Hi i have the circuit as follows, the opamp is used to maintain contant current through the sense resistor. how is it possible to get very high current at low voltage...but the current shouldnot exceed 24 amp.
the problem is that if i use very low RDSon and sense resistor less than 10[m Ohms], then I need at least 2-3 volts to drive the gate, when the voltage across the sense resistor is 2-3,, and value of resistance is 10 [mOhms], the current is massive as much as 200-300[A], if i use low votage, the mosfet will not turn on,

I am just a beginner electrical student. ..please hit me with some clues and ideas, Thanks in advance.

 

constant current when the pulse is on..i mean that

 

Hi i have the circuit as follows, the opamp is used to maintain contant current through the sense resistor. how is it possible to get very high current at low voltage...but the current shouldnot exceed 24 amp.
the problem is that if i use very low RDSon and sense resistor less than 10[m Ohms], then I need at least 2-3 volts to drive the gate, when the voltage across the sense resistor is 2-3,, and value of resistance is 10 [mOhms], the current is massive as much as 200-300[A], if i use low votage, the mosfet will not turn on,

I am just a beginner electrical student. ..please hit me with some clues and ideas, Thanks in advance.
View attachment 47746

Use a higher power supply voltage on the op amp.

 

if i use high voltage eg....4 V then the current on the sense resistor 20 m ohm would be as much as 200A and flow in opposite direction...isn't it....towards 0.5..any suggestion...comments

 

Its just Ohms law mate, you want 20A at 20-milli ohms so thats 400-milli volts.

So alter your voltage and resistance to give you 20Amps.

20A = V/R

 

You have an op-amp circuit with negative feedback. The op-amp will drive to whatever level is required to hold its inverting input at the same voltage as its non-inverting input.

So, as DodgyDave said ... its just ohms law. The sense resistor should be sized as: Rs = Vin / 20A.

Don't forget to calculate the power dissipation in your resistor and choose an appropriately sized resistor.

Also, make sure to do a thermal calculation on your MOSFET to make sure it won't get too hot. Just calculate P = I^2R = 20A^2 x RDS(on), then multiply this by the junction-to-ambient thermal resistance of the MOSFET as given in the datasheet. This will give you the temperature rise in the MOSFET. If its more than the allowed maximum junction temperature of your MOSFET, then you'll need to use a heatsink. Be conservative because the power dissipation you calculated above is only the conduction loss in the MOSFET (I^2R). There is also a switching loss that occurs because the MOSFET does not turn on/off instantly. There is a small amount of time when it is transitioning from on/off or off/on and during this time it has voltage across it and current through it, therefore it is dissipating power - this is called switching loss.

 

You have an op-amp circuit with negative feedback. The op-amp will drive to whatever level is required to hold its inverting input at the same voltage as its non-inverting input.

So, as DodgyDave said ... its just ohms law. The sense resistor should be sized as: Rs = Vin / 20A.

Don't forget to calculate the power dissipation in your resistor and choose an appropriately sized resistor.

Also, make sure to do a thermal calculation on your MOSFET to make sure it won't get too hot. Just calculate P = I^2R = 20A^2 x RDS(on), then multiply this by the junction-to-ambient thermal resistance of the MOSFET as given in the datasheet. This will give you the temperature rise in the MOSFET. If its more than the allowed maximum junction temperature of your MOSFET, then you'll need to use a heatsink. Be conservative because the power dissipation you calculated above is only the conduction loss in the MOSFET (I^2R). There is also a switching loss that occurs because the MOSFET does not turn on/off instantly. There is a small amount of time when it is transitioning from on/off or off/on and during this time it has voltage across it and current through it, therefore it is dissipating power - this is called switching loss.

The MOSFET could have volts across it, depending on the load. The power dissipation is 20A x Vds, not 20A^2 x RDS(on). The MOSFET can get very hot in a very short period of time, possibly being destroyed unless low duty cycle and/or adequate heatsinking are employed.

 

thanks for the help...a lot...learning lot of things now

 

Ron is correct about the conduction loss in the MOSFET.

I design a lot of switching power supplies and the conduction loss in that case is I^2 x RDS(on) because the MOSFET is driven hard on. Thus you use the "on" resistance. This is what I was thinking.

On this case the mosfet is not fully on, and thus you can't use the "on" resistance. This MOSFET is being used for a current source and may well have volts across it!

Thanks Ron for straightening this out.

 

One more thing ... if your input to the op-amp is really a square wave as shown in the sketch, then you need to take the duty cycle into account when calculating the power dissipation in the mosfet.

 

One more thing ... if your input to the op-amp is really a square wave as shown in the sketch, then you need to take the duty cycle into account when calculating the power dissipation in the mosfet.

And the frequency, if it is really low (or high).

Also, frequency compensation is generally required when using large MOSFETs, due to the pole caused by the gate capacitance in conjunction with the op amp Rout.

 

Hi JMac... thanks for all the advice..i want to show you the following multisim simulation...i havenot made choice of the mosfet and the sense resistor...how do i calculate the maximum power dissipation on the condition like this...and how to determine max frequency or dutycycle...or which one should i give the priority

thankyou very much for all the tips..

 

and one more question to make the above questions clear....how can i calculate the total energy dissipation so that i can let the user choose the dutycycle and the frequency...of the square wave input

 

What is your load? The inductor in your diagram, is that a motor winding? If it is, you need a reverse diode across it to dump the energy when the switch turns off.

The maximum frequency is going to be limited by how fast the MOSFET can be turned on/off. MOSFETs have significant gate capacitance and the op-amp has to charge up this capacitance to get the gate up to the desired level. Pick a MOSFET with a low gate charge Qg. If you find that the max frequency is not fast enough, you can add a gate driver between the op-amp output and the MOSFET gate.

Power dissipation is just the power dissipated when the MOSFET is on, P=(VDS*I), multiplied by the duty cycle.

 

Once you get a power dissipation number calculated, multiply it by the junction-ambient thermal resistance in the MOSFET datasheet to get the temperature rise. Add the temperature rise to the maximum ambient temperature your circuit will operate in. This is the max temperature your MOSFET will see. For reliable operation, make sure this is well below the specified max junction temperature given in the MOSFET datasheet, usually 150C. As a rule, I never let anything get over 100C, and try to stay well below this.

If your calculation shows the MOSFET is too hot, you can do a couple things:

(1) Choose a MOSFET in a package with a lower thermal resistance

(2) Heatsink the MOSFET and/or add a fan to cool it

 

What is your load? The inductor in your diagram, is that a motor winding? If it is, you need a reverse diode across it to dump the energy when the switch turns off.

If the inductor is really 30nH, flyback voltage will not be an issue, given the slew rate possible with this circuit.

 

Probably right. But if I were laying out a PCB for this design, I'd include a spot for the diode just in case.

Be sure to test this by putting a scope on the drain of the MOSFET and checking the max voltage.

 

the inductor is supposed parasatic inductance....i wanted to check if the slew rate 5A/us was possible or not...since di/dt=5A/us=V/L...thanks for the suggestions...if i have some problems again in future...im looking forward for help...thanks again...

 

sorry for a lot of questions... but what i have been stuck is...@ calculation max energy dissipation...
To provide some more detail about the circuit.
The worse case situation is 24[A]@20[V]. So, my peak power can be as much as
480[W], but if I keep the pulse width and the frequency maintained...i am not sure but i think that the mosfet can dissipate that amount of power....now for eg....if i limit the user to dutycycle of max 1% at 100 Hz or 10% at 1kHz, mosfet could withstand dissipation...so I want to calculate the max energy dissipation and let the user calculate the dutycyle and frequency at variable voltage and current case...!!

 

What is your load? In the drawing you don't show any load other than the parasitic inductance.

To calculate power dissipation you need to cmultiply the current (24A) by the voltage drop across the mosfet (VDS). You need to know your load to calculate the voltage across the MOSFET when the circut is sinking 24A.

No, your mosfet certainly won't be able to dissipate 480W! You determine what a mosfet can handle by making sure its junction temperature stays below the specified limit - as I described in a previous post. Lets take a common TO-220 mosfet, the IRF640 with thermal resistance=40degC/W. At 480W, the temperature rise is Trise=(480W)(40C/W)=1920C and the mosfet is a puff of smoke

So you really need to properly calculate the power dissipation by figuring out VDS, multiplying it by the current (24A), applying the duty cycle, then multiplying it by the thermal resistance of your mosfet.