What type of load is driven by this MOSFET? What type of MOSFET You use in this circuit?

I am using SiHP15N60E 600v, 15A MOSFET. I am using this MOSFET in Boost converter circuit. 

Please Sreeanth,

could you show us the input and the output wave forms of the MOS transistor.As the output voltage on the MOS transistor is 20V at one ampere current. it means that the MOS transistor driver circuit  is weak and it must be made much stronger. In order to change the conduction  state of the MOS transistor you have to charge and discharge its gate source capacitance as rapid as possible. This is accomplished by decreasing the driver resistance  to the required limit such that it does not slow the rise time and fall times of the pulses. Since the bipolar transistor has large current capability, the are used as drivers for power MOSfets.

In conclusion you must strengthen your driver to switch on and off the power mosfet with high speed.

wish you success.

Thank you Artis sir and zekry sir for your replies.

I am attaching figures of driver circuit diagram, waveforms after opto coupler and waveforms after IR2110 driver in both normal condition and distorted condition.

scope2_1, scope_2 gives output waveforms after IR2110 in both conditons. 

scope2_5, scope_7 gives output waveforms after 6N137 in both conditons. We can clearly get information from scope_2 & scope_7 that under distortion case both wave forms amplitudes decrease from 15V and distortion occurring .

This distortion starts when voltage across MOSFET is greater than 20V or current through it more than 1A.

Zerky sir, I even tried with reduced resistance from driver output to Mosfet gate, but situation remained same.

IR2110 has high- and low-side drivers.

For standard Boost converter circuit low-side driver would be used.

MOSFET that you are using is specified for V_GS=10V, so supply for IR2110 should be higher (Vcc for low-side driver), usually 15V.

Driver itself can sink or source gate current typ. 2A, but driver supply should be capable of supplying that much.  

Capacitors in gate circuit will not help. If driver supply is stabilized (and for best results it should) you should not use zener. Serial resistor between driver output and gate is preferred (usually couple tens of ohms), and value will influence rise and fall time, but not steady-state. 

Check connection between Vss and COM. 

If you are using high-side switch than it would be the best to use separate (isolated) power supply between V_B and V_S. 

I would like to thank Dr. Kurtaj for his answer. From your circuit diagram and the attached wave-forms and after some analysis it is the optocoupler  which i not able to derive the IR2110 driver chip. It outputs less than 10V while it must output 15 V as in the case without distortion in scope 2-5.

Therefore you have to strengthen the output at pin 6 of the optocoupler. This can be best done by decreasing the pull up Resistance connected between pin 8 and 6 of the optocoupler. If this not helping one can can put a logic driver stage at the output of the optocoupler. This will be sure the the solution.

wish you success

I suppose that error in connection from opto-coupler (OC) to pin 13 is only in drawing and not in reality, where it should go to pin 12 (LIN). 

One thing to fix (but since it is working without load not only one) is supply voltage for 6N137 that is supposed to work with 5V (7V Absolute Maximum Ratings for 1 minut maximum). Easiest solution would be to use different OC, or for testing maybe to use additional 5V supply for OC and for logical part of IR2110 (V_DD pin). 

Input current LIN pin according to specification is expected to be max 40 uA, and 3K3 that you are using should do the job. 

One thing to take care of is to separate signal from power circuit. 

Check by voltmeter supply voltage directly at IR2110 pins for some lose connection. 

Thank you kutraj sir and zekry sir,

opto-coupler (OC) pin 6 will  go to pin 12 (LIN). I drawn wrongly. I will check ckt once again 

1. Using different opto coupler

2. using different supplies for both ICs.

@zekry sir, what is meant by "put a logic driver stage at the output of the optocoupler. "

? and how to implement it?

I would propose before 

1. Using different opto coupler

2. using different supplies for both ICs

please try to decrease the pull up resistance ,whose value in your circuit is 2.3 kohm, to increase the ability of the optocoupler to push more current in the input of  IR2110 . Other wise use a cmos line driver such as TC4468 chip. This chip contains 4 drivers and every driver is composed of two stages an AND stage followed by a line driver. You can connect the the two inputs of the AND gate to the output of the optocoupler and the output  to the input of IR2110.

For the data sheet of TC4468 please refer to ww1.microchip.com/downloads/en/DeviceDoc/21425b.pdf

wish you success.

Srekanth, do connect a 100uF 25V electrolytic capacitor across the supply pins of the IR2110 along with a 0.1uF ceramic disc capacitor in parallel. The charging current and discharging current peaks from the MOSFET gate have to be supplied and absorbed respectively since your power supply incoming wires have stray inductance, which will create dips in your power supply itself. Have you checked the power supply voltage waveform on an oscilloscope?  Also connect a 1kohm 0.25W resistor across the gate and source terminal pins of MOSFET.

@Sujit K. Biswas

Sir can you tell me the adv. of using resistor across the gate  to source of MOSFET. 

How to fix the value for it..

Regards

The gate-source resistor tends to discharge the gate input capacitance. In case of high dv/dt across the drain-source, the transient current flow through the d-s capacitance will also charge up the d-g and g-s capacitances. If the g-s capacitance voltage increases above the gate threshold voltage, the MOSFET will turn on. The g-s resistance prevents this. Also, if other transients arrive, the gate voltage cannot rise up easily by charging the gate capacitance due to the resistance.

The resistance value is to be decided in conjunction with the g-s capacitance. Its usual value lies around 1kohm to 4.7kohm per MOSFET. Since it acts as a voltage divider along with the series gate resistance for the gate applied voltage, very low value will reduce actual voltage reaching the gate and hence may not permit driving the MOSFET in the saturated region, causing additional power loss and temperature rise in MOSFET. Larger value will make the resistance ineffective.

@Sujit K. Biswas

Thank you for replay

Sir I am using IRFP460 MOSFET and driving the gate with opto IC with 15V at its collector and my MOSFET is driving a relay coil form 110V battary. Its continuous current is 2A . So what would be the gate to source resistance 

1kohm is good enough. Remember that there will be an additional load of 15mA on the opto output during its high condition. This should be acceptable. Otherwise, increase the value to 2.2kohm for 6.8mA loading.

 I think thant your  circuit will never work at more than 20V(and have high losses between 15 and 20V because low voltage in the gate). The source of the MOSFET needs to be pulled to ground from time to time to recharge the 1 uF bootstrap capacitor. So you have to add another (small) MOSFET to the other output of the IR2110 and swith it on when the main MOSFET is off. By the way, It seems that there are errors in your diagram, for exemple the output of the optocoupler is not connected to the input pin.

Francesc, which circuit are you referring to, that of Bharath or of Sreekanth ?

To the Sreekanth one. As an example of what I said here is a buck circuit. At full load D5 conducts every cycle and the source of Q1 goes to ground, charging C5, but in discontinous operation D3 may not conduct and Q2, Q3 do the job.

By the way, it is simpler to use the gate driver shown in the diagram that half an IR2110.