Conditioning a Switch-Mode Power Supply Current Signal Using TI OP Amps

1 Introduction

Switch-mode power supplies almost always require knowledge of the switching current, often sensed on the primary side of the power transformer. Current Sensing Solutions for Power Supply Designers⁽¹⁾ reviews many methods for sensing the switched current. In lower-power switch-mode power supplies, the method most often used employs a sense resistor. The type of sense resistor used is often a high power, low inductance resistor that can add significant cost and power dissipation to the power supply design. This circuit is shown in Figure 1-1 (a).

Figure 1-1 Typical R_Sense Method and Proposed Op-Amp Method for Sensing Primary Side Switch Current

To overcome the cost and power dissipation of such power resistors, the circuit of Figure 1-1 (b) is proposed. Using a differential amplifier made up of a low-power op amp and discrete resistors can result in several advantages including lower power dissipation (efficiency), noise immunity, cost, and programmability.

1. Current Sensing Solutions for Power Supply Designers, Bom Mammano, Unitrode Power Supply Seminar SEM-1200, 1997.

2 Circuit Design

A switch-mode power supply often switches current on the primary side of a transformer through a MOSFET and measures the primary current with a sense resistor (R_Sense) as shown in Figure 1-1 (a). The pulse width modulator IC (PWM) usually requires a current-sense signal (V_S) in order to provide short-circuit protection or for use in current mode control, or for both protection and control. The peak value of V_S depends on the PWM IC used, but it is typically 1 volt.

The value of the sense resistor R_Sense in Figure 1-1 (a) is chosen based on the peak value of the primary-side current (I_Peak) and the required value of V_S. Therefore, R_Sense is determined by:

$R s e n s e = \frac{V s}{I p e a k}$

The power dissipation in R_Sense is based on the RMS value of the primary-side current (I_rms), which depends on the peak value as well as on the waveshape and the duty cycle. The power dissipated is:

$P s e n s e = {I r m s}^{2} \times R s e n s e$

As an example, let:

I_Peak = 6.67 A

I_rms = 4 A

V_S = 1 V

These values result in an R_Sense of 0.15 Ω, and a power dissipation in R_Sense equal to 2.4 W. Typically, a 5-W-rated resistor would be used in this application.

The circuit of Figure 1-1 (b) can be used to significantly reduce the cost and power dissipation of R_Sense. First, let us review how the circuit of Figure 1-1 (b) operates. This op-amp circuit is configured as a typical differential amplifier. The circuit operates by multiplying the differential sense signal (V_Sense) by the differential gain of the op-amp circuit. If Rf = R3 and Ri = R2, this gain is:

$G a i n = \frac{R f}{R i} a n d V s = V s e n s e x \frac{R f}{R i}$

Using the previous example, assume that the design goal is to use a lower-power sense resistor with a standard value, such as a 0.01-Ω resistor rated at 0.5 W, and to limit the dissipation of this resistor to no more than 0.25 W. From this information, the gain of the circuit can be calculated.

$P s e n s e = {I r m s}^{2} \times R s e n s e = {(4 A_{r m s})}^{2} \times 0.01 Ω = 0.16 W$

$V s e n s e = I p e a k \times R s e n s e = 6.67 A_{p k} \times 0.01 Ω = 66.7 m V$

$G a i n = \frac{V s}{V s e n s e} = \frac{1 V}{66.7 m V} = 15$

Based on these results, let Rf = R3 = 15 kΩ and Ri = R2 = 1 kΩ in Figure 1-1 (b).