Audio Amplifier, Current Limiting Voltage Supply problem

Hello everyone,

In the analog electronics project, we are required to build an audio amplifier. The restrictions are:

- IC's will not be used.
- At least 6 watts will be delivered to 8 ohms speaker.

I have made 2 designs, one class ab and one class d amplifiers, in simulation everything works fine, i can get pure sinusoids and upto 100 watts on speaker, with +-25 double supply rail and, bridge configuration to double the speaker voltage if necessary.

However there is one little problem: The power supply that i use is Agilent E3631A, which gives max + - 25 volt limited by 1 A current for each of supplies. This limited current here means that (i suppose): The power cannot give more than 1 Amps in any instant of time, which will clip the current seen on power supply, and on the rest of the circuit at 1 Amps, if the project is not designed considering this fact.

Therefore, in order not to clip the audio signal, I need to pass at most 1 amps sinusoid across speaker, which will give (1/sqrt(2))^2*8=4 watts.

Now since i use +-25 supplies the 1 amp dc means 50 watt is driven from the supply. Where the rest of power goes is : 1. the driven current is not dc but ac with lower effective value, which means the driven power is much less. 2. in class ab the practical efficiencies are around 60-70% if and the rest is dissipated on the rest of circuit.

I think class ab cannot do much about this current limiting. I am not sure about class d.

I would be very thankful if you could have some comments. Am i correct on these, and may there be anyway to get through these? In addition, i cannot model the power supply such that it limits currents over 1A, in spice. How can i do that?

Thank you for reading

I think you pretty much understand your problem. Basically, you can't get anymore than 50 watts out of your amp because of the power supply. That's just physics. Your class D amp will be much more efficient than AB, but you still won't get 50 watts out of it, maybe 45W.

As far as spice modelling a current-limited voltage source, that's an interesting question. I know you can specify an internal resistance for a source, but that's not quite the same thing.

Barry

Thank you for your comments,

In spice, i connected a 25 ohms to the 25 volts supply, however it ruined the circuit operation.

For the problem, i don't think that i can get more than 4 watts at speaker, since i cannot get more than 1A peak current flowing through it

You are right; I was thinking about available POWER not current. If you want, say, 32 watts, peak from your speaker, you'll need a power supply that can put out 2Amps at 16V . I=sqrt(32W/8ohms)

I had the same current-limit problem in a simulation and came up with an approximate solution.

To do a 1A current limit use a 1A current source with a parallel diode across it (anode to current output) in series with the voltage source. At currents below 1A the output will then appear as a voltage (≈0.7V higher than the voltage source or about 0.4V if you use a Schottky diode) since any excess current from the current source is being shorted by the diode. Above 1A the diode becomes reverse biased and no current above that value can flow.

crutschow said:

I had the same current-limit problem in a simulation and came up with an approximate solution.

To do a 1A current limit use a 1A current source with a parallel diode across it (anode to current output) in series with the voltage source. At currents below 1A the output will then appear as a voltage (≈0.7V higher than the voltage source or about 0.4V if you use a Schottky diode) since any excess current from the current source is being shorted by the diode. Above 1A the diode becomes reverse biased and no current above that value can flow.

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Nice, I like this! I'll keep this in mind for future reference.

efevi, getting back to your problem, here are some fundamentals: If you want 50 W from an 8 ohm speaker your power supply must have the following (minimum) requirements (assuming 100% efficiency)

1) Imin=sqrt(50W/8)=2.5A
2) Vmin=sqrt(50*8)=20V (or +/- 10)

You can see why they often use transformers in amplifiers.

You can also build a more exact I/V limited source using behavioral simulation (if present in your Spice version) or use active circuits to get a better approximation. But the suggested diode limiter should work for most practical requirements, I suppose.

Regarding the original question, the output power limitation for a linear amplifier working from a 1A current limited supply has been correctly calculated. You can increase the undistorted sine output by adding large capacitors.

- Class-D could achieve higher output power because it involves a voltage conversion.
- audio power transformer is a principle option, but not really suitable, I think

FvM said:

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- Class-D could achieve higher output power because it involves a voltage conversion.
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For a given supply voltage the theoretical peak output voltage is the same for Class-D as for a linear amp so the maximum output power would be the same. Class-D just does it more efficiently.

Don't understand what you mean by Class-D involves a voltage conversion. :?:

As for current limiting i found this https://www.electronics-lab.com/blog/?p=18400 circuit. I played with it and result is a supply with 15 volts limited to 1 amps:

**broken link removed**
**broken link removed**

Of course not practicali, just for modeling.. Negative supply version have pnps where there are npns and diodes reversed. The two diodes have a total voltage of 14.4 break voltage. I could not model it with just diodes or resistors. This circuit gave the sharpest regulation and limiting. I hoped in spice this could be much easier by just setting a limit in the properties window of voltage supply.

As for how i can have more watts than possible in this situation, neighter class can build more than 4 watts on speaker in this situation, unless i think, i put a transformer just in front of the speaker. The ratios has to be selected in order to increase current whereas also considering the voltage swing will fall. There will be an optimised point. For example when using +-25 volts 1A supply, lets assume we can have a voltage swing of +-20 volts at output and 20/8=2.5Amp_pp. However in practice since we have 1A limiting we can't have this voltage swing. When we put a 2:1 transformer before speaker, i think, the maximum current will double to 2A, where as voltage swing will fall to 10 volts_pp and limiting the current to 10/8=1.25 Amp_pp. For optimised case20/x)=(1*x)*8 ==> x=1.58 turn ratio ==> than thw voltage wing falls to 20/1.58=12.65Vpp and current becomes 1*1.58=1.58Amp_pp. ==> Voltage swing/ current swing =12.65/1.58=8ohms load. Therefore we have optimised swing and we have (I^2*R)/2=~10 watts power at max at output. I don't know if i am correct and this is applicable..


In class D design, in simulation, i was facing very high shoot-through currents reflected to supply. When i modeled the supply to its max current value, my circuit operation is ruined. Therefore i added large caps accross supplies, hoping that in nanosecond shoot through instances, these previously charged caps can discharge instantly providing necessary current. With a rough calculation if shoot-through time is 30ns and if we have 470u caps, than Q=C*V=470U*15=7e-3 ==> I=Q/t=~230kAmps. Meaning they can discharge up to providing 230kAmps to circuits. And this trick seems to be working with the circuit. However, i don't think that these caps (no matter how big) can sustain more current than supply can give continously, but it can only work with these shoot-through impulse currents.

For a given supply voltage the theoretical peak output voltage is the same for Class-D as for a linear amp so the maximum output power would be the same. Class-D just does it more efficiently.

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No doubt about it. The problem of this thread is current limit rather voltage limit.

Don't understand what you mean by Class-D involves a voltage conversion.

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The class-D involves a voltage down-conversion (buck operation). It can convert 25 W supply power (25V * 1A) to 10V*2.5A matching the 4 ohm load.

Re-editing my last reply:

I did not think that these caps can be used in class ab stage to provide much longer discharge currents to the amp. However considering that supply does not give current for one half cycle(charging the capactior fastly), and since that at the other half cycle the capactior will begin its duty only after source cannot give 1A, this idea is very reasonable. I made the roughest estimates and reasonable sizes of capacitors can do the job. And in the simulation it does not cause problems for audio frequencies(20-20k).
By using these caps, i can give more than 2 times maximum current of supply to the load.

FvM,
For now i don't now much about the voltage down conversion in class d, you are mentioning, but it sounds like a transformer operation to me. I will look up to that when i have time..

FvM said:

........................The class-D involves a voltage down-conversion (buck operation). It can convert 25 W supply power (25V * 1A) to 10V*2.5A matching the 4 ohm load.

Click to expand...

Hadn't though about that aspect of a PWM driving a speaker (inductance) but that should work. You just need to keep the modulation (volume) below the point that it exceeds the 1A limit of the power supply to avoid the severe distortion that would produce.