albertd200

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About albertd200

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  1. Thank you very much @Ledlighter , you’ve been very helpful! Asking to Digilent people is just useless .. 3 times I asked something .. not a clear response from them. Have a very nice Christmas! Albert
  2. Well, this post doesn’t answer the question. Which is the exact plug do I need for the Analog Discovery 2 power supply input? Thank you. Albert
  3. Hi! I need to know the connector type and size to modify a 5v 2.5A power supply to connect it to an Analog Discovery 2. Thank you. Albert Valiente
  4. Hello, Some days ago I installed the new Waveforms software to update from Waveforms 2015. I have found that when running a loopback to check the Network Analyzer, the signals going to both scope inputs, randomly, stop the data adquisition, resulting in a bad representation of the bode plot. It happens in a random fashion, and when it fails, happens at the beggining of the sequence. I tryied to reinstall an older version of Waveforms and it works flawlessly. Checking the latest version in other computers, it fails too. Best regards Albert
  5. Thanks for your answer, but. As I said in my previous post, I would use the maximum output from the AWG that is 5Vp or 10Vpp ... Could I use the potentiometer or not in that case? The link provided doesn't answer my question. Thanks in advance. Albert
  6. Hello, My question is about the output current of the AD2, AWG. I know the current Max. Output is 10mA, but, I can't find if that current is expressed in Irms or it is Ipeak. I need to know that, because I want to build a box with a potentiometer inside, in order to control the output amplitude in an analogic fashion, setting the AWG out (sinewave) to 10Vpp (5Vp or 3.53Vrms) and set the potentiometer to taste to achieve the voltage needed. Trying to figure out what potentiometer I should use, doing the math tells me that a 50ohm pot is way too low to load the AWG output, taking the 50ohm output impedance from the BNC board and adding the 50ohm potentiometer, it gives 50ohm + 50ohm = 100ohm then 5Vp/100ohm = 50mA ... five times the Max Rated current of the AWG output if the 10mA is expressed in peak current, or 50mA x 0.707 = 35.35mA thre and a half times if it is expressed in rms current. So, doing the inverse math, with a maximum current of 10mA, and the maximum voltage coming out of the AWG, it gives : 5Vp / 10mA = 500ohm potentiometer or 3.53Vrms / 10mA = 353ohm (not a standard value). Could I use a 500ohm potentiometer safely to control the amplitude of the AWG output with no drop of the 5Vp? If I'm at the maximum output voltage and current, will it present more THD at this output? Thanks in advance Albert