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OPA140: A High-Voltage Bidirectional Current Source

Part Number: OPA140
Other Parts Discussed in Thread: INA149

Hello there,

I have a question regarding this application note:

It introduces a bidirectional current source, which can accept voltages over +- 100 V. As an example, a 10k resistor is used and the current set for it is 10 mA. In the end circuit in Fig 4:

An OPA140 op amp is used in voltage follower configuration. I dont understand the following thing:

In order for 10 mA to flow through the 10k resistor, a voltage drop of 100 V is needed across the 10k resistor. OPA140 maximum voltage range is way below that. How is this circuit functioning, what am I missing? Lets assume VEE = GND = 0 V and VCC >= 100 V. Also, is this voltage follower really necesarry in the circuit, since the INA149 Ref A and Ref B inputs are high impedance?

I would appreciate all help.

  • Lukasz,

    It is correct that 10mA through 10k resistor implies 100V drop across the resistor and for this reason the supplies, Vcc2 and Vee2 used to power OPA140 and INA149, must be moving in tandem with the output signal self-adjusting for varying output current - see below.

    The whole idea behind this application circuit is to dramatically increase the compliance voltage range of difference amplifier-based bi-directional current sources. By employing a unique type of difference amplifier such as the INA149, the circuit can be bootstrapped to very high supply voltages such as +/-145V allowing dynamically drive up to ±10 mA into a 10 kΩ load - see below.

    Attached please find the Tina_TI schematic you may use for your own simulations.

    A High-Voltage Bidirectional Current Source.TSC

  • Hello Marek,
    Well explained, thank you for help.