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[SOLVED] Hold up capacitor design using MOSFET

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aru_bes

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Hi,

I have to implement the 20ms ,50W Hold up circuit in my design.The voltage is 12V.Based on this, I implemented the MOSFET with 220mF capacitor based Hold up circuit.While simulating the circuit,the cathode of D1 waveform is different from capacitor output.Altenatively,If I replace the diode instead of MOSFET,Im getting the proper output.

Can you pls explain the reason behind this?
Attached the circuit and waveform.

Thanks-Arumugam
 

Attachments

  • Holdup_Waveform.png
    Holdup_Waveform.png
    20.3 KB · Views: 109
  • Capacitor_Hold_up.png
    Capacitor_Hold_up.png
    9.4 KB · Views: 108

Hi,

your description is confusing.
If you are talking about two setups, then post both setups, so one can directely refer to the schematics.

Usually with 5 Ohms loadn amd 220mF capacitor you get a tau of 5 x 220 ms = 1.1s

In your given circuit the power supply shuts down with 10mOhms to GND.
Then the current flows back from capacitor via Mosfet to power supply.

Klaus
 

Hi Klaus,

Thanks for your reply.I have to implement 20ms hold up time at load.The 20ms is amount of time taken to reach from 12V to 9V.Hence I used 220mf capacitor.I used 5ohms to provide the minimum possible charging time for 220mF capacitor and which gives the charging time as 1.1s.
My question here is,when the power supply is turning of ,the current flow from capacitor will go to RL as well as power supply.Is there any other way to avoid current flow back to the power supply when the power supply is turned off condition.
I used the below methods to avoid this:

1.Added the diode instead of MOSFET( solved the problem.But unnecessary ~2W PD in normal working condition).
2.Immediate turning of the MOSFET when the power supply is turned OFF by properly controlling the Gate voltage.Not solved the issue that Im facing now
3.Added the additional MOSFET in series with this MOSFET.Not solved.
Any other method to address this issue
Thanks-Arumugam
 

Your circuit isn't well considered. The PMOS transistor will stay on as long as V(output) is above Vth and discharge the capacitor to input voltage source.
 

Hi,

use a schottky diode, then the power dissipation should be below 1W.

then you just need a capacitor of 18uF. Connect it in parallel to the load. No resistor required.

Klaus
 

Hi,
As per the below formula,I believe the 43.28mf capacitor is needed.Can you pls explain ,how 18uF capacitor solve my requirement.
Total Capacitance = 2 x P x Δt / (η x (V122 – V222))

where :
P = power at the load
Δt =hold-up time required
η = converter efficiency
V1 = charged capacitor voltage before power drop out
V2 = final input voltage before power supply shut down.


Given data:
P 50 W
Δt 2.00E-02 S
η 0.75
V12 12 V
V22 9 V

C 0.042328042 F

Thanks-Arumugam

- - - Updated - - -

Hi,

I solved the issue by adding the NMOS.Modified.png.

Thanks for the support
 

Hi,

C = I x t / dV

U_charged = 11.5 V (12V minus diode voktage drop)
U_discharged = 9V
t = 20ms
I = 11V / 5 Ohms (average from 11.5 to 9.0V plus some headroom)= 2.2A

C = 2.2A × 0.02s / (11.5V - 9.0V) = 17.6mF

Two simple devices: 1 schottky diode, 1 capacitor. Nothing else.

Klaus
 
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