How to calculate output current on secondary of flyback, when switch turns off - Page 1
Forum section: electronics / beginners

author: MacIntoshCZ
author_rank: Regular Contributor
timestamp: January 29, 2025, 09:29:08 pm
content: Hi,i have some gap in theory i guess. What is aprroach to solve this circuit if i know only maximum value of magnetizing current on L1 (550mA), input voltage and inductances?Can you help me do understand that?I want to get to the values of IL2 -> 170mA, IL3 -> 75mA.Thanks

author: MathWizard
author_rank: Super Contributor
timestamp: January 30, 2025, 02:11:42 am
content: IDK either, I haven't really worked on these circuits yet. Have a look at this

author: Konkedout
author_rank: Frequent Contributor
timestamp: January 30, 2025, 04:49:46 am
content: Let's work this a little bit at a time...1) Did you intend to omit output rectifiers and output filter capacitors?2) The transformer (or call it a coupled inductor) cannot change ampere turns instantaneously. The total ampere-turns ramps according to dI/dt=V/L3) The inductance in a given winding is proportional to the square of the turns. I cannot see your schematic as I type this, but I think you included a 1 mH winding and a 5 mH winding. The 5 mH winding would have 2.24x the turns of the 1 mH winding.Let's start with the above and move on from there....maybe me or maybe someone else....

author: MacIntoshCZ
author_rank: Regular Contributor
timestamp: January 30, 2025, 04:53:59 pm
content: 1) Yes, its not "real" circuit, which i have on the bench. This is just simulation, what i have made to learn what is going on. So calling it "flyback" could be missleading.2) Yes3) Yes, know this so far.I know that if turn ratio is 1:1 and i have only one secondary winding then peak current on sec. starts from exact point when it end on primary. This is true if resistance is low enough, if i increase resistance then it drops. If i have different turns ratio, then current is Imax(primary) * N1/N2.I have also seen that if i add another secondary with exact inductance and exact load, then its splits peak current by half. So I would like to get into the point when i will know how it splits current if all inductances is completely differents and we have multiple outputs.

author: Picuino
author_rank: Super Contributor
timestamp: January 30, 2025, 05:40:07 pm
content: First you need calculate the number of turns (N) of every coil:N1 = 100 Turns (for example)L3 = L1 = -> N3 = N1 = 100 TurnsN2^2 = 5 · N1^2 -> N2 = 224 Turns (2.24 ratio)Second, convert impedances of secondary to primary:R2' = 1000/5 = 200 OhmsR3' = 1000/1 = 1000 OhmsBoth in parallel:R23' = 166.7 OhmV_R23' = V_L1 = R23' · 550mA = 91.67 VoltsNow, convert primary voltage to every secondary and calculate every secondary current:Vout = V_R2 = 91.67 * 2.24 = 205 VoltsI_R2 = 205 mAVout2 = V_R3 = 91.67 * 1 = 91.67 VoltsI_R3 = 91.67 mAI am not sure why the same values have not come out, maybe because of the coupling coefficient or losses or that I have made a mistake in something, but the procedure is this one I have written.

author: Konkedout
author_rank: Frequent Contributor
timestamp: January 30, 2025, 05:48:25 pm
content: Well OKIn my discussion BELOW I am interchanging power and energy. Understand that every time M2 turns off, the inductor (flyback transformer) has a fixed amount of energy to discharge. At any moment there is an amount of power being provided to each of your two load resistors. Energy is power integrated over time.We know that if you have 10V across a 10 ohm resistor, you get V/R = 1 Amp and V^2/R = 10 Watts.Now if you double the turns you will get double the voltage 20V so V/R = 2 Amps and V^2/R = 40 Watts. In other words the voltage scales with the turns and the power into a resistor scales with the turns squared. The inductance also scales with the square of the turns. So the power out of the winding will be proportional to the inductance of that winding.Therefore, your 1000 ohms connected to a 5 mH winding will dissipate 5 times as much power as your 1000 ohms connected to a 1 mH winding. So it will dissipate the same amount of power as if it were a 200 ohm resistor connected to a 1 mH winding. Does this get you going further?

author: MacIntoshCZ
author_rank: Regular Contributor
timestamp: January 30, 2025, 06:56:00 pm
content: Quote from: Picuino on January 30, 2025, 05:40:07 pmThanks, i have tried it with different values and it gets me result with like 5% accuracy.

author: MacIntoshCZ
author_rank: Regular Contributor
timestamp: January 30, 2025, 06:59:03 pm
content: Quote from: Konkedout on January 30, 2025, 05:48:25 pmThanks, this is good aproach!

author: MacIntoshCZ
author_rank: Regular Contributor
timestamp: January 30, 2025, 08:26:40 pm
content: Thanks, guys. I get it i guess.

author: MacIntoshCZ
author_rank: Regular Contributor
timestamp: January 30, 2025, 09:33:57 pm
content: Seems i celebrate to early.Stucked at Voltage inducted on primary.Imag = 0,47A; L1 =1m; L2 = 4m; L3 = 16m; R2 = 50; R3 = 500Rref2 = R2*(L1/L2) = 25Rref3 = R3*(L1/L3) = 31,25Rref = Rref2 II Rref3 = 13,89Voltage on primary -> Imag * Rref = 0,47 * 13,89 = 6,53VI dont know what i did before that it works, probably just an accident.

author: Picuino
author_rank: Super Contributor
timestamp: January 31, 2025, 07:25:16 am
content: The problem is that it is not a real Flyback circuit and part of the current is being transferred in the transformer during the first cycle, not in the second cycle (as it should happen in a Flyback).If you place a diode between L2 and R2 and another diode between L3 and R3, the calculation method I put above will work.

author: MacIntoshCZ
author_rank: Regular Contributor
timestamp: January 31, 2025, 02:56:35 pm
content: Quote from: Picuino on January 31, 2025, 07:25:16 amActually it will not. As you can see peak mag. current is same as before, so is inducted voltage on primary (there is little change, but it does not match with results from equations, change is probably due my meas. error and Vfdiode drop i guess)

author: Picuino
author_rank: Super Contributor
timestamp: January 31, 2025, 03:21:08 pm
content: What happens if you delete "K3 L2 L3 1" ?

author: MacIntoshCZ
author_rank: Regular Contributor
timestamp: January 31, 2025, 03:35:52 pm
content: I have gigaAmps and megavolts as the result. K3 should stay there there

author: Picuino
author_rank: Super Contributor
timestamp: January 31, 2025, 03:39:07 pm
content: Are you operating in continuous or discontinuous conduction mode?That is, does the current in the coils reach zero at the end of the cycle?

author: MacIntoshCZ
author_rank: Regular Contributor
timestamp: January 31, 2025, 04:17:26 pm
content: Its just one pulse. What you see here is after transistor turns off

author: MacIntoshCZ
author_rank: Regular Contributor
timestamp: February 01, 2025, 07:19:58 pm
content: Solution provided by Picuino was right, there was something wrong with simulation. I think i removed parasitic resistance in inductors and it started working.So, thanks very much everyone for help