Why current sensing resistor is so tricky ?

[Abdelwakil]

Hi
I didn't understand why when i short the current sensing resistor , the psu/adapter become unstable (voltage dropping and noise) and forever, even if i restore the original resistor !!

 

[DPG]

More information is required here. Ideally diagrams or photos too

 

[Abdelwakil]

More information is required here. Ideally diagrams or photos too

I just remove the low resistance resistor (the current sensing one) and replace it with a shunt ,that make the psu useless forever .

 

[DPG]

Can you give more detail. What value are the resistor and the shunt? How are you connecting them, and what are they powering?

 

[Abdelwakil]

Can you give more detail. What value are the resistor and the shunt? How are you connecting them, and what are they powering?

I talk in general , any type of psu , the resistor is from 4 to 1 ohm for most psus and it is connected between the emitter of the transistor and the GND .

 

[DPG]

Post a schematic diagram. Why are you shorting the current sensing resistor?

 

[Abdelwakil]

I replace it to improve output current ?

 

[DPG]

By shorting out the sense resistor you are losing the control. There is no feedback (which is usually a proportion of the output voltage) to the drive.

 

[Abdelwakil]

By shorting out the sense resistor you are losing the control. There is no feedback (which is usually a proportion of the output voltage) to the drive.

But my experience is not reversible (restoring the original resistor doesn't restore the psu to its normal state)

 

[DPG]

Probably killed the FET or the bipolar transistor. Test them and see.
You won't magically increase current output by shorting the sense resistor. You would need to re-design the power supply and uprate some of the components.

 

[static zap]

I worked previously on Switch mode PSUs ,
early attempts we called Self destruct mode PSUs ,
Detecting that the transformer core is saturated is an important event !
(similar to old self oscillating 12V caravan Fluorescents )

 

[marconi]

I didn't understand why when i short the current sensing resistor , the psu/adapter become unstable (voltage dropping and noise) and forever, even if i restore the original resistor !!

Abdelwakil: I will try to help you understand. IRF840 is a power MOSFET working in the enhancement mode - which means it requires a gate to source voltage Vgs to turn it on. If it was a depletion mode MOSFET it would require a Vgs voltage to turn it off. The way to recognise enhancement mode is the dashed vertical line to which the arrow points. Depletion mode has a solid vertical line.

Enhancement-mode N-Channel MOSFET

Broadly speaking, the E-mode MOSFET can operate in three regions; Cut-off - fully off, Ohmic and Saturation - fully on. In switching applications the MOSFET operates in the Cut-off and saturation regions to minimise Ohmic heating in the transistor. In you application this is achieved by applying a sinusoidal voltage feedback signal (-MOSFETs are voltage driven devices-) derived from the 3 turn winding of the transformer. This voltage is applied to the gate and source of the MOSFET. During the positive half cycles of this feedback signal Vgs is high enough to cause the MOSFET to turn on and operate in the saturation region. During the negative half cycles Vgs is less than 0V and so the MOSFET is cut-off and not conducting. The MOSFET then is switching on and off in response to the polarity of the feedback signal. Current impulses - the drain current Id - flow during every quarter cycle of the voltage waveform across the upper 15 + 15 turns coils, regularly exciting oscillations of the LC circuit. The LC circuit will resonate at a particularly frequency determined by the values of L and C and will present a resistive load to the switch at this frequency. (You need to remember that the voltage in the 3 turn coil is proportional to dId/dt so if you draw out the Id sinusoid, then overlay dId/dt on it - a cosinusoid waveform - you will see that there is only one quarter cycle (or 90degree conduction angle if you prefer) when the feedback signal which provides Vgs is positive and this turns on the MOSFET. The MOSFET only conducts for positive Vds.

derivative of a sine wave - Google Search - https://www.google.com/search?q=derivative+of+a+sine+wave&rlz=1C1GCEA_enGB822GB822&sxsrf=ALeKk01a_i-cet0HcO_CpPrzG5l4y2XxzQ:1582806890817&tbm=isch&source=iu&ictx=1&fir=j1Ga93ncxJSp7M%253A%252CnXdPWRuou0ZzwM%252C_&vet=1&usg=AI4_-kTHzVBtHtukDsIhRwMIVGnbRYbJcw&sa=X&ved=2ahUKEwjmquaT3_HnAhVxlFwKHU0jD6EQ9QEwGHoECAUQCQ#imgrc=j1Ga93ncxJSp7M

The only thing that limits the current through the MOSFET is the resistance of the 30+30 winding and of the MOSFET. To safeguard the MOSFET by limiting the drain source current two components are added - the 16-18V Zener diode Zd and the 0R22 source resistor r. As the drain source current increases the voltage rises across 0R22. This voltage rise acts to reduce the Vgs potential difference. If the reduction of Vgs is high enough the MOSFET will begin to turn off ie enter the Ohmic region and thus the effective resistance to the flow of drain source current rises too acting to reduce the current - this is a form of feedback to limit the peak drain current. The Zener limits the Vgate to ground voltage to 16-18V; assuming say a Vgs of 1V, 16 = Vgs + Vr, so 16 = 1 + (Is x 0.22); Is max is of the order (16-1)/0.22 = 68A. Any more current than this and Vgs becomes low enough for the MOSFET to begin to turn off or enter the Ohmic region.

Now if you short out r, Is max is no longer sensed to produce a voltage Vr which backs-off against the applied Vg to ground voltage to reduce Vgs and thence limit Is by turning off the MOSFET/decreasing the conductance of the drain-source path. There is no limit on the drain source current so the MOSFET risks being operated beyond its rating. I suspect this is what has happened to IRF840 - damaging it.

Or something like this. This gives you some idea why 0R22 is necessary and reducing its value was a bad idea.

 

[DPG]

Marconi's excellent reply has made my previous post look rather poor.

 

[marconi]

Marconi's excellent reply has made my previous post look rather poor.

Recalling what Newton said, 'if I have seen further it is because I stood on the shoulders of giants such as you'.

 

[DPG]

Recalling what Newton said, 'if I have seen further it is because I stood on the shoulders of giants such as you'.

I believe that is printed on the border of two pound coins.