55:141 Advanced Circuit Techniquess-iihr64.iihr.uiowa.edu/MyWeb/Teaching/ece_55141... · A. Kruger 55:141: Advanced Circuit Techniques The University of Iowa, 2013 Linear Regulators,

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


Voltage Regulators

Highly stable, temperature compensated

reference ~ 1.2 V

Negative feedback, frequency

compensation…

Power


Voltage Regulators

Why Darlington transistor?

?=OV

What type of feedback: series-series,…?

21

2

RRRVV OREF +

=

+=

2

11RRVV REFo


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


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

∆



Reverse polarity protection

Reverse polarity protection

+ -



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


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


Example


Recap of How Voltage Regulators Work


NPN Regulator

Quiescent current is the lowest

IQ


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


Quasi-LDO (Quasi Low Dropout Regulators)

Quiescent current is in between NPN and true LDO

IQ


Comparison


Protection

Short circuit

Constant Current

Thermal

Current foldback

Reverse Polarity

Input Output

Low voltage indicator


Polarity Protection

Some but not all regulators are protected if the input is accidentally reversed.

If not protected, add a Schottky diode.


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.

+

+


Protection

Constant current


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)


Proper Ground Connection

Make sure the regulator senses at the load. This is especially important at high currents.


Constant Current Source

Regulator maintains 5 V here


Limit Current on Bus

Voltage on buss is almost equal to 𝑉𝑉𝑂𝑂𝑂𝑂𝑂𝑂 (LDO) and the circuit uses the regulator’s built-in current limiting.


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


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.


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.


Rectifier Circuits

𝑉𝑉𝑝𝑝

We will call this 𝑉𝑉𝑝𝑝 but note that this is 2𝑉𝑉𝐷𝐷 less than what comes out of the transformer.


Conduction Angle, Ripple Voltage

𝑉𝑉𝑝𝑝 = 𝑣𝑣𝑠𝑠 = 2𝑉𝑉𝐷𝐷

𝑡𝑡

We will ignore this



𝑉𝑉𝑝𝑝

𝑡𝑡



𝑉𝑉𝑝𝑝

𝑡𝑡

−𝑉𝑉𝑃𝑃cos 𝜔𝜔𝑡𝑡



𝑉𝑉𝑝𝑝

𝑡𝑡

Capacitor supplying output current 𝐼𝐼𝐿𝐿

Transformer & diodes are supplying 𝐼𝐼𝐿𝐿 and replenishing smoothing capacitor.



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.



𝑉𝑉𝑝𝑝

𝑡𝑡

𝑉𝑉𝑟𝑟

B

A

𝑉𝑉𝑟𝑟 is the ripple voltage


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



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


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


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⁄


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)


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.


Inrush Current

SPICE simulation results for worst-case.

Peak current through 𝐷𝐷1 is almost 80 A!

Load current 2 A


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.


NTC Thermistors

Examples of NTC thermistors in a circuit

Though-hole

SMD


LM117 Adjustable Regulator

𝑅𝑅1

𝑅𝑅2

𝑉𝑉𝑜𝑜 = 1.2 1 +𝑅𝑅2𝑅𝑅1

+ 𝐼𝐼𝐴𝐴𝐷𝐷𝐴𝐴𝑅𝑅2 𝐼𝐼𝐴𝐴𝐷𝐷𝐴𝐴


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



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



Digitally Selectable Output Voltage

Better to replace BJTs with MOSFETS (Why?)

microcontroller

Switch in different resistors to set output voltage.



50 mA Constant current battery charger

Adjusting multiple on-card regulator with singe control



High-Gain Amplifier

𝑅𝑅𝑜𝑜



Power Follower

𝑅𝑅𝑜𝑜


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


Simulating LM117

LM117 with slow start

Circuit with slow start


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


Documents

55:141 Advanced Circuit Techniquess-iihr64.iihr.uiowa.edu/MyWeb/Teaching/ece_55141... · A. Kruger 55:141: Advanced Circuit Techniques The University of Iowa, 2013 Linear Regulators,