the DAC0808 circuit need a 8bit digital input connector to works ? right ?
but without digital input, my circuit already have a voltage between 2-4V, then i tried to connect the 8bit digital input connector, it's still the same.
what is the operating voltage of your opamp which you have used in your circuit. That could be the only reason that you are not getting correct voltage levels
Where are your checkpoints voltage ?
add them on the schematic ...
"diviser pour mieux regner !"
1) test first without the buffer AOP ..
see schematic test of datasheet ..only one resistor on output of the DAC
don't forget that A1 is MSB bit !
test output with all inputs of DAC grounded .. to get 0V as output
after use only A1 (MSB bit) conneted to +5V (logical level 1) ..
you must get 5V if our Vref is 10V (half range output)
what is the operating voltage of your opamp which you have used in your circuit. That could be the only reason that you are not getting correct voltage levels
Where are your checkpoints voltage ?
add them on the schematic ...
"diviser pour mieux regner !"
1) test first without the buffer AOP ..
see schematic test of datasheet ..only one resistor on output of the DAC
don't forget that A1 is MSB bit !
test output with all inputs of DAC grounded .. to get 0V as output
after use only A1 (MSB bit) conneted to +5V (logical level 1) ..
you must get 5V if our Vref is 10V (half range output)
is there any problem with the circuit from the photo in post #4, the circuit i get from the DAC0808 datasheet.
i am using 5000k ohm resistor, is that the right one ? there are 2 resistor using 5000k ohm.
Where are your checkpoints voltage ?
add them on the schematic ...
"diviser pour mieux regner !"
1) test first without the buffer AOP ..
see schematic test of datasheet ..only one resistor on output of the DAC
don't forget that A1 is MSB bit !
test output with all inputs of DAC grounded .. to get 0V as output
after use only A1 (MSB bit) conneted to +5V (logical level 1) ..
you must get 5V if our Vref is 10V (half range output)
there is a point after 5 value => 5 K ohms !
maybe the additional 000 is to tell accuracy needed here . because 5K is not a standard value.
4,7K, 5,6K
except when uing accurate resistor at 1% or 0,1% of accuracy.
To check , you can use 4,7K + 330 ohms
there is a point after 5 value => 5 K ohms !
maybe the additional 000 is to tell accuracy needed here . because 5K is not a standard value.
4,7K, 5,6K
except when uing accurate resistor at 1% or 0,1% of accuracy.
To check , you can use 4,7K + 330 ohms
yes, that resistor make me confuse, i already change it to 5k6 ohm. and the problem of output voltage is solved. now about how the DAC works, it not even work when i give 8bit input to DAC..