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.

Register Log in

SMPS switching Transistor continues current

Status
Not open for further replies.
B

baby_1

Advanced Member level 1
Joined
Dec 3, 2010
Messages
415
Helped
1
Reputation
2
Reaction score
1
Trophy points
1,298
Activity points
4,277
Hello
as you know for SMPS we use a power transistor as electronic switch.i choose SSP4N60AS.as i see the continues current is 4A.but if my input voltage (VDS) is 325 volt is turn on the mosfet the current that draw is about 130Amper(325/2.5)(2.5 Rds).i see a lots of smsps circuit that they use a resister between .2 to .68 in Source to grand of mosfet.but if you calculate again the current now is 93Amper.

what is my fault? how can we protect our switching mosfet ?
 

baby_1 said:
if my input voltage (VDS) is 325 volt is turn on the mosfet the current that draw is about 130Amper(325/2.5)(2.5 Rds).
Click to expand...
Hi

The current will not be so high. Remember the inductor is between the MOSFET and the 325V VDS.

When the MOSFET switched on, there is very little voltage across it - most of the voltage is across the inductor. The current through the inductor and MOSFET will rise until it is a little more than 4A, then the MOSFET switches off again.
 
  • Like
Reactions: baby_1

    B

    baby_1

    Points: 2
    Helpful Answer Positive Rating
Thanks
i see that the inductor of transformer does have a low resister, doesn't it? so we have a low voltage drop on inductor.
 

In a SMPS the power transistor are continuously switching on and off with high frequency in order to provide the transfer of electric energy via energy storage components (inductors and capacitors). So in general the power transistor and the inductor (transformer) are placed in series, and the transformer accumulates energy in the magnetic field as current flows through it, then transfers all or a portion of this energy into another circuit during the alternate part of the switching cycle. By varying duty cycle, frequency or a relative phase of these transitions an average value of output voltage or current is controlled.
The transformer inductance value determines circuit current, not the resistors.
 
  • Like
Reactions: baby_1

    B

    baby_1

    Points: 2
    Helpful Answer Positive Rating
Vl=L\[\frac{ di}{dt }\]

dt=1/switching frequency
L=is define

VL and di is unkown how can we calculate them?
 

VL and di is unkown how can we calculate them?
Click to expand...
How about assuming VL = Vdc (input voltage) and ignoring all small voltage drops (Rds, Rshunt, Rprim) in a first attempt?
 

Thanks FVM
if i put VL=325 then we have
L=100micro Henry
so we have
\[\frac{ di}{dt }\] =3.25MegaAmper/second !!!! wow
 

3.25 A/µs sounds more handy, still a bit high, but you didn't talk about specifications (power, switching frequency).
 

if my frequency be 40khz how many current does we have?
 

Status
Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top