Upload
others
View
3
Download
0
Embed Size (px)
Citation preview
A. Kruger Linear Regulators, Slide 1 55:141: Advanced Circuit Techniques The University of Iowa, 2013
55:141 Advanced Circuit Techniques
Linear Power Supplies
Material: Lecture Notes & Handouts
A. Kruger Linear Regulators, Slide 2 55:141: Advanced Circuit Techniques The University of Iowa, 2013
Voltage Regulators
Highly stable, temperature compensated
reference ~ 1.2 V
Negative feedback, frequency
compensation…
Power
A. Kruger Linear Regulators, Slide 3 55:141: Advanced Circuit Techniques The University of Iowa, 2013
Voltage Regulators
Why Darlington transistor?
?=OV
What type of feedback: series-series,…?
21
2
RRRVV OREF +
=
+=
2
11RRVV REFo
A. Kruger Linear Regulators, Slide 4 55:141: Advanced Circuit Techniques The University of Iowa, 2013
Three-Terminal Regulators Vin Vout
GND
Iload Iin
IQ
+3.3 V, +5 V, +12 V, + 15 V, .. -5 V, -12 V
+=
2
11RRVV REFo
• Fixed voltage: R2 internal • Adjustable: R2 external
A. Kruger Linear Regulators, Slide 5 55:141: Advanced Circuit Techniques The University of Iowa, 2013
Three-Terminal Regulators Issues
Drop-out voltage & LDO
Reverse polarity protection Capacitor ESR
Temperature compensation
Load regulation: %100)(
)()( ×−
NLO
FLONLO
VVV
Max output current and still regulate
Output resistance:
O
Oof I
VR∆∆
−=
Short circuit protection (current
limiting)
V∆
Thermal protection Capacitors are often
crucial for stability. Carefully read datasheets
Quiescent current
IQ
Ripple rejection (at some f) Line regulation: %100×∆
I
O
VV
VVI
∆
A. Kruger Linear Regulators, Slide 6 55:141: Advanced Circuit Techniques The University of Iowa, 2013
Three-Terminal Regulators Issues
Reverse polarity protection
Reverse polarity protection
+ -
A. Kruger Linear Regulators, Slide 7 55:141: Advanced Circuit Techniques The University of Iowa, 2013
Three-Terminal Regulators Issues
Capacitor ESR
Capacitors are often crucial for stability. Carefully read datasheets
How big?
Dipped tantalum: few ohms or less do to a few mΩ ($$)
General purpose aluminum will often lead to trouble
Larger values have lower ESR
A. Kruger Linear Regulators, Slide 8 55:141: Advanced Circuit Techniques The University of Iowa, 2013
Loop Gain, Bode Plots etc. Example of the magnitude of the loop gain for a three-terminal regulator
With the proper/improper phase shift the loop can be unstable regulator is unstable.
Thus, the output capacitor is crucial.
If the ESR is too high can become unstable
If the ESR is too low can also become unstable
Consult and follow manufacturers guidelines
A. Kruger Linear Regulators, Slide 9 55:141: Advanced Circuit Techniques The University of Iowa, 2013
Example
A. Kruger Linear Regulators, Slide 10 55:141: Advanced Circuit Techniques The University of Iowa, 2013
Recap of How Voltage Regulators Work
A. Kruger Linear Regulators, Slide 11 55:141: Advanced Circuit Techniques The University of Iowa, 2013
NPN Regulator
Quiescent current is the lowest
IQ
A. Kruger Linear Regulators, Slide 12 55:141: Advanced Circuit Techniques The University of Iowa, 2013
LDO (Low Dropout Regulators) In the low-dropout (LDO) regulator, the pass transistor is a single PNP transistor. The big advantage of the LDO is that the PNP pass transistor can maintain output regulation with very little voltage drop across it. Full-load dropout voltages < 500 mV are typical. At light loads, dropout voltages can fall as low as 10 to 20 mV.
Quiescent current is the highest
IQ
A. Kruger Linear Regulators, Slide 13 55:141: Advanced Circuit Techniques The University of Iowa, 2013
Quasi-LDO (Quasi Low Dropout Regulators)
Quiescent current is in between NPN and true LDO
IQ
A. Kruger Linear Regulators, Slide 14 55:141: Advanced Circuit Techniques The University of Iowa, 2013
Comparison
A. Kruger Linear Regulators, Slide 15 55:141: Advanced Circuit Techniques The University of Iowa, 2013
Protection
Short circuit
Constant Current
Thermal
Current foldback
Reverse Polarity
Input Output
Low voltage indicator
A. Kruger Linear Regulators, Slide 16 55:141: Advanced Circuit Techniques The University of Iowa, 2013
Polarity Protection
Some but not all regulators are protected if the input is accidentally reversed.
If not protected, add a Schottky diode.
A. Kruger Linear Regulators, Slide 17 55:141: Advanced Circuit Techniques The University of Iowa, 2013
Polarity Protection
When the input is turned off/shorted, and the output capacitor is still charged, and the load is light, then significant currents can flow back into the regulator.
Modern regulators have built-in protection against this, but check the data sheet of your part. You may have to add a discharge diode.
+
+
A. Kruger Linear Regulators, Slide 18 55:141: Advanced Circuit Techniques The University of Iowa, 2013
Protection
Constant current
A. Kruger Linear Regulators, Slide 19 55:141: Advanced Circuit Techniques The University of Iowa, 2013
Protection
Current Foldback
If the maximum current is exceeded, the regulator dials down the output current. It will not supply a large current until the short is removed (hysteresis)
A. Kruger Linear Regulators, Slide 20 55:141: Advanced Circuit Techniques The University of Iowa, 2013
Proper Ground Connection
Make sure the regulator senses at the load. This is especially important at high currents.
A. Kruger Linear Regulators, Slide 21 55:141: Advanced Circuit Techniques The University of Iowa, 2013
Constant Current Source
Regulator maintains 5 V here
A. Kruger Linear Regulators, Slide 22 55:141: Advanced Circuit Techniques The University of Iowa, 2013
Limit Current on Bus
Voltage on buss is almost equal to 𝑉𝑉𝑂𝑂𝑂𝑂𝑂𝑂 (LDO) and the circuit uses the regulator’s built-in current limiting.
A. Kruger Linear Regulators, Slide 23 55:141: Advanced Circuit Techniques The University of Iowa, 2013
Increasing Output Voltages
One can use an %-V regulator such as the LM7805 to regulate at higher voltages.
𝑉𝑉𝑜𝑜𝑜𝑜𝑜𝑜 = 5 + 𝐼𝐼𝑄𝑄 + 𝐼𝐼𝑅𝑅1 𝑅𝑅2 = 5 + 𝐼𝐼𝑄𝑄𝑅𝑅2 +5𝑅𝑅1𝑅𝑅2
= 1 +𝑅𝑅2𝑅𝑅1
5 + 𝐼𝐼𝑄𝑄𝑅𝑅2
The problem with this method is that 𝐼𝐼𝑄𝑄 can be large and varies with loads and temperature. To combat this, use small values for 𝑅𝑅1 and 𝑅𝑅2 so large current flow through them swamp 𝐼𝐼𝑄𝑄
Regulator maintains 5 V here
A. Kruger Linear Regulators, Slide 24 55:141: Advanced Circuit Techniques The University of Iowa, 2013
Adjustable Voltage Regulators
Adjustable regulators are very similar to fixed voltage regulator, but don’t have the internal voltage setting resistors. The 𝐼𝐼𝑄𝑄 currents are designed to be small and constant
𝑉𝑉𝑜𝑜𝑜𝑜𝑜𝑜 = 1 +𝑅𝑅2𝑅𝑅1
𝑉𝑉𝑅𝑅𝑅𝑅𝑅𝑅 + 𝐼𝐼𝑄𝑄𝑅𝑅2 Now 𝐼𝐼𝑄𝑄 is small and much more constant than with fixed voltage regulators.
A. Kruger Linear Regulators, Slide 25 55:141: Advanced Circuit Techniques The University of Iowa, 2013
Rectifier Circuits We will analyze full-wave bridge rectifier. Extension to half-wave is straightforward.
Power Transformer
60 Hz 120 Hz 120 Hz
Note that because the full-wave rectification, there is 120 Hz ripple present. Thus, many components related to power supplies (capacitors, regulators) are tested and specified at 120 Hz. For example, regulator will have a ripple-rejection specified at 120 Hz.
A. Kruger Linear Regulators, Slide 26 55:141: Advanced Circuit Techniques The University of Iowa, 2013
Rectifier Circuits
𝑉𝑉𝑝𝑝
We will call this 𝑉𝑉𝑝𝑝 but note that this is 2𝑉𝑉𝐷𝐷 less than what comes out of the transformer.
A. Kruger Linear Regulators, Slide 27 55:141: Advanced Circuit Techniques The University of Iowa, 2013
Conduction Angle, Ripple Voltage
𝑉𝑉𝑝𝑝 = 𝑣𝑣𝑠𝑠 = 2𝑉𝑉𝐷𝐷
𝑡𝑡
We will ignore this
A. Kruger Linear Regulators, Slide 28 55:141: Advanced Circuit Techniques The University of Iowa, 2013
Conduction Angle, Ripple Voltage
𝑉𝑉𝑝𝑝
𝑡𝑡
A. Kruger Linear Regulators, Slide 29 55:141: Advanced Circuit Techniques The University of Iowa, 2013
Conduction Angle, Ripple Voltage
𝑉𝑉𝑝𝑝
𝑡𝑡
−𝑉𝑉𝑃𝑃cos 𝜔𝜔𝑡𝑡
A. Kruger Linear Regulators, Slide 30 55:141: Advanced Circuit Techniques The University of Iowa, 2013
Conduction Angle, Ripple Voltage
𝑉𝑉𝑝𝑝
𝑡𝑡
Capacitor supplying output current 𝐼𝐼𝐿𝐿
Transformer & diodes are supplying 𝐼𝐼𝐿𝐿 and replenishing smoothing capacitor.
A. Kruger Linear Regulators, Slide 31 55:141: Advanced Circuit Techniques The University of Iowa, 2013
Conduction Angle, Ripple Voltage
𝑉𝑉𝑝𝑝
𝑡𝑡
Capacitor supplying output current 𝐼𝐼𝐿𝐿
𝐼𝐼𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝
𝑡𝑡 𝐼𝐼𝑝𝑝𝑎𝑎𝑝𝑝 Δ𝑡𝑡
Diodes conduct only for a short time. Peak current is much higher than 𝐼𝐼𝐿𝐿
Transformer & diodes are supplying 𝐼𝐼𝐿𝐿 and replenishing smoothing capacitor.
A. Kruger Linear Regulators, Slide 32 55:141: Advanced Circuit Techniques The University of Iowa, 2013
Conduction Angle, Ripple Voltage
𝑉𝑉𝑝𝑝
𝑡𝑡
𝑉𝑉𝑟𝑟
B
A
𝑉𝑉𝑟𝑟 is the ripple voltage
A. Kruger Linear Regulators, Slide 33 55:141: Advanced Circuit Techniques The University of Iowa, 2013
Ripple Voltage with Constant Load 𝑰𝑰𝑳𝑳
Thus, one can solve for −𝑉𝑉𝑝𝑝 cos 𝜔𝜔𝑡𝑡 = − 𝐼𝐼𝐿𝐿 𝐶𝐶⁄ 𝑡𝑡 + 𝑉𝑉𝑝𝑝 to determine the time the capacitor supplies the output current and from this determine 𝑉𝑉𝑟𝑟.
Since 𝐼𝐼𝐿𝐿 is a constant current, 𝐴𝐴𝐴𝐴 is straight line. Slope = − 𝐼𝐼𝐿𝐿 𝐶𝐶⁄
𝑣𝑣 𝑡𝑡 = − 𝐼𝐼𝐿𝐿 𝐶𝐶⁄ 𝑡𝑡 + 𝑉𝑉𝑝𝑝
For the rectified (green) wave 𝑣𝑣 𝑡𝑡 = −𝑉𝑉𝑃𝑃 cos(𝜔𝜔𝑡𝑡).
𝑇𝑇𝐶𝐶 = 1 2𝑓𝑓⁄ 𝑉𝑉𝑟𝑟
𝑉𝑉𝑝𝑝
A
B
𝑡𝑡
For simplicity, we will assume 𝑉𝑉𝑟𝑟 ≪ 𝑉𝑉𝑝𝑝, which means that the capacitor supplies power for most of 𝑇𝑇𝐶𝐶 = 1 2𝑓𝑓⁄ and 𝑉𝑉𝑟𝑟 ≈ 𝐼𝐼𝐿𝐿 𝐶𝐶⁄ 𝑇𝑇 = 𝐼𝐼𝐿𝐿 2𝑓𝑓𝐶𝐶⁄ .
−𝑉𝑉𝑝𝑝cos 𝜔𝜔𝑡𝑡
A. Kruger Linear Regulators, Slide 34 55:141: Advanced Circuit Techniques The University of Iowa, 2013
Conduction Angle, Ripple Voltage
𝑉𝑉𝑝𝑝
𝑡𝑡
𝑇𝑇𝐶𝐶 = 1 2𝑓𝑓⁄ 𝑉𝑉𝑟𝑟 A
B
−𝑉𝑉𝑝𝑝 cos 𝜔𝜔𝑡𝑡
Δ𝑡𝑡
We assume the conduction angle is small so we can use cos 𝜔𝜔Δ𝑡𝑡 ≈ 1 −12𝜔𝜔Δ𝑡𝑡 2
𝜔𝜔Δ𝑡𝑡 = 2𝑉𝑉𝑟𝑟 𝑉𝑉𝑝𝑝⁄
From the diagram it is clear that 𝑉𝑉𝑟𝑟 = 𝑉𝑉𝑝𝑝 − 𝑉𝑉𝑝𝑝 cos 𝜔𝜔Δ𝑡𝑡
Combining gives Δ𝑡𝑡 =1𝜔𝜔
2𝑉𝑉𝑟𝑟 𝑉𝑉𝑝𝑝⁄ and
A. Kruger Linear Regulators, Slide 35 55:141: Advanced Circuit Techniques The University of Iowa, 2013
Average and Peak Diode Currents 𝑉𝑉𝑟𝑟 =
𝐼𝐼𝐿𝐿2𝑓𝑓𝐶𝐶
Ripple voltage Conduction time Δ𝑡𝑡 =1𝜔𝜔
2𝑉𝑉𝑟𝑟 𝑉𝑉𝑝𝑝⁄
During the conduction interval Δ𝑡𝑡, the transformer/diodes supply the charge supplied by the capacitor (C𝑉𝑉𝑟𝑟) and the charge to the load (𝐼𝐼𝐿𝐿Δ𝑡𝑡):
𝑄𝑄𝐷𝐷 = 𝐶𝐶𝑉𝑉𝑟𝑟 + 𝐼𝐼𝐿𝐿Δ𝑡𝑡 = 𝐼𝐼𝐿𝐿2𝑓𝑓
+ 𝐼𝐼𝐿𝐿Δ𝑡𝑡
Average diode current is 𝑄𝑄𝐷𝐷 Δ𝑡𝑡⁄ : 𝐼𝐼𝐷𝐷(𝑝𝑝𝑎𝑎𝑝𝑝) =𝐼𝐼𝐿𝐿2𝑓𝑓
1Δ𝑡𝑡
+ 𝐼𝐼𝐿𝐿 ⇒ 𝐼𝐼𝐷𝐷(𝑝𝑝𝑎𝑎𝑝𝑝) = 𝐼𝐼𝐿𝐿 1 + 𝜋𝜋 𝑉𝑉𝑝𝑝 2𝑉𝑉𝑟𝑟⁄
The peak current is right at the onset of the conduction interval:
𝐼𝐼𝐷𝐷(𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝) = 𝐶𝐶𝑑𝑑𝑉𝑉𝑐𝑐𝑑𝑑𝑡𝑡
+ 𝐼𝐼𝐿𝐿 𝐼𝐼𝐷𝐷(𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝) = 𝐶𝐶𝑑𝑑 −𝑉𝑉𝑝𝑝 cos 𝜔𝜔𝑡𝑡
𝑑𝑑𝑡𝑡+ 𝐼𝐼𝐿𝐿 = 𝐶𝐶𝜔𝜔𝑉𝑉𝑝𝑝 sin 𝜔𝜔𝑡𝑡 + 𝐼𝐼𝐿𝐿
Evaluating this at 𝑇𝑇𝐶𝐶 − Δ𝑡𝑡 or equivalently at Δ𝑡𝑡 gives 𝐼𝐼𝐷𝐷 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 = 𝐶𝐶𝜔𝜔𝑉𝑉𝑝𝑝 sin 𝜔𝜔Δ𝑡𝑡 + 𝐼𝐼𝐿𝐿
Using sin 𝑥𝑥 ≈ 𝑥𝑥 for small 𝑥𝑥 𝐼𝐼𝐷𝐷 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 ≈ 𝐶𝐶𝜔𝜔𝑉𝑉𝑝𝑝 𝜔𝜔Δ𝑡𝑡 + 𝐼𝐼𝐿𝐿 = 𝐶𝐶𝜔𝜔𝑉𝑉𝑝𝑝 2𝑉𝑉𝑟𝑟 𝑉𝑉𝑝𝑝⁄ + 𝐼𝐼𝐿𝐿
Simplifying leads to 𝐼𝐼𝐷𝐷 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 = 𝐼𝐼𝐿𝐿 1 + 2𝜋𝜋 𝑉𝑉𝑝𝑝 2𝑉𝑉𝑟𝑟⁄ Note the factor 2
A. Kruger Linear Regulators, Slide 36 55:141: Advanced Circuit Techniques The University of Iowa, 2013
Average and Peak Diode Currents
𝑉𝑉𝑟𝑟 =𝐼𝐼𝐿𝐿2𝑓𝑓𝐶𝐶
Ripple voltage
Conduction time Δ𝑡𝑡 =1𝜔𝜔
2𝑉𝑉𝑟𝑟 𝑉𝑉𝑝𝑝⁄
Average diode current during conduction interval 𝐼𝐼𝐷𝐷(𝑝𝑝𝑎𝑎𝑝𝑝) = 𝐼𝐼𝐿𝐿 1 + 𝜋𝜋 𝑉𝑉𝑝𝑝 2𝑉𝑉𝑟𝑟⁄
Peak diode current 𝐼𝐼𝐷𝐷 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 = 𝐼𝐼𝐿𝐿 1 + 2𝜋𝜋 𝑉𝑉𝑝𝑝 2𝑉𝑉𝑟𝑟⁄
Average diode over whole period 𝐼𝐼𝐷𝐷(𝑝𝑝𝑎𝑎𝑝𝑝) = 𝐼𝐼𝐿𝐿 1 + 𝜋𝜋 𝑉𝑉𝑝𝑝 2𝑉𝑉𝑟𝑟⁄Δ𝑡𝑡𝑇𝑇
= 𝐼𝐼𝐿𝐿 1 + 𝜋𝜋 𝑉𝑉𝑝𝑝 2𝑉𝑉𝑟𝑟⁄1𝜔𝜔
𝑉𝑉𝑟𝑟 2𝑉𝑉𝑝𝑝⁄
A. Kruger Linear Regulators, Slide 37 55:141: Advanced Circuit Techniques The University of Iowa, 2013
Example
A transformer’s secondary RMS voltage is 12.3 V, 60 Hz. A bridge rectifier with diodes with 𝑉𝑉𝐷𝐷 = .7 V rectifies the ac , and the filter capacitor is 𝐶𝐶 = 400 𝜇𝜇F . The load current is 𝐼𝐼𝐿𝐿 = 0.1 A. Determine the ripple voltage, conduction time, the average diode current, and peak diode current.
𝑣𝑣𝑠𝑠 = 2 12.3 = 17.4 V Transformer secondary 𝑉𝑉𝑝𝑝 𝑉𝑉𝑝𝑝 = 17.4 − 2 × 0.7 = 16 V
𝑉𝑉𝑟𝑟 𝑉𝑉𝑟𝑟 =𝐼𝐼𝐿𝐿2𝑓𝑓𝐶𝐶
=0.1
2 60 440 × 10−6= 1.89 V
Conduction time Δ𝑡𝑡 =1𝜔𝜔
2𝑉𝑉𝑟𝑟 𝑉𝑉𝑝𝑝⁄ =1
2𝜋𝜋 603.78 16⁄ = 1.3 m𝑠𝑠 ≅ 16%
Average diode current during conduction 𝐼𝐼𝐷𝐷(𝑝𝑝𝑎𝑎𝑝𝑝) = 𝐼𝐼𝐿𝐿 1 + 𝜋𝜋 𝑉𝑉𝑝𝑝 2𝑉𝑉𝑟𝑟⁄ = 0.1 1 + 𝜋𝜋 4.233 = 0.75 A
Peak diode current 𝐼𝐼𝐷𝐷 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 = 𝐼𝐼𝐿𝐿 1 + 2𝜋𝜋 𝑉𝑉𝑝𝑝 2𝑉𝑉𝑟𝑟⁄ = 0.1 1 + 2𝜋𝜋 4.233 = 1.4 A
Note that the peak diode current is 14 × larger (1.4 A) than the load current (0.1 A)
A. Kruger Linear Regulators, Slide 38 55:141: Advanced Circuit Techniques The University of Iowa, 2013
Inrush Current Another consideration for power supplies is the issue of inrush current. Note the capacitor’s low ESR of 26 mΩ, which is what one would expect for a good, large value capacitor intended for use in a power supply. A quick calculation for the circuit shows that in the steady state the current through the load is about 2 A, and the peak current though diode D1 is about 10 A.
When the power is first turned on, however, the capacitor is discharged, and the ESR of the smoothing capacitor and the dynamic resistance of the diodes limits the current, and large currents can flow.
The best case is when the power is turned on during the zero crossing of the source, and the worst case is when the power is turned on at the peak.
A. Kruger Linear Regulators, Slide 39 55:141: Advanced Circuit Techniques The University of Iowa, 2013
Inrush Current
SPICE simulation results for worst-case.
Peak current through 𝐷𝐷1 is almost 80 A!
Load current 2 A
A. Kruger Linear Regulators, Slide 40 55:141: Advanced Circuit Techniques The University of Iowa, 2013
Inrush Current Protection using NTC
Much smaller inrush current. As the current flows through thermistor, its resistance decreases and load current increases.
Load current increases to 1.8 A in steady after 25 s
Negative Temperature Coefficient (NTC) thermistor. When current flows through it, it heats up, and resistance decreases, which lets more current flow, …. Eventually, it reaches its minimum resistance.
A. Kruger Linear Regulators, Slide 41 55:141: Advanced Circuit Techniques The University of Iowa, 2013
NTC Thermistors
Examples of NTC thermistors in a circuit
Though-hole
SMD
A. Kruger Linear Regulators, Slide 42 55:141: Advanced Circuit Techniques The University of Iowa, 2013
LM117 Adjustable Regulator
𝑅𝑅1
𝑅𝑅2
𝑉𝑉𝑜𝑜 = 1.2 1 +𝑅𝑅2𝑅𝑅1
+ 𝐼𝐼𝐴𝐴𝐷𝐷𝐴𝐴𝑅𝑅2 𝐼𝐼𝐴𝐴𝐷𝐷𝐴𝐴
A. Kruger Linear Regulators, Slide 43 55:141: Advanced Circuit Techniques The University of Iowa, 2013
Applications of 3-Terminal Regulators
Electronic shutdown
When the TTL input goes high, the transistor turn on, grounding the ADJ terminal, and the output voltage goes to 5V
Precision current limiter
A. Kruger Linear Regulators, Slide 44 55:141: Advanced Circuit Techniques The University of Iowa, 2013
Applications of 3-Terminal Regulators
Slow turn-on 15 V regulator
At turn-on, the 𝐶𝐶1 is uncharged, and the BJT’s collector and base are connected, and the BJT is turn on hard, effectively shorting 𝑅𝑅2. The output voltage is 1.2 V.
Then 𝐶𝐶1 charges starts to turn the BJT off. Eventually, the BJT is completely off and the output voltage is
𝑉𝑉𝑜𝑜 = 1.2 1 +𝑅𝑅2𝑅𝑅1
+ 𝐼𝐼𝐴𝐴𝐷𝐷𝐴𝐴𝑅𝑅2
A. Kruger Linear Regulators, Slide 45 55:141: Advanced Circuit Techniques The University of Iowa, 2013
Applications of 3-Terminal Regulators
Digitally Selectable Output Voltage
Better to replace BJTs with MOSFETS (Why?)
microcontroller
Switch in different resistors to set output voltage.
A. Kruger Linear Regulators, Slide 46 55:141: Advanced Circuit Techniques The University of Iowa, 2013
Applications of 3-Terminal Regulators
50 mA Constant current battery charger
Adjusting multiple on-card regulator with singe control
A. Kruger Linear Regulators, Slide 47 55:141: Advanced Circuit Techniques The University of Iowa, 2013
Applications of 3-Terminal Regulators
High-Gain Amplifier
𝑅𝑅𝑜𝑜
A. Kruger Linear Regulators, Slide 48 55:141: Advanced Circuit Techniques The University of Iowa, 2013
Applications of 3-Terminal Regulators
Power Follower
𝑅𝑅𝑜𝑜
A. Kruger Linear Regulators, Slide 49 55:141: Advanced Circuit Techniques The University of Iowa, 2013
Simulating LM117 To simulate a circuit, we don’t need to simulate every component. Rather, we need to capture the essence of what we are interested in
LM117 – regulates so that there is always 1.2 V between ADJ and Output Pins
This circuit, using a voltage-controlled voltage source, will regulate so there is always 1.2 V between Vo and VADJ
A. Kruger Linear Regulators, Slide 50 55:141: Advanced Circuit Techniques The University of Iowa, 2013
Simulating LM117
LM117 with slow start
Circuit with slow start
A. Kruger Linear Regulators, Slide 51 55:141: Advanced Circuit Techniques The University of Iowa, 2013
Simulating LM117
Don’t calculate operating point
Sometimes you don’t want to calculate operating point of a circuit, so turn this off in the Transient Analysis
A. Kruger Linear Regulators, Slide 52 55:141: Advanced Circuit Techniques The University of Iowa, 2013