Pass Transistor Boosts Current from Negative Linear Regulator

Jan 24 2003

103.00K

Abstract

The addition of a pass transistor to the circuit of Figure 1 allows the linear regulator (LDO) to deliver more current to the load. A detailed power dissipation analysis is included to assist circuit developers in choosing the proper power rating of each component. Furthermore, lab data shows that the device is stable across temperature, line, and load.

Adding four components to a negative linear regulator (U1 in Figure 1) increases the allowable load current by 60%. Cost for the components is less than $0.17 in 1k quantities.

Figure 1. A pass transistor and associated resistors boosts load current in this negative linear regulator by 60%.

Connecting the SET terminal to ground sets U1's output voltage to -2.5V. U1's maximum load current is 200mA, and the extra components (Q₁, R₁, R₂, and R₃) draw another 120mA maximum from the load, producing (without degrading the output regulation) a total maximum load current of 320mA.

R₁ reduces the power dissipated in Q₁, prevents thermal runaway in Q₁, and provides momentary protection against short-circuited outputs. It also prevents oscillation by limiting gain in the Q₁ loop. Current flowing through U1 from OUT to V_SS produces a voltage drop of V_R2 across R₂ and R₃, and thereby allows Q₁ to conduct load current as V_R2 approaches the base-to-emitter threshold of Q₁. The threshold (V_BE) is approximately 0.7V at room temperature.

Choose the values of R₁, R₂, and R₃ to ensure that R₂, R₃ and Q₁ dissipate maximum power at the maximum load current (320mA in this case). At 320mA, U1 conducts 200mA and Q1 conducts 120mA. Component power dissipation at maximum load is as follows:

P_R1 = I_R1² × R₁ = 120mA² × 9.1Ω ≈ 131mW
P_Q1 = V_Q1 × I_Q1 = (V_SS - V_R1 - V_OUT) × I_Q1 = (5V - 1.1V - 2.5V) × 120mA ≈ 168mW
P_R2 = I_R2² × R₂ = 100mA² × 18Ω ≈ 180mW
P_R3 = I_R3² × _R3 = 100mA² × 18Ω ≈ 180mW
P_U1 = V_U1 × I_U1 = (V_SS - V_R2- V_OUT) × I_U1 = (5V - 1.8V - 2.5V) × 200mA ≈ 140mW

To provide higher load current, you can easily modify the circuit by increasing the power-dissipation ratings of R₁, R₂, R₃, and Q₁. Table 1 details the components shown for 320mA load current. For power dissipation, the circuit board should have ample copper connected to the leads of power-dissipating components. Heat then conducts through the component leads to the circuit board, spreads into the copper areas, and is removed from the board by convection.

Table 1. Figure 1 Components
Component	Manufacturer Part Number Description	Package	Power Dissipation	Allowable Power Dissipation at +85°C
R₁	KAMAYA, INC. RMC18-9R1JB 9.1Ω ±5% Resistor	1206	250mW derate 4.55mW/°C above +70°C	181.75mW
R₂, R₃	KAMAYA, INC. RMC18-18RJB 18Ω ±5% Resistor	1206	250mW derate 4.55mW/°C above +70°C	181.75mW
Q₁	Central Semiconductor Corp. CMPT2222A NPN Transistor	SOT23-3	350mW derate 2.8mW/°C above +25°C	182mW
U₁	Maxim Integrated MAX1735EUK25 200mA Negative LDO	SOT23-5	571mW derate 7.1mW/°C above +70°C	464.5mW

Figure 2a.

Figure 2b.

Figure 2c.

Figure 2d.

Figure 2e.

Figure 2f.

Figure 2f.

Figure 2. Curves and waveforms characterize the output of Figure 1: output voltage vs. load current (a), output voltage vs. supply voltage (b), output voltage vs. temperature (c), line transient response (d), load transient response (e), and shutdown response (f).

A similar version of this article appeared in the November 25, 2002 issue of EDN magazine.