Hi guys,
I have got confused with class D BTL configuration. In fact, the output voltage after LC should be VCC*2. What in this case happens to the current? Should it get 2 times as small or stay the same as for a single-ended connection?
What happens to a PBTL configuration in this case? What‘s the current value? I also can’t understand how to calculate a maximum peak duration pulse for the power source for the first and the second case and what will the equation be.
Thank you, Olga
I have got confused with class D BTL configuration. In fact, the output voltage after LC should be VCC*2. What in this case happens to the current? Should it get 2 times as small or stay the same as for a single-ended connection?
What happens to a PBTL configuration in this case? What‘s the current value? I also can’t understand how to calculate a maximum peak duration pulse for the power source for the first and the second case and what will the equation be.
Thank you, Olga
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If you mean a single-supply BTL amplifier, with a single Vcc, how can you get 2*Vcc at the output? 🙂
At least to my logic, the maximum peak current will be Vcc / R (R being the load resistance). Well, Vcc minus whatever voltage drop there is across the output devices, but at least for the sake of argument, that can be ignored.
In SE configurations, the max current will actually be (Vcc / 2) / R. That is, once again, assuming it's a single-supply amplifier, where the outputs idle at half the supply voltage. That's why you'll see big DC-blocking capacitors on those outputs.
PBTL is just the same, except that the amplifier(s) can supply about twice the current, before hitting the current-limit trip point (if we're talking amp chips).
At least to my logic, the maximum peak current will be Vcc / R (R being the load resistance). Well, Vcc minus whatever voltage drop there is across the output devices, but at least for the sake of argument, that can be ignored.
In SE configurations, the max current will actually be (Vcc / 2) / R. That is, once again, assuming it's a single-supply amplifier, where the outputs idle at half the supply voltage. That's why you'll see big DC-blocking capacitors on those outputs.
PBTL is just the same, except that the amplifier(s) can supply about twice the current, before hitting the current-limit trip point (if we're talking amp chips).
Thank you, Khron 🙂 Actually, I mean a single supply full-bridge differential BTL and PBTL outputs.
Could you please suggest me how to calculate a maximum peak duration pulse for the power source?
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Umm... Real sorry, but i'm afraid the laws of physics won't allow that 😕
You can't have a voltage across the load, that's GREATER than the supply voltage.
IF we're talking about a full-bridge amplifier supplied with +/-12V, THEN what you're saying is true 🙂
I mean, just draw up a sketch, you should be able to easily see that "best case", in a single-supply bridged amplifier, you'll have the load connected between Vcc (12v, for example) and ground - that's 12Vpp right there.
What you might often see, is amplifiers having X number of watts in SE mode, for 4 ohm loads, and 2*X watts in BTL, for *8* ohm loads (so the current stays the same). Remember Ohm's law?
And regarding the peak current for the power supply... Doesn't that depend on the content of the signal present at the amplifier input? And then there's also the capacitance present on the amplifier board and the power supply to keep in mind...
PS: See if anything over here gets you closer to the answer(s) you're seeking: https://www.eetimes.com/document.asp?doc_id=1274915
You can't have a voltage across the load, that's GREATER than the supply voltage.
IF we're talking about a full-bridge amplifier supplied with +/-12V, THEN what you're saying is true 🙂
I mean, just draw up a sketch, you should be able to easily see that "best case", in a single-supply bridged amplifier, you'll have the load connected between Vcc (12v, for example) and ground - that's 12Vpp right there.
What you might often see, is amplifiers having X number of watts in SE mode, for 4 ohm loads, and 2*X watts in BTL, for *8* ohm loads (so the current stays the same). Remember Ohm's law?
And regarding the peak current for the power supply... Doesn't that depend on the content of the signal present at the amplifier input? And then there's also the capacitance present on the amplifier board and the power supply to keep in mind...
PS: See if anything over here gets you closer to the answer(s) you're seeking: https://www.eetimes.com/document.asp?doc_id=1274915
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