Understanding a flyback circuit PWM controller - Page 1
Forum section: electronics / beginners

author: gkmaia
author_rank: Frequent Contributor
timestamp: February 10, 2019, 11:19:27 pm
content: I got a power supply with feedback circuit that does not feature an optocoupler and has been reverse engineered on my coloured sketch.Attached there's another circuit, similar that makes use of a TOP100. I my case my circuit uses a Panasonic MIP 2E3DMY PWM Switch.On 2E3DMY Control voltage is 10V.If on both cases when the voltage on the control capacitor exceed the PMW control voltage the current will flow into the control pin that will set decrease the duty cycle.With that in mind I have a question. What would happen if I replace my PMW by one with a control voltage off 8v?

author: Wimberleytech
author_rank: Super Contributor
timestamp: February 10, 2019, 11:30:28 pm
content: Your schematic makes my head hurt.

author: gkmaia
author_rank: Frequent Contributor
timestamp: February 11, 2019, 12:54:36 am
content: Quote from: Wimberleytech on February 10, 2019, 11:30:28 pmthere you go!

author: Yansi
author_rank: Super Contributor
timestamp: February 13, 2019, 02:01:51 am
content: Yes, the schematic is childish.But, to answer your question: Replacing it with an IC with lower reference would (surprise!) result in lower output voltage, proportionally.But... do NOT! swap those ICs, as they may and will have different peak current limits and frequency of operation.

author: gkmaia
author_rank: Frequent Contributor
timestamp: February 13, 2019, 02:18:35 am
content: Quote from: Yansi on February 13, 2019, 02:01:51 amThe reason the schematic looks like that is it because it is just an x-ray overlay of the current circuit made with Sketch (UI tool). Colours are just to help identifying the nodes. I could then translate it to a proper schematic using a proper EE tool but is it worth the time?!No I wont swap those. I was just trying to understand if my logic was right.When you say different frequencies what you mean? They both output at 100Khz. Are you talking about the restart frequency which is 2 and 1.2hz?Thanks for the help!

author: Yansi
author_rank: Super Contributor
timestamp: February 13, 2019, 02:54:43 am
content: I am talking SWITCHING FREQUENCY. Restart frequency is a completely different matter. (read the datasheet to see what it is used for).You can also check and compare the peak currents of both devices.There is no need for graphical computer overlay tool with fancy colors or EE schematic suite program. Good old pencil and paper would be very likely much quicker for this purpose. Do not complicate yourself the job with unnecessary computerization of tasks

author: gkmaia
author_rank: Frequent Contributor
timestamp: February 13, 2019, 03:09:01 am
content: The oscillators on TOP100 (or any TOP) and MIP clock at 100khz.In terms of current the TOP 225 will be more suitable.The thing is the MIP shunt at 6.3 and the TOP at 5.7. I could not find anything with the same package to shunt at lower voltages. Anyway... I guess it was just a learning exercise to understand better primary feedback regulation.I understand your point of view in terms of tools. What I like about doing it that way is that it is much harder to miss anything and much easier to zoom in and check traces, SMDs, etc... and I can do it sitting on my couch while I have some snacks. But that is up to preference I guess. A pencil and a magnifier would do as well.

author: Yansi
author_rank: Super Contributor
timestamp: February 13, 2019, 02:48:19 pm
content: The current is "not suitable" but must be the correct one, for which the transformer was designed.Too low peak turn off current will not provide enough output power, too high current may saturate the transformer core and blow the shit out of your IC.Only smaller current replacement is acceptable - in that case frequency will need to be increased to obtain same output power.The theory behind operation of flyback is just that with each period of the switching cycle, you push some small amount of energy through to the secondary.P = 0.5 * f * Lp * Ipeak2Where L is the primary magnetizing inductance and Ipeakthe peak switch off current.