HI

The data sheet of AD9665 shows the graph of Iout according to Vout.

The higher the Vout, the worse the linearity of Iout.

The AD9665 has a load of only LD in my circuit.

How can you lower it?

Best regards

HI

The data sheet of AD9665 shows the graph of Iout according to Vout.

The higher the Vout, the worse the linearity of Iout.

The AD9665 has a load of only LD in my circuit.

How can you lower it?

Best regards

Hi JinoL

I tested it with the circuit in figure 14 of the datasheet.

When the resistor is exist and when the resistor is not exist, the LD response is changed.

so I removed the resistor.because nonlinearity occurs according to Vout,(Figures 10, 11, and 13 in the datasheet) I wonder how to set Vout to 2V.

Can you give me a glimpse of how wrong it is to calculate the input current?

Assuming the circuit in Figure 14, if 4V is applied to Vin_WR1 I wonder if it is correct to calculate 4V / (4.32k + 200) as input current.

If this calculation is correct, my board which is driven with a pulse of 4ns is outputted without being amplified by the gain of each channel.best regards

Hi Eunki,

Apologies for the late response.

AD9665 is a current source and measuring the voltage at the output will serve no meaning. However, this can be done by putting a small resistor between the output and the ground, Vout = Iout *RL. This can be clearly illustrated in figure 14 of the datasheet.

Regards.