From: Alan Neel [neel@edgetech.com] Sent: Thursday, July 31, 2008 7:32 AM To: Massion, Gene Cc: Mohammed Sanhaji; John Spruance Subject: RE: Two more questions Gene, TP1 and 2 are connected to the input terminals of the amp, TP7 is the amp enable input. Next, TP6 and 4 are the inputs to the power stage. The signals at this point have the full gain applied. From there to TP3 and 5 which are connected to the output terminals, the gain is approximately 1 or slightly less. The black TP8 and 9 are the same as the negative 48 volt supply and is also the return for the power amp enable signal. The input terminals are transformer isolated immediately after TP1 and 2. As a precaution, you can't ground any of the white test points with the scope ground clip since this will short internal stages and cause possible damage, all the TP7, 6, 4, 3, and 5 are referenced to the ground TP8 or 9. Remember, if you seem to see some distortion on one side, it may not be real since the signals are differential may be misleading. They can also be viewed by placing a scope probe on each signal pair and perform a channel 1 minus channel 2 display function. The output of the DAC should be fine but you may use an RC low pass filter on the input with fc set to about 2 or 3 X the highest frequency if you see objectionable frequency content in the output. Remember, the input impedance of the amp can in the range of 150 to about 1K ohms which will load your filter. Alan -------------------------------------------------------------------------------- From: Massion, Gene [mailto:magene@mbari.org] Sent: Wednesday, July 30, 2008 8:02 PM To: neel@edgetech.com Subject: Two more questions Alan Well things are looking pretty good, I’m just about to fire up the amplifier with our signal generator in the test tank here at MBARI. Is it possible to find out what the various test points are on your amplifier card? I’m particularly interested in looking at my input signal as it passes through the various stages of your amp. I’m also curious what you think the effect of a signal generated by a 100 kHz digital to analog converter will be on the amp. I’ve attached a scope shot with no filtering on the signal out of the DAC and you can clearly see the signal steps every 10 usec. Cheers Gene Massion Monterey Bay Aquarium Research Institute 7700 Sandholdt Road Moss Landing, California 95039 831 7751922 (tel) 831 7751646 (fax) -------------------------------------------------------------------------------- From: Alan Neel [mailto:neel@edgetech.com] Sent: Wednesday, July 02, 2008 4:30 PM To: Massion, Gene Cc: John Spruance; Mohammed Sanhaji Subject: RE: PS data sheets Gene, Sorry for the delay, I was out of town. The ESR is unimportant for the cap bank. The amp has its own low ESR caps to filter the high frequencies. Alan -------------------------------------------------------------------------------- From: Massion, Gene [mailto:magene@mbari.org] Sent: Thursday, June 26, 2008 6:09 PM To: neel@edgetech.com Cc: John Spruance; Mohammed Sanhaji Subject: RE: PS data sheets Alan Thanks for the equations. One more question. Can you estimate how low the ESR of the capacitors should be? Does it make sense to build up the 32,000 uF with a bunch of smaller caps in parallel to get the ESR down? Regards Gene Massion Monterey Bay Aquarium Research Institute 7700 Sandholdt Road Moss Landing, California 95039 831 7751922 (tel) 831 7751646 (fax) -------------------------------------------------------------------------------- From: Alan Neel [mailto:neel@edgetech.com] Sent: Wednesday, June 25, 2008 4:55 PM To: Massion, Gene Cc: John Spruance; Mohammed Sanhaji Subject: RE: PS data sheets Gene, These calculations have several simplifying approximations which are not stated: The numbers are from the assumptions given below. Power supply energy to cap during pulse: I = 6V / 1 ohm / 2 = 3 Amps average, E = .05 sec * 3 A * (52V + 46V) / 2 = 7.35 Joules Energy supplied to amp during pulse = 200W * 1 / .6 (percent eff) * .05 sec = 16.65 J Cap = 2 * E / ( Vi ^ 2 - Vf ^ 2 ) = 2 * ( 16.65 - 7.35) / (52 ^ 2 - 46 ^ 2 ) = 31,000 uf Average power dissipated in resistor = approx (6V / 1 ohm) ^ 2 * 1 ohm * .05 duty cycle = 1.8 Watts ( or 10% duty = 3.6 Watts) Alan -------------------------------------------------------------------------------- From: Massion, Gene [mailto:magene@mbari.org] Sent: Wednesday, June 25, 2008 7:27 PM To: neel@edgetech.com Cc: John Spruance; Mohammed Sanhaji Subject: RE: PS data sheets Many thanks Alan, this is a big help. Would you be willing to pass on the calculations you did that gave you the 32,000 uF? Thanks again Gene Massion Monterey Bay Aquarium Research Institute 7700 Sandholdt Road Moss Landing, California 95039 831 7751922 (tel) 831 7751646 (fax) -------------------------------------------------------------------------------- From: Alan Neel [mailto:neel@edgetech.com] Sent: Wednesday, June 25, 2008 4:14 PM To: Massion, Gene Cc: John Spruance; Mohammed Sanhaji Subject: RE: PS data sheets Gene, I should have added a schematic, see attached. Alan -------------------------------------------------------------------------------- From: Alan Neel [mailto:neel@edgetech.com] Sent: Wednesday, June 25, 2008 6:47 PM To: 'Massion, Gene' Cc: John Spruance; Mohammed Sanhaji Subject: RE: PS data sheets Gene, I have done some calculations with the following assumptions: Output power = 200W Input voltage = 52V ( this will affect your power supply selection or you will have to adjust the voltage up) Max duty cycle = 5%, 10% is probably ok but you will have to heat sink the amp well in both cases. Capacitor droop = 6V (during pulse, see comment below) Series resistor = 1 ohm, 5 watt (if needed for stability as we discussed) Pulse width = 50 ms You will need about 32,000 UF of capacitance for energy storage (63 Volt caps is a good choice). The droop will not affect your pulse fidelity if the peak to peak output voltage (between each output pin and the minus power rail (usually ground)) is about 42 Volts maximum. This will give you a total peak to peak output voltage of 84 volts. There are many ways to do this but going with a larger droop makes the cap bank much smaller (by factor of about 4 over a 3 volt droop and a 48 V power supply). A power supply of at least 200 watts also makes life easier. The amp should have a heat sink that doesn't let the aluminum bar get any hotter than about 80 degrees C. This will be directly proportional to duty cycle. The heat sink will need to dissipate about 400 watts * .05 (duty cycle) = 20 watts in the worst case where the load is highly reactive. A more reasonable value may be more like 10 watts. Good Luck, Alan Neel -------------------------------------------------------------------------------- From: Massion, Gene [mailto:magene@mbari.org] Sent: Wednesday, June 25, 2008 4:13 PM To: neel@edgetech.com Subject: PS data sheets Neel Sorry I forget these in the initial e-mail. I’m looking at the Vicor VI-MU or the Lambda PFE500. Thanks for your help Gene Massion Monterey Bay Aquarium Research Institute 7700 Sandholdt Road Moss Landing, California 95039 831 7751922 (tel) 831 7751646 (fax)