⎛ 100
⎞
= 10k ⎜
− 1⎟
⎝ 2.495
⎠
Or
= 390.8k
R1 = 390k
to the nearest E12 value
and within 0.21%.
Determine IR3.
IR3 = IL + 0.5mA
= 5.5mA
Next, determine R3 and VZ, R4 and R5.
Worst case conditions are full load current and minimum input voltage.
Hence,
R3 =
=
VIN (min) − VOUT
IR3
15
0.0055
= 2.73k⍀
R3 = 2.7k
to the nearest lower value
in E12.
With a VOUT of 100V and the need to keep VKA within 36V, it is evident that VZ can not be less than
64V. Therefore a VKA value of 30V may be adopted for the design. This means VZ will have to be 70V.
Like most electronic components, zener diodes are only available in certain preferred voltage values,
usually from the E24 preferred values list. Therefore a search for the nearest preferred value to 70V
needs to be carried out.
The two nearest values to 70V from the E24 table are 68V and 75V of which 68V is the nearer of the
two. This will make VKA 32V which is still within 36V.
VZ = 68V
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Figure 7 below shows the circuit with all circuit values.
Vin
I R3
R3
IL
2.7K
Vout
IKA
ZD1
VZ
68V
R1
VREF
390K
AP431
C1
REF1
0.1uF
R2
10K
GND
Figure 7 100V shunt regulator using one AP431 and a zener diode
High input worst case analysis
I R 3(max) =
=
VIN (max) − VOUT
R3
135 − 100
2700
= 12.96mA
Less than 200mA as
required.
Power ratings
Power in AP431:
P( AP 431) = VKA ⋅ I R 3(max)
= 32 × 0.013
P(AP431) = 416mW
Power in ZD1:
This is too much power for
the SOT23, Therefore a
bigger package will be
required. Alternatively, the
power could be reduced by
reducing the voltage
dropped by it.
P( ZD1) = VZ ⋅ I R 3(max)
= 68 × 0.013
P(ZD1) = 884mW
Power in R3:
A zener diode with a power
rating of at least twice this
is recommended.
P( R 3 ) = VR 3(max) ⋅ I R 3(max)
= 35 × 0.013
P(R3) = 455mW
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A resistor with a power
rating of at least twice this
is recommended.
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This circuit in Figure 7 was simulated and the graphs below show the results
120
100
V
80
60
40
Vin
Vout
Vz
I(R3)
I(KA)
I(L)
20
0
18
16
14
mA
12
10
8
6
4
2
-0
0
0.2
0. 4
0.6
0. 8
1
1.2
1.4
1. 6
1.8
2
Time/ mSecs
2.2
2.4
200uS ecs/div
Figure 8 Reference/Zener circuit without load
120
100
V
80
60
40
Vin
Vout
Vz
I(R3)
I(KA)
I(L)
20
0
18
16
14
mA
12
10
8
6
4
2
-0
0
0.2
0. 4
0.6
0.8
1
1.2
1. 4
1. 6
1.8
Time/ mSecs
2
2.2
2.4
200uSecs/div
Figure 9 Reference/zener circuit with 20k load
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Conclusion
The operating voltage range for references can be easily extended by using any of a number
methods shown above. In most cases without compromising the accuracy or other parameters
of the device.
Recommended further reading
AN58 - Designing with Shunt Regulators - Shunt Regulation
AN59 - Designing with Shunt Regulators - Series Regulation
AN60 - Designing with Shunt Regulators - Fixed Regulators and Opto-Isolation
AN62 - Designing with Shunt Regulators - Other Applications
AN63 - Designing with Shunt Regulators - ZXRE060 Low Voltage Regulator
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The circuits in this design/application note are offered as design ideas. It is the responsibility of the user to ensure that the circuit is fit for
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written approval of the Chief Executive Officer of Diodes Incorporated. As used herein:
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1. are intended to implant into the body
or
2. support or sustain life and whose failure to perform when properly used in accordance with instructions for use provided in the
labeling can be reasonably expected to result in significant injury to the user.
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