Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronic 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.

Register Log in

Load Regulation in Boost Converter

Status
Not open for further replies.

mengghee

Full Member level 3
Joined
Nov 29, 2005
Messages
164
Helped
8
Reputation
16
Reaction score
5
Trophy points
1,298
Location
United Kingdom
Activity points
3,112
In the theoretical design of the boost converter, if i am not wrong, changing the load current wouldn't affect the output voltage. but in practical design, it does ... can anybody tell me what causes the variation of output voltage ? thank you.


regards,
mengghee
 

rts

Newbie level 4
Joined
Feb 24, 2006
Messages
6
Helped
4
Reputation
8
Reaction score
0
Trophy points
1,281
Activity points
1,356
as for all converters variation in output voltage
causes :
For DC :
= duty cycle
= load current
= input voltage
For AC :
= all above
= switching frequency

Because real elements contains conduction loss as minimum (serial resistors connected with ideal element) and paracitic inductance and capacitance.

For example your mosfet contain drain to source
resistance in open state. This resistance create voltage drop that impact on output voltage.
the same story with diode voltage drop,
inductance dc resistance.

Please provide more info about your question.
 

    mengghee

    points: 2
    Helpful Answer Positive Rating

mengghee

Full Member level 3
Joined
Nov 29, 2005
Messages
164
Helped
8
Reputation
16
Reaction score
5
Trophy points
1,298
Location
United Kingdom
Activity points
3,112
thank you,

my question was ... what causes the vout to vary if i change my load resistor ? in the equation vout = vin / 1- D . and through the equation. it has nothing to do with RL. thank you

regards,
mengghee
 

rts

Newbie level 4
Joined
Feb 24, 2006
Messages
6
Helped
4
Reputation
8
Reaction score
0
Trophy points
1,281
Activity points
1,356
yes for ideal case vout= vin/1-D , but for real case
when your inductance has DC resistance RL connected in series with it formula will be :
vout = [vin/1-D]*[1/(1+RL/(R*(1-D)^2))]
 

    mengghee

    points: 2
    Helpful Answer Positive Rating

mengghee

Full Member level 3
Joined
Nov 29, 2005
Messages
164
Helped
8
Reputation
16
Reaction score
5
Trophy points
1,298
Location
United Kingdom
Activity points
3,112
thanks rts,

do you have any reference that i can see ? i am just wondering how that is derived


regards,
mengghee
 

v_c

Advanced Member level 2
Joined
Oct 11, 2005
Messages
514
Helped
113
Reputation
226
Reaction score
37
Trophy points
1,308
Location
Chicago, IL, USA
Activity points
8,229
take a look at page 30 (output voltage) and page 31 (efficiency) of this document
http://ece-www.colorado.edu/~pwrelect/book/slides/Ch3slides.pdf

The expressions show how the output voltage \[V\] and the efficiency \[\eta\] are influenced by the diode forward drop \[V_D\], diode on resistance \[R_D\], inductor series resistance \[R_L\], MOSFET on resistance \[R_{on}\] and load resistance \[R\]. Note that \[V_g\] is the input voltage, \[V\] is the output voltage, \[D\] is the duty ratio and \[D\prime=1-D\].

By the way, if you take the expressions for the output voltage and efficiency and set \[V_D\], \[R_D\], \[R_L\], \[R_{on}\] to zero, you will arrive at the theoretical values of \[V=V_g/D\prime=V_g/(1-D)\] and \[\eta=1\]=100%.

Best regards,
v_c
 

    mengghee

    points: 2
    Helpful Answer Positive Rating
Status
Not open for further replies.
Toggle Sidebar

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Top