Transformer understanding and turns calculation. - Page 1
Forum section: electronics / beginners

author: axawire
author_rank: Contributor
timestamp: September 14, 2021, 03:26:39 am
content: Hello,I am having some problems with understanding the following to apply it in different situations:QuoteSource:https://ludens.cl/Electron/Magnet.htmlDoes the 0.000003 seconds correspond to the rise time of the PD across the primary and does 0.000006 seconds correspond to the drop time of the PD across the primary or is it the time that must be "dead" where the primary has no current and so the magnetic flux contributed by the primary is currently zero for at least 0.000006 while the secondary is taking the magnetic flux that was stored and the dead time is any time where the PD is constant, either constant high or constant off? I do not understand how this rise time and drop time is controlled, is it purely a feature of the switching transistor or is it the fact of just using the time in the calculation means the number of turns selected forces that to be the charge and discharge times and as long as the times used fall with in the Frequency it would be clocked at we are good to go or something else?How do I extended this concept when it is a full H bridge type topology? I understand I would have a division by 4 but if I state I want a 0.000003 second charge time and 0.000006 discharge time can I still control it like this with a full H bridge, is the rise time in this situation the time to go from complete magnetization in one direction all the way to complete magnetization in the other direction and the discharge is the complete swing starting from the other side over the 0.000006 seconds or is the 0.000006 the time their must be no magnetic flux from the primary side, so no current flow through primary for 0.000006 seconds before the signal continues on the rest of the way to recharging the magnetic flux in the opposite direction to where it came from? I would like to take advantage of the situation where the voltage change is higher than the turns ratio or does this only work with a flyback topology and only step down?I hope that makes sense.Thanks in advance.

author: MIS42N
author_rank: Frequent Contributor
timestamp: September 14, 2021, 04:43:03 am
content: Confusing, yes. Here the magnetic core is an energy storage device. The impulse on the primary puts energy in by moving the magnetic field from zero to some amount. When the supply is cut off, the magnetic field collapses and finds the easiest way out. Because the primary is blocked, the easiest way out is the secondary. It doesn't matter if the secondary is one turn or 100 turns, if the primary is truly blocked the energy will come out the secondary even if it has to generate 10,000V. It is the way the spark is generated in a car, establish a field then break the primary, the secondary produces thousands of volts for a very short time.The ratio of windings establishes a relationship between the voltages experienced on the primary and secondary. In the case of the car ignition coil, the ratio may be 10 turns of the primary to 1000 turns of the secondary. When the primary current stops, the voltages are in proportion so while the secondary is delivering 20,000V the primary is trying to deliver 200V. But it is blocked by an open switch so the discharge goes out the secondary. While the primary switch is closed, the primary is receiving 12V from the battery and if you measured the secondary you would find it has 1200V on it. But that is not enough to cause a spark so the energy goes into the magnetic field.In the example you give, the ratio is the other way round. While the primary is delivering a charge at 300V the secondary will have -30 and it is blocked by the rectifier. Open the primary, the secondary rises to whatever voltage allows the rectifier to conduct (15V) and dumps the charge into the load at 15V, with the primary showing -150V.I was trying to think of an easy way to explain the time it takes to build up and collapse the magnetic energy, but I can't think of one. Your reference explains it. The dead time is because you have to be sure all the magnetic energy is transferred to the load before putting another pulse in. If that were not done, the energy would build up each pulse and eventually the core saturates and the primary starts to behave like a short circuit, or the extra energy is dissipated as losses and things get hot.I don't think it is applicable to an H bridge. If current flows in the primary at the same time as in the secondary then it is a standard transformer and the turns ratio determines the voltage ratio. So yes, this is only applicable for flyback situation, but can be step up (car ignition) or step down (example given).Hope that helps.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 14, 2021, 07:56:23 am
content: Use the correct voltage and time for either case. It is equivalent, indeed equal because the fluxes must balance.300V * 3us = 150V * 6usIn a full-wave (H-bridge) forward converter, you will have symmetrical (bipolar, i.e. it goes above and below 0) flux. The peak flux is given by the applied voltage and the time taken to rise from zero to peak. An on-period goes from negative peak to positive peak flux, so the correct time is half, or: Phi_pk = V * t_on / 2 = V / (4 f_sw).Tim

author: axawire
author_rank: Contributor
timestamp: September 14, 2021, 10:17:38 pm
content: Thanks everyone