Push pull transformer  - Page 1
Forum section: electronics / general

author: user1991
author_rank: Newbie
timestamp: October 17, 2022, 01:54:42 pm
content: Hi,Please, can you help me to solve these questions. I didn’t work with push pull converter before so I am trying to understand the concept of how it works but still I couldn’t figure out how to answer the below questions. (The figure of the circuit is attached)What is the maximum current that we can supply with this design?What is the maximal and minimal turns ratio of the transformer to make sure that the power is reliable?

author: jonpaul
author_rank: Super Contributor
timestamp: October 18, 2022, 06:15:14 am
content: Find mfg app note transformers specifiedread transformers spec sheetSuggest that you read mfg app note, read power converter topologies book or app note.Jon

author: Stray Electron
author_rank: Super Contributor
timestamp: October 18, 2022, 01:53:23 pm
content: I think this is your transformer. It's looks like a 1 to 1.7 ratio. <https://www.digikey.com/en/products/detail/w%C3%BCrth-elektronik/760390013/3831261>.it looks like this PS is made to supply 150Ma at 3.3 volts. That's what's shown on the output; and the same amount of current in shown on the input of the last IC but at at 5 VDC.You REALLY need to look up the data sheets for the transformer and all of the ICs to see exactly what's going on but the voltages and currents shown in the schematic should be correct.What is the maximum current that we can supply with this design?Look up U25 and see what the data sheet says. But the other parts of the circuit may also limit the amount of current.What is the maximal and minimal turns ratio of the transformer to make sure that the power is reliable?It's not that simple! If you want "reliable" stay with the circuit as designed.

author: jwet
author_rank: Frequent Contributor
timestamp: October 18, 2022, 02:39:56 pm
content: The transformer driver is 1 amp at 3.3 in, so 3.3W input. Efficiency is pretty high so secondary will be 5.5 v and best current will be about 550 mA. The first linear downstream is 5v and is only rated for 150 mA, this drives the second linear 3.3 too. The total that can be drawn for the 5v and 3.3 is 150 mA. Say 75 mA each. The 5v regulator will be dropping very little, the 3.3 will be dropping 1.7v.A couple of other observations- this circuit is riddled with issues. Only one of the two drivers on the left is used- either the Maxim or TI part. The PN for the TI is incorrect, it should be SN6505- typo. Also this is a misapplication of the Maxim part, it designed for HV input 8-36v, it won't work at 3.3v - they make other parts that will work at 3.3v input (253,845 and others)The 1:1.7 ratio of the transformer doesn't look right either- it should be more like 1:2 or 1:2.2. This 1.7x would take the 3.3v and boost to 5.5 which is not quite enough for diode drops (.3v), regulator dropout (140 mV- full load) and resistive losses in the secondary. (probably 150 mV). The 5v regulator needs some headroom for good noise rejection. (see its data sheet). The noise is specified with 1v of headroom across the regulator. Near dropout- it doesn't do anything. One final problem is that the abs max of the linear is 6v which doesn't give you any margin. Also if you boost much higher, you'll increase the power dissipated in the 5v linear, lower efficiency, etc. Basically not a well designed circuit. Won't work.As others have said, look at the data sheets for details. The transformers for these parts are made by Halo and others for Maxim or TI when the IC was defined. They're somewhat specialized and have other specs besides just turns ratio that make them special- imbalance, leakage, etc that are important in this bang bang circuit.These little supplies are for galvanic isolation and are designed to run a usually small isolated bit of circuitry on the far side. They are commonly available in modules and unless your volumes are very high, this will be cheaper than sourcing all the parts and rolling your own.