10/23/2023 0 Comments Pulltube maxI just misinterpreted your post but wanted to check and make sure I understand this. I always thought that the current ratings were RMS, based on a 60Hz sinusoid input and resistive load so that the entire winding of the transformer is always conducting into a linear load. It makes perfect sense now that you elaborate. I worked out all of these factors for a power systems class in Engineering school 25 years ago (University of Michigan - Go Blue!) but since then I've just used the cheat sheet like everyone else. So when I said that the common current derating was 62%, I was assuming the configuration of the Oddwatt KT88 power supply. There is actually a very good cheat sheet on the Hammond website which shows the basic rectifier configurations and the idealized derating factors on voltage and current for each one. Luckily, the resultant deratings for the most common rectifier and smoothing circuits have been worked out for generations. It is possible to use vector diagrams to sort it all out but you really need to use s-plane analysis to get the right answers. All of this gets very complicated very quickly. cos(theta)) into play (across multiple harmonic frequencies) before the rectifiers and introduces the complexities of charge/discharge cycles on inductors and capacitors after the rectifiers. Second, the voltage and current waveforms are no longer pure sinusoids and are no longer in phase with each other. This has serious implications for both the transformer windings and the magnetic currents in the transformer core. First, the load on all or part of the transformer secondary may become a pulsed load operating over conduction angles anywhere from 0 to 360 degrees (depending on the rectifier and filter configuration). However, when you begin to hang rectifiers and complex impedances across the transformer secondary, a couple of things begin to happen. The total power rating for the transformer is 25VA (but not necessarily 25 Watts). As such a 250v 100mA transformer will develop 250vrms across a 2500 ohm load and deliver 100mArms current. The voltage, current, and power ratings on a power transformer are the maximum AC sine wave voltage, current, and power ratings assuming a purely resistive load. I skipped a bunch of theory because of the specific nature of the configuration in question.Īs far as power transformer ratings go, transformers are AC devices and this is how they are specified. The fact the current is 62% and not half implies to me more efficient use of the transformer with higher power handling capability when using a 4 diode full wave bridge and leaving the CT disconnected. If the power remained the same, the current would be 50% of what it was. I’ve always thought the voltage output using a full wave bridge would be double the voltage obtained by grounding the CT and using a diode on each half coil as a full wave rectifier. You also seem to be implying that the transformer has a higher power rating when the entire coil is used as opposed to only half. Leaving the CT disconnected and using a full wave BRIDGE rectifier will require derating the transformer output current to 62% of the specified value, but the power output will actually increase? (yes/no?) This is a subject I’ve often wondered about but I tremendously oversize my power trannys so I’ve never really worried about it.Ĭorrect me if I’m wrong, but what you seem to be saying is that the current rating stated for a center tapped power transformer is the maximum DC current that the transformer can supply when the center tap is grounded and a 2 diode full wave rectifier is used. Under these conditions the transformer is derated to ~62% of its rated output current. In the 4 diode full wave bridge, the secondary conducts on both halves of the cycle due to the opposed diodes hanging off of each side of the primary. The accepted convention is that a transformer secondary will meet it's rated output current only if each half of the secondary only conducts for a maximum of one half of the cycle.
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