2010-12-20 12:14 AM
How much current can the DAC output?
2011-05-17 05:18 AM
Hi zeng,
In the datasheet, it is mentioned: ''When the buffer is OFF, the Minimum resistive load between DAC_OUT and VSS to have a 1% accuracy is 1.5 MOhm'' so the current is 3.3V (max output of DAC) / 1.5MOhm = 2.2 µA (if you need 1%of accuracy). That is my own understanding and I don't know if it is acceptable for you.HerzlichMCU Lüfter
2011-05-17 05:18 AM
I'll post your advice to my manager, Thank you so much!
2011-05-17 05:18 AM
Hi Seven,
Thanks to let me know about the response because I am curious to know and I am not an expert, just an MCU fan.Herzlich,MCU Lüfter
2011-05-17 05:18 AM
You should also consider the load on your ''RC filter'' (from whatever is connected to sense its voltage). Why don't you use the on-chip buffer? Either way, I think you will need some active buffering if you want effective filtering and DC accuracy.
2011-05-17 05:18 AM
Thanks John F., the load on the RC filter is an Op amp, the on-chip buffer is not adapted just because the minimum voltage is 0.2V while the buffer is on, which is unacceptable.
The technicla support just called me, and showed me that the impedance output is 15K, so 1% accuracy can be accquired while the load is 1.5M. so i'm sorry to tell 'lowpowermcu' that your opinion is not so exact. Now we still have another issue, just after POR, the DAC output pin is float because we can not pull it down with a resistor thanks to the huge inpedance output, but that is also unacceptable, how can we solve this problem? Your answer is so appreciated!2011-05-17 05:18 AM
It's always a problem to set up initial conditions safely whether digital or analogue and your circuit / system design must solve the problem. In this case, you could perhaps consider using an Op-Amp buffer with an enable output and put the pull down resistor on the Op-Amp buffer. Use a GPIO pin to enable the Op-Amp when the DAC output is enabled and stable.
2011-05-17 05:18 AM
I'm afraid no more pin is available, but it seems that there is no better way so far, I'll suggest them as u said, many thanks!