OPA548: E/S current flow clarification

Hi Art,

The OPA548 datasheet conforms to Positive conventional current flow into the input terminals as datasheet note (2) states. Therefore, regarding your questions:

Does that mean that 70uA flows out of ES when it is at 0.8V and 65uA flows out of ES when it is at 2.4V?

The Electrical Characteristics table indicates that for IE/S LOW (output disabled) that the  E/S pin in low state current is typically  –70 µA. Therefore, as the sign indicates the E/S pin will source current out in the low state. It also indicates that the pin will source –65 µA when in the high state. Therefore, the E/S pin typically sources (outputs) about –65 µA to –70 µA in the two states indicating it doesn't change much between them.

If the ES-I/O-Pin on the OPA548TG3 is at 0.8V, how much current can it sink before it is pulled up to 2.4V?

Internally, the OPA548 E/S pin is the base of a PNP transistor and a 15 kilohm resistor that pull up to an internal bus that is biased at a voltage higher than the voltages than the 0.8 V, or 2.4 V levels. The PNP transistor is ON when the input level is low, and OFF when the input level is high. Hence, the small increase in the current sourced by the input in the low state. The 15 kilohm resistor accounts for the rest of the E/S pin input current. Therefore, as the datasheet indicates the E/S pin only sources current, and it doesn't sink current.

Regards, Thomas

Precision Amplifiers Applications Engineering