Please help me.
I made a circuit to amplify the signal from the DAC. I need to get Vout = 10V at Vin = 3.3V. Diagram attached. As a result, Vs + = 10V when Vin = 3.3V Vout = 7B get Rg = 47k. G=3
Why Vout <10V ?
AD623 will never reach an output voltage of 10V even though it's rail-to-rail because it is restricted by the output voltage swing of +VS - 0.5V (max) which in your case is 9.5V.
If you want to have an output of 10V, then you may replace AD623 with AD8221 or AD8422 which are precision in amps that have also configurable gains of 1 to 1000 and requires a single supply voltage of up to 36V.
The fact is that when Vin = 3.3V, Vout=7V < 9.5V?
From Figure 21 of the AD623 datasheet, the maximum output voltage that you can have for a input voltage of 3.3V is around 2V. So, if you have a +Vs=10V that would be around 7V. In order to get roughly 10V output, use an input voltage between -3V to 2V (for gain=>10) or ~3.1V to 1V (for gain=1).
For more details on this, please refer to Figure 20 to 23 of the datasheet here - http://www.analog.com/static/imported-files/data_sheets/AD623.pdf
Thanks for the answer.
But no description of the relevant characteristics of my conditions. Vs = +10 V, 10> G> 1 (3.12).
Or tell me how I get it?
Next come the results that I got on my target board. Diagram is shown in post first.
Legend: all voltages measured with respect to GND2 (4,5 pins):
U1 voltage 1 pin
U2 voltage 2 pin
U3 voltage 3 pin
U6 voltage 6 pin
U8 voltage 8 pin
In this case the input pin 3 through the resistor 4,7kOm DC voltage was applied, with the following levels
(0,05; 0,13; 0,37; 0,54; 0,78; 1,6; 2,42; 3,23)(V);
In this case the following results in the attachment.
At the moment, I can not understand why the voltage at pin 1> 0.6 V?
Retrieving data ...