Hi Dave, Target voltage is 9VDC at 1A. output voltage + regulator overhead + diode drop(s) + ripple voltage 9 + 2 + 1.2 + ripple voltage. The question here is what defines the ripple voltage. Using Olin's formula, it's the difference between the desired voltage and the peak voltage from the transformer. In the above example, the desired voltage is 9V but there is the diode and regulator and what we want into the input of the regulator is 2V higher than the 9V => 11V. Next there is that pesky diode voltage drop through the bridge. Another 1.2V => 12.2V therefore in reality the _desired_ RMS transformer voltage is 12.2V So, if we had a rock solid DC supply capable of providing 100A with 1mV ripple current then the 9V regulator only needs 12.2V. The input capacitor is used to simulate that rock solid supply but provide current during the times that the transformer voltage drops below the required input of 12.2V. Many transformers appear to have a 12.6V secondary and the peak out of that transformer is 17.8V, so we need to have a maximum droop from 17.8V to 12.2V = 5.6V. I.e with no load the cap will charge up to 17.8V and with 1A load the cap shouldn't discharge to more than 12.2V. OK. Using Olin's formula. 8.3 mCoulombs / 5.6V = 1.482mF or 1400uF. So throw in a 2000uF capacitor for good measure. Looks like the 1000uF per Amp is wrong by about 50% And of course there's no provision for component tolerance using 12.2V but by using a 2000uFd cap, there's enough headroom now. Let's do the same exercise with your 5V supply. 5V + 1.2V + 2.0V = 8.2V which is surprisingly close to the 8VAC value used in the 10A to 30A S100 power supplies at the beginning of the computer hobby revolution. So say I had a 30A power supply. What size filter would I need? 30A * 8.3ms = 249 mCoulombs 249mCoulombs / 8.2V = 30.5mF or around 30,000 uFd. Since the caps are notorious for the +20/-80% tolerance ratings and they also age I'd double the value and use around 60,000. Seems however, I've underestimated what others have done. I dusted off and opened the covers on three different S100 computers with 30A 5VDC power supplies. They all have an 8V rail for the 5V regulators and use 120,000 uFd. Using the larger transformers and smaller capacitors will work but is terribly inefficient and creates a much higher dissipation across the regulator. Back in the S100 days, with as many as 4 of the 1A 7805 regulators on an 8K byte static RAM card, each volt above the 2V regulator headroom added lots of heat and I have discoloured boards that demonstrate that. So back to Tal's request. For a 9V power supply a 12VAC transformer will be fine with 2000uFd filter but if he's interested in really clean power for analogue, I stand by my original suggestion to use an 18VAC transformer, create 12VDC with that and then create 9VDC from that. All the heat and voltage fluctuations are handled by the 12V regulator and the 9V supply will be much more stable. Signed Confused in Canada. > > > John -- > > > I'm a bit confused here. > > You sure are! > > > If the formula is for calculating the droop between peaks > then shouldn't > > the peak voltage of the transformer be used? > > Yes... > > > i.e. 9VAC is 9*1.414 = 12.76V peak. If we are drawing 1A during the > > 8.3ms then we get 8.3mCoulumbs/12.7cV = .652mF or 652 uF. > I would think > > that using 1000uF per 1Amp does work out nicely. Then even with the > > diode drops it works out with the tolerance of the capacitor. > > Where's your allowance for the output voltage? > > The peak transformer voltage needs to be: > > output voltage + regulator overhead + diode drop(s) + ripple voltage > > which in this case (5V, 1A, 1000uF/A) works out to > > 5.0 + 2.0 + 1.2 + 8.3 = 16.5 V > > This means that the RMS value needs to be at least 11.67 V. Hence, the > selection of a 12VAC transformer. > > -- Dave Tweed > > -- > http://www.piclist.com hint: To leave the PICList > mailto:piclist-unsubscribe-request@mitvma.mit.edu > > > > -- http://www.piclist.com hint: To leave the PICList mailto:piclist-unsubscribe-request@mitvma.mit.edu