Electronics > Projects, Designs, and Technical Stuff
Texas Instruments 3116D (ClassD) amp chip
TimFox:
To your quick question: to change the ratio of voltage to current (i.e., the load resistance) without changing the DC supply and active devices, a transformer is needed to keep the same power. A "step-up" transformer will give more volts and less amps into a higher resistance, with the same power (ignoring transformer losses). If "N" is the turns ratio, then V2 / V1 = N = I1 / I2 (note the opposite order for V and I), R2 / R1 = N2 , and P2 = P1.
Solid-state designs avoid transformers, and are optimized for one load impedance, although they will function over a reasonable range of load. I remember a solid-state linear amplifier from 1969 that did have a manual switch on the power supply to change the optimal load from 4 to 8 ohms.
Power amp specs are often un-trustworthy. RMS power is a misnomer for mean power (the M in rms).
My question about your equipment was not an insult: I noticed that you had not replied to my suggestions for simple measurements. If you don't have access to or experience with such equipment, there's no further point in my suggesting measurement techniques.
DW1961:
--- Quote from: TimFox on August 17, 2020, 05:56:27 pm ---To your quick question: to change the ratio of voltage to current (i.e., the load resistance) without changing the DC supply and active devices, a transformer is needed to keep the same power. A "step-up" transformer will give more volts and less amps into a higher resistance, with the same power (ignoring transformer losses). If "N" is the turns ratio, then V2 / V1 = N = I1 / I2 (note the opposite order for V and I), R2 / R1 = N2 , and P2 = P1.
Solid-state designs avoid transformers, and are optimized for one load impedance, although they will function over a reasonable range of load. I remember a solid-state linear amplifier from 1969 that did have a manual switch on the power supply to change the optimal load from 4 to 8 ohms.
Power amp specs are often un-trustworthy. RMS power is a misnomer for mean power (the M in rms).
My question about your equipment was not an insult: I noticed that you had not replied to my suggestions for simple measurements. If you don't have access to or experience with such equipment, there's no further point in my suggesting measurement techniques.
--- End quote ---
I didn't take it as an insult. The best way to do this is like you suggested, get out the equipment an just test it, done! (And, BTW, test it for THD at different levels too, and the type of distortion, and all the rest!) I don't have that equipment nor the knowledge to run it.
So when doing calculations with a digital power supply, like the power brick included with a class D amp, you get the amps and voltage that the equation spits out and that's it?
If the power brick is 24v, and the amp needs 28V to get 50 watts per channel, how do you get 100 watts total RMS out of that amp using x2 3116D2 chips?
So how can we tell what the power of this amp is, generally, at 8 ohms, without actually measuring it--don't we have enough information to guesstimate it pretty closely?
Also, so it can put out 36 watts RMS x2 with two 3116D chips, it can do the same with one chip--so again, what's the benefit of two chips vs one chip at 8 ohms?
TimFox:
If you know the voltage and maximum current of the power supply, and you understand the mode of operation of the active amplifier and its losses, then they determine the optimal load resistance and the power into it. Usually, the power supply is treated as a constant voltage output with a maximum current capability, but the voltage will fall with higher current and excessive current could smoke the supply in a bad design.
The details of that calculation are important, and you should consult the relevant textbooks. I have used Class-D, but am not an expert on your chipset.
At the optimal load, the peak current (of the sinusoidal waveform) into the load (maximum possible from the circuit) will require the maximum voltage that the amplifier can put out. With a higher load resistance, the maximum peak voltage into the load requires less current and the power is lower. With a lower load resistance, the maximum peak current requires a lower peak voltage and the power is lower. Note the usage here of maximum (what the circuit can produce) and peak (what the load will do with the voltage or current supplied to it). In any of these three cases, if you try to exceed either the maximum voltage or current (whichever is hit first as you increase the signal), the output waveform will show clipping. Here, "peak" means the highest voltage (or current) in the waveform, half the "peak-to-peak" value.
What complicates all of this is that it is usually uneconomical (for home audio use) for the DC power supply to maintain its voltage for a long time at the maximum current. Therefore, if you do a continuous measurement, the DC voltage probably "droops" after a short time and limits the power below that which could be delivered for a brief moment (as in normal music).
rsjsouza:
--- Quote from: DW1961 on August 17, 2020, 06:17:46 am ---
--- Quote from: rsjsouza on August 17, 2020, 03:15:30 am ---"If I understood correctly your amplifier has two devices in PBTL (mono) mode and each is driving its own 2\$\Omega\$ loudspeaker. If so, figure 22 tells me that, at 24V, it can go as high as 150W each device, rendering your 24V/5A rather inadequate to supply this "beast". To get absolute maximum continuous power, you would need to have a minimum of 300W + efficiency losses. "
--- End quote ---
I dunno. That's why I'm trying to figure out if how the magic in these chips works.
The single 3116D2 chip runs in BTL mode, which tells me it is really 4 amps on one chip. In BTL mode, you need two chips running mono. However, the 3116D chip runs in stereo. So if it is running in BTL mode all of the time, and it is stereo, then it must have 4 amps on the chip. That's how they can run x2 3116D chips in mono and get stereo in the amp I have, I assume. Otherwise, if it were just two amp chips, they would be only getting a mono amp out of it. So the way I understand it, to get stereo from a BTL system, you need 4 amps.
So the amp I have must really have 8 amps on it (x2 3116D2 chips @ 4 amplifiers each), but they are running the 3116D in mono mode, so they need another one to do stereo.
How they are connecting them together, I have no idea! PBTL? I just dunno.
--- End quote ---
Section 7.4.1 shows the PBTL (mono) connection.
It is important to keep the nomenclature identical to the datasheet - in this case the authoritative source of information. You are correct the device internally has four amplifiers connected two-by-two in BTL that could be bridged yet again to form a single mono amplifier, to which the datasheet calls it "PBTL mode".
--- Quote from: DW1961 on August 17, 2020, 06:17:46 am ---
--- Quote from: rsjsouza on August 17, 2020, 03:15:30 am ---"So, why are you talking about 2\$\Omega\$? Sorry, I really don't get exactly the circling around. "
--- End quote ---
To get 100 watts RMS from one 3116D, it has to run at 2 Ohms in mono.
--- End quote ---
Nope. As Kai's reply on e2e also clarifies, to get 100W RMS from one 3116D you can run it in BTL (stereo) at 4 ohms, which yields 50W per channel or 100W total, or in PBTL (mono) at 2 ohms, which yields 100W on a single channel. According to figures 14 and 22, you can get all this power using "only" 19V.
Keep going up in voltage and you will get up to 170W from a single 3116D, either in BTL or PBTL modes.
--- Quote from: DW1961 on August 17, 2020, 06:17:46 am ---
--- Quote from: rsjsouza on August 17, 2020, 03:15:30 am ---"24V * 5A = 120W. Where did the 72W come from? "
--- End quote ---
Using this calculator: https://www.rapidtables.com/calc/electric/watt-volt-amp-calculator.html
Add resistance at 8 ohms: 8 ohms/3 amps/24v/72 watts. You could increase current, but then you would need to increase voltage too. I guess Ohms law?
--- End quote ---
Well, a resistor would consume 72W, which is equal to 242/8, but that is where you need to consider the amplifier efficiency. The datasheet tells you that 42W per channel is the maximum under these load and supply conditions. As TimFox mentioned, the amplifier will draw whatever it needs from the power supply - in other words, the power supply will passively provide whatever is within its limits, compensating with a reduction in voltage in case the amplifier consumes more than it will provide (thus clipping the output voltage).
cv007:
>What is the benefit of running two of these when the power stays the same, and one amp chip is capable of putting out the same wattage using the same driver for both chips?
You also have a set of mosfets that have to dissipate power (8x120 milliohm 'resistors', 4 active at a time). If a single ic is 'heat happy' putting out 50Wx2 (BTL, 100W total), then its probably also happy putting out 100Wx1 (PBTL, 100W total). Since there is a voltage limit, the only way to get to the 100W is by decreasing the impedance. The PBTL mode is simply a way to share the heat. Since there is no perfect mosfet, I imagine heat is the monster that lurks everywhere in the datasheet.
If you are using 8ohm speakers, you cannot take advantage of the possible extra power (because of voltage limit), but you still get the benefit of sharing the heat among 2 ic's. That may mean you do not need a heatsink, or could mean the wimpy heatsinks used in an enclosed little case will not cause thermal shutdown.
You need to look at any seller of audio gear the same way you look at laundry detergent that says '2x the cleaning power'. Its a waste of time trying to sort out the claims being made. Just look beyond the sales pitch, and look at the ingredients to figure out what you have.
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