Linear regulator current boost circuits

Voltage Regulator Current Boost Circuits

Voltage regulators, such as the LM317, 78xx series, or other linear regulators, are commonly used for providing a stable output voltage. However, these regulators have a maximum current limit, often in the range of 1A to 1.5A for typical ICs. To increase the current handling capability, external pass transistors are added in a current boost configuration.

How Current Boost Circuits Work

The circuits shown in the image use NPN power transistors (2N6133, 2N6124) to handle additional current while the regulator maintains voltage regulation.

1. First Circuit (Single NPN Pass Transistor)

   • The voltage regulator drives the base of Q1 (2N6133).
   • The transistor acts as a current amplifier.
   • R1 limits the base current to Q1, ensuring proper operation.
   • The regulator supplies only a small portion of the total current, while Q1 handles most of the load current.

2. Second Circuit (Darlington Configuration with Current Limit)

   • Uses an additional PNP transistor (Q2, 2N6124) to improve control.
   • R_SC sets a current limit by dropping voltage based on load current.
   • The Darlington configuration provides higher current gain, reducing the current demand on the voltage regulator.
   • The current limit feature protects against overcurrent conditions.

Why This Works

• Transistor as a Current Amplifier: The external transistor bypasses excess current, preventing the regulator from exceeding its rated output.
• Voltage Follows Regulator: The collector-emitter voltage of the transistor follows the regulator output, maintaining regulation.
• Higher Current Supply: This method allows regulators like LM317 or 78xx to supply several amps instead of being limited to their built-in capacity.

---

Negative Aspects of Current Boost Circuits

While these circuits provide higher current, they come with some drawbacks:

1. Heat Dissipation

   • The pass transistor dissipates power proportional to (Vin - Vout) × I_load.
   • Without proper heat sinking, the transistor may overheat and fail.

2. Reduced Voltage Regulation Accuracy

   • The voltage drop across the transistor's V_BE (~0.7V) may cause a slight deviation in the output voltage.
   • Load regulation may degrade under heavy loads.

3. Current Sharing Issues

   • In circuits with multiple transistors, current sharing may not be perfect, leading to imbalanced load distribution.
   • Matching β (gain) values of transistors helps improve performance.

4. Slower Response Time

   • The external transistors introduce additional lag, affecting transient response.
   • This can lead to voltage dips or overshoots under fast load changes.

5. Complexity and Additional Components

   • The design requires extra transistors, resistors, and heat sinks.
   • A simple switching regulator (buck converter) may be a more efficient alternative.
    

   Here’s a detailed mathematical analysis of voltage regulator current boost circuits and how it would differ when using an LDO (Low Dropout Regulator).

   ---

   Mathematical Analysis of Current Boost Circuits

   1. Single NPN Pass Transistor Configuration

   For the first circuit, the pass transistor (Q1, NPN) handles most of the current while the voltage regulator maintains the output voltage.

   Key Equations:

6. Regulator’s Output Current:
   The regulator itself supplies I_REG, which is the base current of Q1:

   $$I_{\text{REG}} = \frac{I_O}{\beta}$$

   where β is the current gain of Q1.

7. Total Output Current:

   $$I_O = \beta I_{\text{REG}}$$

   This shows that for a given regulator current (I_REG), the transistor boosts the total available current.

8. Base Resistor (R1) Selection:
   To ensure proper transistor operation, R1 limits the base current:

   $$R_1 = \frac{V_{BE(Q1)}}{I_{\text{REG}}}$$

   where $V_{BE(Q1)}$ is the base-emitter voltage (~0.7V for silicon transistors).

9. Power Dissipation in Q1:
   The transistor dissipates heat based on the voltage drop across collector-emitter (V_CE):

   $$P_{\text{Q1}} = (V_{\text{in}} - V_O) \times I_O$$

   This heat dissipation requires a proper heat sink.

---

2. Darlington Configuration with Current Limit

In the second circuit, a PNP transistor (Q2) adds a current limit feature using R_SC.

• Current Limit Resistor (R_SC) Selection:

  $$R_{SC} = \frac{0.8}{I_{SC}}$$

  where I_SC is the desired current limit (in amperes).

• Base Current for Darlington Pair:

  $$I_{\text{base(Q1)}} = \frac{I_O}{\beta_1 \times \beta_2}$$

  where β1 and β2 are the current gains of Q1 and Q2.
  Since Darlington transistors multiply their gains, they require very little base current, making them effective for high-current applications.

---

What If We Use an LDO Instead?

LDO (Low Dropout Regulator) with Current Boost

An LDO regulator is a linear voltage regulator that operates with a small input-to-output voltage difference (low dropout).

Key Differences:

1. Lower Voltage Drop

   • Standard regulators like LM317 require at least Vin ≥ Vout + 3V.
   • LDOs work with Vin ≥ Vout + 0.3V to 0.6V.
   • This reduces power dissipation.

2. Current Boost Works Similarly

   • The external transistor still handles most of the current.
   • However, V_BE of Q1 must be considered because the LDO has a low output voltage.

3. Impact on Dropout Voltage

   • The external pass transistor adds an extra V_BE drop (~0.7V).
   • Some LDOs cannot regulate properly if Vout drops below their internal reference.
   • A PNP transistor or MOSFET might be better than an NPN.

4. Efficiency Considerations

   • Using an NPN transistor in a high-side configuration can increase dropout voltage.
   • A PNP transistor allows better dropout performance but needs proper drive circuitry.

   Using High Current in Short Pulses with Two TIP32 Transistors

   If your circuit needs to handle high currents in short pulses rather than continuously, the design considerations change. Instead of worrying primarily about steady-state power dissipation, you need to focus on:

5. Peak current handling capability
6. Thermal transients (short-term heating effects)
7. Capacitor selection for transient response
8. Switching speed and response time of TIP32

---

Key Considerations for Pulsed High-Current Operation

1. TIP32 Pulse Current Handling

• The continuous current rating of TIP32 is ~5A (with proper heat sinking).
• The peak pulse current rating is much higher (easily 10A+ for a few milliseconds).
• Ensure the pulse duration is within the Safe Operating Area (SOA) of TIP32 to avoid thermal failure.

2. Thermal Considerations (Junction Heating in Pulses)

• Unlike steady-state operation, short pulses do not cause immediate overheating.
• However, if pulses are frequent, the junction temperature can rise over time.
• Use the thermal resistance equation to estimate junction heating:

$$\Delta T = P \times R_{\theta JC}$$

where:

• $P$ = Pulse Power Dissipation $(V_{\text{CE}} \times I)$
• $R_{\theta JC}$ = Junction-to-Case Thermal Resistance (~2°C/W for TIP32)
• $\Delta T$ = Temperature rise per pulse

If pulses are short (<10ms) and infrequent, the transistor does not reach thermal runaway.

3. Capacitor Selection for Fast Transient Response

Since the regulator and transistors cannot instantly respond to large pulse currents, bulk capacitors are critical.

Capacitor Types and Values:

Capacitor Type	Role	Value Recommendation
Electrolytic (Aluminum/Tantalum)	Provides bulk charge storage	4700µF+ (depends on pulse width and current)
Ceramic (MLCC, X7R)	Handles high-frequency transients	1µF - 10µF
Film (Polypropylene)	Reduces voltage spikes	100nF - 1µF

Use a combination of electrolytic + ceramic capacitors near the load for best results.

4. Base Drive Considerations (Speed of TIP32 Turn-On/Off)

• TIP32 is a BJT, so it is slower than MOSFETs.
• For fast pulse currents, use a low-value base resistor (to ensure fast transistor turn-on).
• A small capacitor (~10nF) across the base resistor can help improve switching speed.

---

Mathematical Analysis for Pulsed Operation

1. Current Sharing Between Two TIP32 Transistors

If total pulse current = 10A, each transistor ideally carries 5A.

Base Current Calculation for Pulsed Operation

Assume:

• $\beta$ (hFE) = 30 (for TIP32 at high current)
• Pulse current per transistor $I_{O}/2 = 5A$

$$I_B = \frac{I_O}{\beta} = \frac{5A}{30} = 167mA$$

If your regulator cannot supply this, consider using a small NPN driver transistor (e.g., 2N2222).

Base Resistor (R_B) Calculation for Fast Switching

$$R_B = \frac{V_{\text{reg}} - V_{BE}}{I_B} = \frac{5V - 0.7V}{167mA} = 26Ω$$

Use a standard 27Ω resistor per TIP32.

---

Practical Example: Pulsed Load with 10A for 1ms

Case 1: No Capacitors → Poor Performance

• If the regulator and transistors cannot react in 1ms, the voltage drops significantly.
• This could cause the load (e.g., a microcontroller or motor driver) to reset.

Case 2: Proper Capacitor Selection → Stable Voltage

• Assume 4700µF electrolytic capacitor.
• Using Q = CV, the capacitor can supply pulse current:

$$I = C \frac{dV}{dt}$$

For a 1ms, 10A pulse, assuming we want voltage drop <0.5V:

$$C = \frac{I \times dt}{dV} = \frac{10A \times 1ms}{0.5V} = 20,000µF$$

Thus, using at least 20,000µF will prevent a large voltage dip.

---

Comparison: TIP32 vs. MOSFET for Pulsed Currents

If speed is a concern, a MOSFET (like IRF9540 P-channel MOSFET) is better.

Feature	TIP32 (PNP BJT)	IRF9540 (P-MOSFET)
Switching Speed	Slow (Base drive required)	Fast (Gate drive required)
Voltage Drop (Losses)	High (V_BE drop of 0.7V)	Low (R_DS(ON) = milliohms)
Efficiency	Lower (Heat dissipation)	Higher (Less heat)
Base/Gate Drive Requirement	High base current	Low gate current

For fast pulses, a MOSFET is the better choice, but TIP32 still works if properly driven.

 

Figure : regulator current boost circuit ; source : TexasInstruments