# Trying to understand this Class D amp (using IR2113) circuit

I understand the concept of the Class D amp and this circuit, but in this particular case there is one part I just don't get.

What I don't understand is how the low side of the signal should work at the power stage. In case of the high side as I understand when the MOSFET is on, the current can flow from VCC to the load. But at the low side I don't understand how could current flow through the load and the low side MOSFET.

Could you please clarify it? Thanks!

• You should add the schematic here, not just a link - the goal is to make sure that questions remain available for longer the the link you point to. – le_top May 24 '20 at 13:46
• Okay, sure, I included it. – Isty001 May 24 '20 at 13:52
• The other answers pointed out the major flaw, but I'll add that you could mitigate that by making J1 not have its 2nd pin to ground, but to a capacitive divider, or mid-point of two sources. The caps should be large to accomodate the current requirement, and they can serve as a replacement of two sources. Of course, the connector should be explicitely marked as "live"(!). – a concerned citizen May 24 '20 at 16:22

If you dig up the original Great Scott video you'll see that he has put a 1000uF capacitor in series with the speaker. It can be seen at 5:37 in this video: https://www.youtube.com/watch?v=3dQjIeYoIdM

It stops the DC component of the output signal. Without it the speaker would get continuous DC about half of the DC supply voltage. It could still work somehow but the dissipation in the speaker would be unacceptable at least for me.

I have said it previously in other cases: Do not believe YouTube entertainers. Believe even less those who build their show on some YouTube video. You must get to know the things by understanding them. Your question here shows you have already thought it.

• Oh, I've missed that in the video, thanks for pointing it out! – Isty001 May 24 '20 at 14:27
• the capacitor is not presented elsewhere. – user287001 May 24 '20 at 14:33
• Thanks, but maybe I'm terribly missing something about how this IC or the circuit works, but could you tell me about how or where the DC is getting into the output signal? – Isty001 May 24 '20 at 16:10
• The mosfets conduct in turns and connect the inductor of the the output smoothing filter in turns to the ground and to the Vcc. The average (=the DC component) is about Vcc/2. BTW. There's no quarantee that nothing is missed in the wiring of the mosfet driver U2. Checking it is another of your problems. The next problem is the distortion caused by the non-linear triangle wave made by 555. I would skip this circuit and try something from a creditable source, for ex. from a manufacturer of D-class amp IC. – user287001 May 24 '20 at 17:07
• Thanks for the clarification! – Isty001 May 24 '20 at 18:08

It does so badly. This amplifier puts DC through the speaker. Think of the output biased at 1/2 Vcc (50% duty cycle), so that the output range is almost Vcc (close to 100% duty cycle) to 0V (close to 0% duty cycle).

Most Class D amplifiers use an H-bridge driven differentially so that the average voltage across the load is zero.

You can use a half-bridge if you have bipolar supplies with one side of the speaker grounded. Or add a BFC (big "fat" capacitor) in series with the speaker.