Post by thread:[EE]:Power transformer question

Rusty neurons here: How do you calculate peak diode current in a full wave or bridge power supply? I know it's based on the cap size, which determines the ripple voltage, but I don't remember the diode peak current part. -- Dave's Engineering Page: http://www.dvanhorn.org I would have a link to http://www.findu.com/cgi-bin/find.cgi?KC6ETE-9 here in my signature line, but due to the inability of sysadmins at TELOCITY to differentiate a signature line from the text of an email, I am forbidden to have it. -- http://www.piclist.com hint: PICList Posts must start with ONE topic: [PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads

> How do you calculate peak diode current in a full wave or bridge power supply? Fastest dV/dt of the AC waveform times the capacitance after the rectifier, plus maximum cicuit current draw. ******************************************************************** Olin Lathrop, embedded systems consultant in Littleton Massachusetts (978) 742-9014, spam_OUTolinTakeThisOuTembedinc.com, http://www.embedinc.com -- http://www.piclist.com hint: PICList Posts must start with ONE topic: [PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads

At 03:12 PM 9/9/01 -0500, you wrote: >Rusty neurons here: > >How do you calculate peak diode current in a full wave or bridge power supply? > >I know it's based on the cap size, which determines the ripple voltage, but >I don't remember the diode peak current part. For the recurring peak current... you can calculate it from the known point in the cycle where the diode starts to conduct (from the p-p ripple voltage), and the source impedance of the transformer winding (which you can measure by loading it and measuring the voltage change) and the diode resistance and capacitor ESR. You know the dv/dt of a perfect sine wave at the point where the diode starts to conduct (dv/dt of Vo*sin(wt) = w*Vo*cos(wt)). Typical AC waveforms are a bit flattened these days, though. This is typically not limiting, however it's worth noting that the RMS diode current is much higher than the output DC average current, so the power dissipation is higher than you would have with DC. Dividing the average current by fraction of the cycle that the diode is conducting will get you a ball-park figure (though it will be optimistic). The inrush current at startup is much worse- worst case there is not much limiting the current to a reasonable value other than the source impedance- if the voltage is switched on at the peak of the cycle. In one device I designed recently I used a spare PIC output (yeah, I did have one) to control a small relay which shorted out a series resistor. The PIC was also controlling the load so I could charge the capacitor (about the size of a can of Coke) at a reasonable 4A peak to 300V for a second or so, before shorting out the series resistor (waiiiiit for the relay to pull in) and then cranking the power devices up. Or you can just use an NTC, but there is a conflicting requirement in that the NTC stays hot longer than it takes for the voltage on the capacitor to discharge. Best regards, =-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= Spehro Pefhany --"it's the network..." "The Journey is the reward" .....speffKILLspam@spam@interlog.com Info for manufacturers: http://www.trexon.com Embedded software/hardware/analog Info for designers: http://www.speff.com Contributions invited->The AVR-gcc FAQ is at: http://www.bluecollarlinux.com =-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= -- http://www.piclist.com hint: PICList Posts must start with ONE topic: [PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads

An add on to the answers so far. If you have any headroom then a small series resistor in series with the transformer secondary can, rather unintuitively, improve some aspects of performance and reduce required component specifications. This reduces peak currents, and spreads diode conduction angle thereby reducing diode spec, switching noise, capacitor heating and more. Reducing capacitor peak currents can improve lifetime in temperature limited designs. This resistor arguably worsens load affected regulation but is generally insignificant with an active regulator and a smidgin of free head room available. Russell McMahon apptechKILLspamclear.net.nz > Rusty neurons here: > > How do you calculate peak diode current in a full wave or bridge power supply? > > I know it's based on the cap size, which determines the ripple voltage, but > I don't remember the diode peak current part. -- http://www.piclist.com hint: PICList Posts must start with ONE topic: [PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads

At 12:12 PM 9/10/01 +1200, Russell McMahon wrote: >An add on to the answers so far. >If you have any headroom then a small series resistor in series with the >transformer secondary can, rather unintuitively, improve some aspects of >performance and reduce required component specifications. This reduces peak >currents, and spreads diode conduction angle thereby reducing diode spec, >switching noise, capacitor heating and more. Reducing capacitor peak >currents can improve lifetime in temperature limited designs. > >This resistor arguably worsens load affected regulation but is generally >insignificant with an active regulator and a smidgin of free head room >available. I plan to. I have some cap lifetime equations that I've done before, and I'll put that in as well. I developed them for a thermal printer I did. Most illuminating. -- Dave's Engineering Page: http://www.dvanhorn.org I would have a link to http://www.findu.com/cgi-bin/find.cgi?KC6ETE-9 here in my signature line, but due to the inability of sysadmins at TELOCITY to differentiate a signature line from the text of an email, I am forbidden to have it. -- http://www.piclist.com hint: PICList Posts must start with ONE topic: [PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads

David VanHorn wrote: > > Rusty neurons here: > > How do you calculate peak diode current in a full wave or bridge power supply? > > I know it's based on the cap size, which determines the ripple voltage, but > I don't remember the diode peak current part. Put a 0.1 ohm resistor in series and look on the CRO. With all my high current DC supplies I do this anyway, using a good sized resistor in series between the transformer sec and the bridge. This also tends to improve long term reliability when you turn the thing on (and the cap is basically a short circuit). Even dissipating a couple of percent in heat the resistor is always worth the cost. Always try and use a resistor to take the punishment, keep your semiconductors for more subtle roles... :o) -Roman -- http://www.piclist.com hint: The list server can filter out subtopics (like ads or off topics) for you. See http://www.piclist.com/#topics

At 03:03 AM 9/11/01 +1000, Roman Black wrote: >David VanHorn wrote: > > > > Rusty neurons here: > > > > How do you calculate peak diode current in a full wave or bridge power > supply? > > > > I know it's based on the cap size, which determines the ripple voltage, but > > I don't remember the diode peak current part. > > >Put a 0.1 ohm resistor in series and look on the CRO. I'd snap a current probe on it if it existed yet. Unfortunately, I'm in the position of having to develop a spec. It's further complicated by some design variables that take the required current output over a pretty wide range. -- Dave's Engineering Page: http://www.dvanhorn.org I would have a link to http://www.findu.com/cgi-bin/find.cgi?KC6ETE-9 here in my signature line, but due to the inability of sysadmins at TELOCITY to differentiate a signature line from the text of an email, I am forbidden to have it. -- http://www.piclist.com hint: The list server can filter out subtopics (like ads or off topics) for you. See http://www.piclist.com/#topics

David VanHorn wrote: > > At 03:03 AM 9/11/01 +1000, Roman Black wrote: > >Put a 0.1 ohm resistor in series and look on the CRO. > > I'd snap a current probe on it if it existed yet. > Unfortunately, I'm in the position of having to develop a spec. > It's further complicated by some design variables that take the required > current output over a pretty wide range. It's gonna get messy. :o) Power diodes Vf forward drop is not fixed, and unless you know the conduction time and impedances of the xformer and cap and secondary filter (if there is one) it's a nightmare. And the cap impedances will vary from part to part and with device heat. For example a common 1N4007 1A type rectifier diode might drop 0.6v for low If, but at 1A constant they can be as bad as 1.4v Vf. Now in a supply that had 0.8A constant load with a large cap the diode on-period was very short, peak current was a few amps and I think from memory the Vf during the period of actual conduction was a few volts! Adding a tuned series resistor helped a lot, and a second series resistor from C1 to C2 (pi filter) with the bulk of the capacitance in C2 gave much less losses and heat on the diodes, allowing me to use the cheap diode instead of going to larger diodes, and of course improved the reliability of the circuit a lot. Most manufacturers will "cheat" and use a lossy cheap trnasformer which has very low current overhead. In my case that was not possible. -Roman -- http://www.piclist.com hint: The PICList is archived three different ways. See http://www.piclist.com/#archives for details.