OPA548: Strange behavior of disable (E/S-Pin)

Boris Christoffel40

Part Number: OPA548

I'm using the output disable function of the OPA548 like shown in the Datasheet (Fig. 24).
The application is a modified Howland current source which controlls the current of a high power LED.
A DC voltage sets the output current and the output disable is used to pulse the LED.

When I activate the disable, I expect the output to go to zero. This is actually what it does with a resisitive load (R=10Ohm) - but when I connect a LED, the output goes to the negative supply!

Could someone please explain that? Did I miss something?

Thanks, Boris

Load: 10 Ohm resistor (yellow: Load; blue trigger signal)

Load: LED (yellow: Load; blue trigger signal)

the schematic:

over 7 years ago

0 Thomas Kuehl over 7 years ago

TI__Guru* 93105 points

Hi Boris,

The primary function of the OPA548 enable/shutdown is to significantly reduce the amplifier's operating current and power dissipation when output isn't required from it. When in shutdown, the OPA548 output will pull down close to the negative supply rail unless there is a very low resistance connected from the output pin to ground. In that case the output will be pulled close to 0 V, which is what you observed with the 10-ohm load resistor to ground.Increasing the resistance allows the voltage to move closer to V-.

When the output LED in your application is off it appears as an open circuit to the OPA548 output, and the amplifier's output voltage will pull down close to V-, or -5 V. This is seen in the first TINA Spice simulation image shown below. Do note there are dc paths around the circuit loops and current flow even though the OPA548 is in shutdown.

A solution that one of my colleagues suggested was to add a Schottky power diode across the LED as shown in the second TINA Spice circuit. The Schottky diode becomes forward biased and clamps the voltage across the output to about -100 mV.

Regards, Thomas

Precision Amplifiers Applications Engineering