SBOA553A August   2022  – January 2023 OPA593

 

  1.   Abstract
  2. 1Introduction
    1. 1.1 Parallel Output Circuit Design Considerations
  3. 2AC Considerations
  4. 3Bench Test Results
  5. 4Conclusion
  6. 5Revision History

Parallel Output Circuit Design Considerations

Directly hard-wiring the outputs of two power op amps is not a good electrical practice. If the outputs of the two op amps are directly connected together uneven current sharing can result. That happens because each of the two op amps tries to force a slightly different Vout voltage which is dependent on their individual Vos level. This current imbalance can lead to uneven power dissipation and uneven device heating.

Even though the resulting Vout differences between the op amps may be small, it is the electrical equivalent of connecting two very low-impedance voltage sources directly across one another. The op amp output impedances (Zcl) are very low, typically millohms due to the high loop gain at which each op amp operates. The Zcl differ between them if each op amp has a different closed-loop gain setting as can be the case with the leader-follower configuration. When there is nearly zero connection resistance between the outputs, the current can be quite high and affect the basic circuit functionality. The added output ballast resistances do much to minimize the current that can flow between the two outputs.

An OPA593 leader-follower parallel output circuit is shown in Figure 1-3. The DC voltage and current levels result from the 5-Vdc input, and a resulting output voltage Vout of +50 V. The load current I4 is 0.45 A. Higher output voltage is obtained by increasing the power supplies, up to 85 V across the V+ and V– pins. The output current is upwards to 0.5 A for the two OPA593 op amps in this configuration.

The power dissipations of the OPA593 op amps and the load must be taken into consideration because of power and thermal limitations as with any electronic component. Each OPA593 dissipates 1.25 W, while the 111-Ω load resistor power is 22.5 W for this specific example. The two OPA593 power dissipations become exceedingly high and they go into thermal shutdown if the delta between the supply voltage and theri output voltage is large and sustained for too long.

Remember that the intended application for the presented circuit involves a fast ATE test sequence where the dwell times are short fractions of a second. The average power dissipation in the fast ATE test sequence is much lower than when operating in a continuous high-power dissipation condition.

The Leader-follower configuration offers the following advantages:

  • Similar to a single op-amp amplifier it may be connected either as a non-inverting, or inverting amplifier
  • One set of gain resistors establishes the overall voltage gain of the complete amplifier doing away with matching sets of resistors
Figure 1-3 Leader-follower Parallel Output Amplifier

The U1 leader amplifier in Figure 1-3 connected as a non-inverting amplifier has its gain set to +10 V/V. The follower amplifier U2, is a simple unity-gain buffer with a gain of +1 V/V. The two op amps U1 and U2 use individual 3.0-Ω output ballast resistors, designated as Rb1 and Rb2. Though the OPA593 typical room temperature Vos (Ta = 25°C) is ±10 μV, it can increase to an overtemperature maximum of ±350 μV. The potentially higher Vos differences between U1 and U2 across temperature reinforces the need for ballasting to minimize the circulating current flow that can occur between their outputs.

The non-inverting of U1 is driven by a +5-Vdc source in the circuit diagram. Its output voltage when measured on the load side of Rb1 is 10 × 5 V, or 50 V. This occurs because Rb1 is within the feedback loop of U1. The voltage at U1 output adjusts to account for the voltage drop developed across Rb1 as the current through the resistor changes.

The follower amplifier U2 senses the voltage directly at the output of U1, and then the output of U2 follows precisely. If the U1 output voltage rises, the output voltage of U2 also rises and compensates for the voltage drop across Rb2 that occurs in response to an I2 increase. I1 and I2 move in unison each providing one-half the total load current I4, which in this example is 450 mA. The U1 and U2 voltage output moved up to +50.69 V, compensating for the 0.69 V voltage drops across Rb1 and Rb2 for the respective I1 and I2 output currents.

There is a reasonable amount of freedom when selecting the Rb1 and Rb2 ballast resistor (Rbal) values. Here are factors to consider to help establish their resistance:

  • Always set Rb1 and Rb2 equal in value. The output currents I1 and I2 ratio in proportion to the resistance mismatch between them. A 1% resistance mismatch results in a 1% current imbalance.
  • The higher the ballast resistance (Rbal) resistance value, the higher the voltage drop (VDROP) and the higher the resistor power dissipation (Pd) at a particular output current level. The VDROP limits the maximum Vo output swing range if set too high in resistance.
  • When driving a high C-load, the effect of the Rbal resistance is to increase the phase margin when the load connects after the op-amp loop as with U2, or somewhat decrease the margin when the load connects within the loop as with U1. Use a compromise that best suits both the leader and follower circuits. Details are provided in Section 2 - AC Considerations.
  • The 3.0-Ω Rbal used for Rb1 and Rb2 in Figure 1-3 results in a good balance between performance and compromise for this OPA593 0.5 A (maximum) application.
  • Rbal resistance is usually made lower as the output current range of an application increases. The 3.0 Ω Rbal used for each OPA593 op amp in this application is reasonably scaled to one-tenth the value (0.30 Ω ) for a 5-A application.

The complete transfer function for the Leader-follower circuit is surprisingly mathematically involved, but when reduced to a simpler form, Vout is equal to:

Vi = Vin Figure 1-3

Vos1 = Input voltage offset of U1

Equation 1. Vout=Vi R1×Rb1+R1×Rb2+R2×Rb1+R2×Rb2R2×Rb1+R2×Rb2+Vos1R1×Rb1+R1×Rb2+R2×Rb1+R2×Rb2R2×Rb1+R2×Rb2

If Rb1 = Rb2, then

Equation 2. Vout=(Vi+Vos1)1+ R1R2

Something that may not be immediately apparent with the Vout equation is that the U2 voltage offset voltage (Vos2), is absent from the transfer function. The full transfer function has two equal, but opposite Vos2 terms that directly cancel each other when Rb1 equals Rb2.

The fact that Vos2 cancels out does not mean that it does not have an effect on the circuit. In fact, any Vos2 present results in the U1 and U2 output currents being imbalanced. The higher the Vos2 voltage offset, the higher the imbalance will be. Since the OPA593 has very low Vos the imbalance due to Vos2 is minimal and not necessarily of concern. However, it might be more of an issue for higher power op amps having Vos on the order of millivolts, or more.