Understanding Class D Amps

Monicaa

Member
2020-01-31 10:08 pm
I am a novice in designing class d amplifiers. I currently designing a class d amp with group of my classmates and we are confuse about how the amplification occurs. I know we need a gate driver to turn on the high side MOSFETs but when we applied the Pwm inputs to the gate driver the outputs is not a square waveform. After researching we believe that the gate driver will output the same Pam signal but with a higher output voltage and then send this signal to the MOSFETs for amplification.
Are wrong thinking the gate driver outputs a pwm signal?
 
The gate driver outputs a PWM signal.

PWM stands for pulse width modulation. It changes the duty cycle of the square waves produced by the mosfets. So for a negative half cycle of input you will see a narrower pulse than for a positive half cycle.

When to switch and thus the duty cycle is determined by a comparitor that compares a ramped voltage (usually a triangle wave) against the audio input. With no input the comparitor will switch, oscillator style, producing a 50% duty cycle.

When and how often it switches is determined by comparing the audio input to the ramped voltage. It will switch low for a voltage less than the ramp and high for a voltage higher than the ramp voltage. This, of course varies up and down the ramp giving you differing duty cycles that vary with the amplitude of the input signal.

All you need to do to get the audio out of that is to run it through a low pass filter to remove the switching waveform, leaving the audio on to go to the speakers.

More detail and perhaps a clearer explanation is HERE.
 

Monicaa

Member
2020-01-31 10:08 pm
Thanks to you both for replying.

@rayma, I don't think so, we are using the probes for our tektronix digital oscilloscope.

@Douglas Blake, yes I understand that, but we are inputting a pwm (complementary) signal of 50 kHz to the hip4081a gate driver and the output is seen in the image below. This is where we are confuse because we are suppose to get a pwm wave. Maybe is could be our bootstrap circuit, we are using a 10uF and 1n4148.

View attachment IMG_3954-converted.pdf
 

Monicaa

Member
2020-01-31 10:08 pm
Sorry, I didn't know that. Also, yes the low side output set source is grounded.
The MOSFETs full bridge was disconnected from the gate driver because she didn't want to blow the MOSFETs. My teammate decided to do that just to make sure that the gate driver is outputting a pwm signal. We are responsible for different stages.
 
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rayma

Member
2011-04-29 8:37 pm
Both output fets must be connected to the driver IC for the system to function.
When the main output is low, the diode charges the bootstrap capacitor.
When the main output goes high, the diode turns off, and the bootstrap capacitor provides
a floating voltage supply to the high side output fet driver so the high side output fet can be switched.

The center connection of the two output fets is what pulls the high side driver down
so the bootstrap capacitor can charge. No fets -> no bootstrapping action, no switching.
 
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rayma

Member
2011-04-29 8:37 pm
Sorry, I didn't know that. Also, yes the low side output set source is grounded.

Don't try to connect the unbalanced probe to the circuit. The ground clip will cause failure, either in the circuit
or in the scope/probe. If you are using only a small 20VDC-30VDC bus voltage for the output fets, you MAY be able
to safely scope ONLY the lower fet gate/source voltage if you are careful, but only if the lower output fet's source is
truly grounded (to the utility ground), since the probe's ground clip IS so grounded. (If you are using source resistors
in the lower output devices for current control, you cannot even do that, and you MUST use a differential voltage probe.)

Oscilloscope do not have differential inputs, and they measure only wrt the utility ground potential. You must use a differential voltage probe (that is rated at enough voltage to be safe) in place of the standard probe to do otherwise.
 
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Both output fets must be connected to the driver IC for the system to function.
When the main output is low, the diode charges the bootstrap capacitor.
When the main output goes high, the diode turns off, and the bootstrap capacitor provides
a floating voltage supply to the high side output fet driver so the high side output fet can be switched.

The center connection of the two output fets is what pulls the high side driver down
so the bootstrap capacitor can charge. No fets -> no bootstrapping action, no switching.

However the HIP4081A has a separate charge pump so it can work without bootstrapping if there's not too much load on the high side gate pin (just a MOSFET's input capacitance) and you don't switch very fast.

For class D the switching is too fast for just the charge pump, but it does allow you to park the bridge in either state (low or high) indefinitely.

When testing such circuitry start initially with only a very light load on the bridge, to reduce the possbility of blowing things up. The HIP4081A has shoot-through prevention logic which will help a lot. You need to set this correctly for your circuit using the resistors - too little deadtime, BANG!, too much and more distortion to the signal.

Wear eye protection with high voltage/power MOSFET circuits, faults can explode the tops off the packages.

Ensure you have spare chips/MOSFETs, its a learning curve and you will damage components at some point.
 

Monicaa

Member
2020-01-31 10:08 pm
However the HIP4081A has a separate charge pump so it can work without bootstrapping if there's not too much load on the high side gate pin (just a MOSFET's input capacitance) and you don't switch very fast.

For class D the switching is too fast for just the charge pump, but it does allow you to park the bridge in either state (low or high) indefinitely.

When testing such circuitry start initially with only a very light load on the bridge, to reduce the possbility of blowing things up. The HIP4081A has shoot-through prevention logic which will help a lot. You need to set this correctly for your circuit using the resistors - too little deadtime, BANG!, too much and more distortion to the signal.

Wear eye protection with high voltage/power MOSFET circuits, faults can explode the tops off the packages.

Ensure you have spare chips/MOSFETs, its a learning curve and you will damage components at some point.

Actually my pwm is 30 kHz (audio is bass freq) and I'm powering the driver with 12 volts and the MOSFETs with 24 volts. Do you think the bootstrapping circuitry is needed for such a low frequency?
 

Monicaa

Member
2020-01-31 10:08 pm
do.PNG

Above is the schmatic we are using. A few changes we made since we don't have some components available. the 1N5817 diode is being used and the dead-time resistors are both 100K (not variable)
 

rayma

Member
2011-04-29 8:37 pm
The ground for the mosfets and the driver are virtual ground taken from my dual supply of 5 volts
from the modulation stage. The load is a 4 ohm subwoofer speaker.

Can you post a complete, exact schematic of the entire system, including power supplies and their connections?
There is no virtual ground that can be used here.

Also suggest that you use only a resistive load that will draw a reasonable current that is within your design limits.
You will need a floating measurement for the bridge voltage output, regardless.