Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

How SMPS regulates the Voltage

Status
Not open for further replies.

i200yrs

Full Member level 2
Full Member level 2
Joined
Aug 23, 2016
Messages
147
Helped
1
Reputation
2
Reaction score
2
Trophy points
1,298
Activity points
2,365
How SMPS regulates the Voltage?

I have done my research, I always see regulation is done by varying duty cycles...But still could not get to understand fully how the smps power supply regulates the voltage. As an example, how the smps regulates 12Vdc output from high voltage DC at primary side...Really appreciate someone helps....thanks a lot.
 
Last edited:

There are many modes of control for SMPS and also there are different type of control. Read a good book. I will suggest Power Electronics Converter by Robert Erickson will give you full idea about voltage regulation of SMPS.

But if you want a quick idea read the following article
https://www.edn.com/design/power-ma...-isolated-DC-DC-converter-voltage-regulation-

And also give a quick search on current control mode of DC DC converter which is one of the most widely used control method in SMPS.
 
Hi,

there are many different SMPS techniques.

But for most of them the folowing is true:

Example:
Let´s say it is a 100kHz DCM flyback SMPS, input voltage 320V DC, inductor primary inductance of 2mH, output: 12V, 1A

12V, 1A means 12W.
100kHz means 10us cycle time

Now calculate the energy that the load draws within one cycle:
12V x 1A x 10us = 120uWs.
(100.000 times per second the inductance has to move 120uWs of energy form primary side to secondary side.)

Now the energy stored in the inductance can be calculated as 0.5 x I x I x L
so we can calculate I: 120uWs = 0.5 x I x I x 2mH --> I = 346mA
(100.000 times per second the inductance needs to be connected to the 320V input voltage ..... exactely as long that the current just reaches 346mA)

But how long does it take for the current to rise to 346mA:
L = I x t / V; 2mH = 0.346A x t / 320V --> t = 2.16us

This means 100.000 times per second the primary switch needs to be ON for exactely 2.16us.
2.16us every 10us means a duty cycle of 21.6%

(This is an ideal calculation. Reality may differ somehow)
****

Klaus
 
  • Like
Reactions: i200yrs and j33pn

    j33pn

    Points: 2
    Helpful Answer Positive Rating

    i200yrs

    Points: 2
    Helpful Answer Positive Rating
    V

    Points: 2
    Helpful Answer Positive Rating
Status
Not open for further replies.

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top