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?
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.
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.