bipolar transistor circuit with 2 resistors

BIPOLAR TRANSISTOR CIRCUIT WITH 2 RESISTORS PLUS

When you feed the voltage drop back to IC 1 ‘s input through R 5 and R 4, the voltage drop across R 2, representing Q 1 ‘s base current, increases the regulated voltage across R 1 by the ratio of R 5 /R 4. If you short-circuit base resistor R 2, note that any common-mode error that resistors R 5 and R 6 introduce cancels and thus has no effect on Q 1 ‘s base voltage. To understand its operation, think of the circuit as a voltage regulator that delivers a voltage equal to V REF across R 1. You can simplify the circuit and preserve its error-correction properties by combining IC 1 and IC 2 better yet, you can add two resistors to Figure 1 to achieve the same effect.

Using perfectly matched resistors simplifies the output-current equation to I OUT =(V REF /R 1 ). Because the transistor inherently “subtracts” its base current, its collector current and, hence, the output current have no base-current error.įigure 2 Adding base-current error compensation improves the circuit’s performance.

BIPOLAR TRANSISTOR CIRCUIT WITH 2 RESISTORS PLUS

As a result, the regulated voltage across R 1 represents the sum of the desired output current plus the transistor’s base current. Because IC 1 ‘s inverting input connects to current-shunt resistor R 1 ‘s upper end and not to ground, the reference voltage, V REF, determines the error voltage applied to Q 1, preserving output scaling and allowing output-current calculation as V REF /R 1. Resistors R 4 and R 5 scale and sum the error and reference voltages you apply to differential amplifier IC 1. You could compensate for base-current errors by measuring the output transistor’s collector current and introducing a correction factor, but that approach increases circuit complexity and reduces the sink’s output impedance.įigure 2 shows a better approach, which adds a differential amplifier, IC 2, and resistors R 6 through R 9 to measure Q 1 ‘s base current by sampling the voltage across R 2. In other words, even though a bipolar transistor presents higher output impedance at high frequencies, the error from its base current makes it a poor choice for a high-precision current sink. Superbeta power transistors are rare, have typically lower unity-gain-bandwidth frequencies, and are more expensive.

You can use a Darlington transistor to increase the circuit’s current gain and reduce the output error, but few Darlington transistors offer good high-frequency parameters. Its normal output current is I OUT =(V REF /R 1 )-I B. The base current varies with changes in Q 1 ‘s collector current and current gain, which in turn depend on Q 1 ‘s production tolerances, junction temperature, and collector-emitter voltage.įigure 1 This typical quickly responding constant-current sink uses a bipolar transistor but suffers from base-current-induced error. Figure 1 shows a design for a bipolar-transistor-based current sink that unfortunately suffers from accuracy errors due to Q 1 ‘s base current’s flowing into the current-measurement resistor R 1. However, a high-power MOSFET presents high input and output capacitances that reduce the sink’s high-frequency output impedance.Īs an alternative, a low-current-gain, bipolar power transistor presents a much lower output capacitance than does a MOSFET of comparable power ratings. When you design a current sink, you can use a MOSFET for the sink’s pass transistor because of its nearly infinite power gain and low gate current. To do so, you measure the transistor’s base current and add a proportionally scaled error term to the source’s reference voltage. As a bonus, you also compensate for errors that a low-current-gain pass transistor’s base current introduces. You can improve a current sink’s accuracy by at least two orders of magnitude by adding two standard 1%-tolerance resistors.

Bipolar transistor circuit with 2 resistors