Hi
I have an old design schematics. I couldn't able to understand the mode of operation.
Please refer the attached schematics and let me know the mode.
Regards
Srikanth Kacchu
AD9768
Obsolete
The Analog Devices AD9768SD D/A converter is a monolithic current-output converter
which can accept 8 bits of ECL-level digital input voltages and convert...
Datasheet
AD9768 on Analog.com
Hi
I have an old design schematics. I couldn't able to understand the mode of operation.
Please refer the attached schematics and let me know the mode.
Regards
Srikanth Kacchu
Hi,
First, it appears that the diagrams for Figures 1 and 2 are swapped. It looks like the AD9768 is being used in voltage multiplying mode, similar to the schematic in Figure 1 (which again is mislabeled as Conventional AD9768), and described in the VOLTAGE MULTIPLYING MODE section. However, the output is configured differently than all 3 modes described in the datasheet. It is possible that the node on output pin 13 is connected a load which could be similar to that in the datasheet if you have more info on this let me know. Also note that the digital inputs of the AD9768 are essentially tied together by the ECL mux, so they are either all 1’s (full current dependent on the drive voltage at pin 16) or all 0’s (output current 0). So the DAC portion of the AD9768 is really being used as an on / off switch for the output current. You might be able to replace this whole function with an all-analog amplifier circuit that realized the transfer function in Figure 3, with a bandwidth of 250KHz for Vm. The advantage of using the AD9768 for this was to also be able to scale the transfer function using the digital inputs (which is not being used in your design).
I am not aware of another DAC that would replace this part and have better-specified settling time than the AD9708. Although the AD9708 is specified at 35ns to 0.1% while the AD9768 was specified, not sure what your actual requirement is for settling. The AD9114 has a settling time of 11.5ns to 0.1 %, better than the AD9708, but still no 0.2% spec to compare to the AD9768.
Regards,
David
Hi David,
We have connected IOUT(Pin 13) of AD9768 to Pin 8 of AD834 with two parallel resistors(49.9Ohm and 7.87K) connected to ground.
As per datasheet recommendation, we have to use 50 ohm load. But the circuit implementation does not follow the datasheet. How does it works?
Regards
Srikanth Kacchu
Hi David
In datasheet it is found that the output current equation is not depending on digital input. But there is one equation with all digital inputs assumed as 1.
Can you please clarify?
Regards
Srikanth Kacchu
Hi,
If you are referencing the 50 ohm load to ground, you have set Vret to ground. And yes, the full scale current is set by the equation.
Regards,
David
Hi David
Sorry I couldn't understand. Can you please rephrase?
Regards
Srikanth Kacchu
Hi,
The full scale current is set by the equation on page 1:
And if you refer to Figure 1, you have set Vret to ground.
Regards,
David
Hi David
I am looking for the equation for every digital input. Is it available in datasheet?
Regards
Srikanth Kacchu
Hi,
It is an 8 bit DAC, so it will be the input DAC code divided by (2^8-1). But in your design all the digital inputs are tied together, so you will either get full scale or 0. THere is a small offset current when 0, 60uA typical.
Regards,
David
Hi David
Can you please provide the complete equation?
We didn't tie all the inputs together. Whatever the format shown in schematics is one of the method to represent the bus type connection.
Regards
Srikanth Kacchu
Hi,
It is an 8 bit DAC, so it will have 2^8 - 1 =255 steps from the offset current (which is the current for all 0s). So each step would ideally be (1/255)X (Fullscale current) . Please see the full scale current equation above.
Regards,
David