I'm a teen rapper now by asking the purpose of a stage I feel can be bypassed with 2 transformers. If you read from the beginning you would see a humble approach as opposed to yours. If you are going to try and sling insults, at least let them be humorous or even relavant to the point at hand. Yours was neither and only a lame attempt to get an "lol" from your cyber pals and a waste of my time replying. Since your circuits are as solid as ur sense of humor the feel free to post a sketch of a topography no one has seen.
BTW, we still don't know your overall requirements.
What load are you going to drive? 8 ohm?
What output power do you expect?
LM6171 with nominal +/-15V rails gives +/-12V output swing (according to the datasheet), giving us 8.5V RMS at the output. Without additional voltage amplification, you will most likely loose some of this swing on the output follower, ending up with what? Some 8W @ 8 ohm?
What does that mean? What is going to kick it up? 😕
What load are you going to drive? 8 ohm?
What output power do you expect?
LM6171 with nominal +/-15V rails gives +/-12V output swing (according to the datasheet), giving us 8.5V RMS at the output. Without additional voltage amplification, you will most likely loose some of this swing on the output follower, ending up with what? Some 8W @ 8 ohm?
What does that mean? What is going to kick it up? 😕
But how? How are you going to bypass the VAS stage with 2 transformers? Can you please draw the schematic? Nobody understands what you really mean so far. Or is it just me? 😱
Jared please draw a sketch about your idea.
It's really really hard to get it from your writings.
It's really really hard to get it from your writings.
I think he wants to replace the VAS stage with a stepup transformer.
i.e. OpAmp - Stepup trafo - OPS?????????
i.e. OpAmp - Stepup trafo - OPS?????????
You can think of an audio amplifier as a sort of "power supply modulator". The power supply converts the AC mains into two DC rails that you then feed to the amplifier, which "modulates" them with the input signal. For this to be done in an efficient way that gives you the required output power (voltage times current) to drive a loudspeaker, somewhere in the amplifier there has to be a stage that amplifies the input signal, usually about +/-1.5V or so peak-to-peak in voltage amplitude, to something as close as possible to +/- whatever DC rails you have. That's the VAS. Then the output stage provides the required current gain.
This is the "flow" from mains socket to loudspeakers:
- The power supply uses a transformer to convert the mains voltage, 0-120V (AC, 60Hz), to, say, 55-0-55V (AC, 60Hz).
- Then the rectifier and filter convert this to your +/-78V (DC). Not really, it's less than that because of voltage drop in the rectifier diodes, secondary resistance, finite filter capacitance, etc. Note that it isn't that we're "pumping" 55 to 78V: in fact, for AC we usually talk about RMS values, and the peak value of a sinusoid is sqrt(2) times the RMS value (I think that's where you got your 1.4142 -not 1.44- factor), so in fact we're converting a signal with a peak-to-peak voltage of +/-78V to two DC rails of (less than) +/-78V.
- Then the amplifier takes these DC rails and "moves" them up and down following the input signal to give you the output signal.
So, the fact that from a seemingly lower voltage of 55 we get a seemingly higher one of 78 in the power supply is just a consequence of the way a rectifier and filter work: it has absolutely nothing to do with amplification. This means that the power supply transformer cannot be used for any sort of signal amplification, which, if I've understood you correctly (I'm not really sure I have), is what you were trying to do.
On the other hand, it is true that a (different) transformer could be used to provide the required voltage amplification in the amplifier, but, as vzaichenko mentioned above, this is not without problems, and there's a reason why we normally use a transistor to do the job.
As I said I'm not sure I've understood your idea, but it looks like you're confusing what goes on in the transformer with signal amplification and I thought that this basic explanation was in order. Apologies if you already knew all this and the idea is different, in which case I join the choir of members requesting a diagram of what you're trying to do.
Cheers,
Cabirio
This is the "flow" from mains socket to loudspeakers:
- The power supply uses a transformer to convert the mains voltage, 0-120V (AC, 60Hz), to, say, 55-0-55V (AC, 60Hz).
- Then the rectifier and filter convert this to your +/-78V (DC). Not really, it's less than that because of voltage drop in the rectifier diodes, secondary resistance, finite filter capacitance, etc. Note that it isn't that we're "pumping" 55 to 78V: in fact, for AC we usually talk about RMS values, and the peak value of a sinusoid is sqrt(2) times the RMS value (I think that's where you got your 1.4142 -not 1.44- factor), so in fact we're converting a signal with a peak-to-peak voltage of +/-78V to two DC rails of (less than) +/-78V.
- Then the amplifier takes these DC rails and "moves" them up and down following the input signal to give you the output signal.
So, the fact that from a seemingly lower voltage of 55 we get a seemingly higher one of 78 in the power supply is just a consequence of the way a rectifier and filter work: it has absolutely nothing to do with amplification. This means that the power supply transformer cannot be used for any sort of signal amplification, which, if I've understood you correctly (I'm not really sure I have), is what you were trying to do.
On the other hand, it is true that a (different) transformer could be used to provide the required voltage amplification in the amplifier, but, as vzaichenko mentioned above, this is not without problems, and there's a reason why we normally use a transistor to do the job.
As I said I'm not sure I've understood your idea, but it looks like you're confusing what goes on in the transformer with signal amplification and I thought that this basic explanation was in order. Apologies if you already knew all this and the idea is different, in which case I join the choir of members requesting a diagram of what you're trying to do.
Cheers,
Cabirio
Lets start small and easy with 75watt 8 ohm load
Additional voltage will come from another winding off the toroid, so basically you are using 2 voltages in series to push the driver of the LFETS. If that isn't possible, then one of the INAs can provide large rail to rail swings after the LM6171, i need to check which one.
sorry when i use the term kick up, i am referring to increasing the voltage. When i said the second winding of 78 for 4/5 outputs, you said it would be increased to 110, i figured without vas the voltage increase wouldn't be there and said a 55v winding would get it to 78 then.
Open loop resistance is 14 ohms - that is the only impedance stat on the white sheets, what was the impedance issues Val was referring to?
Am i a little more clear now?
yes. that is it. An INA can provide more rail to rail swing for huge ones if using an opamp.
But i am also saying another step up transformer may not be needed - additional windings on the main transformer can be rectified as well to provide a different voltage. i.e. 1kva w 55v secondaries and center tap ground with additional 12v or 15v winding. Am i a little more clear?
Not really... Please draw a schematic even a draft like.
Do you want to use a 2 level power supply at the output stage
with that 55 and 78V? But why? I still dont get the idea...
Do you want to use a 2 level power supply at the output stage
with that 55 and 78V? But why? I still dont get the idea...
A simple Vas stage can be made using a single NPN transistor with a resistor load (RL) from the collector to a positive voltage supply rail. In a more advanced system RL would be a constant current source which increases the dynamic impedance.
The output to the following stage will be taken from the collector of the NPN transistor.
In a single supply rail system the emitter connected will be connected to earth - otherwise in a dual rail supply system the connection would be to the negative supply rail.
The NPN transistor is set up so it will conduct a suitable fixed value of current under static conditions - Class A. The output is an inverted version of the signal input.
When an a.c. signal applied to the base is positive it will cause the transistor to conduct more current causing a larger voltage drop across RL. The effect is the opposite with negative signals. The idea is to generate as large an a.c. signal across RL as needed by this means.
A simple formula for the amplification is Vout/Vin =gm*RL
where RL = k.Ohms and gm = millisiemens. gm will be approximately 40 per milliamp of bias current.
The value of RL will be reduced by the input impedance reflected by following power stages. The value of this might be a little less than 3 k.ohms.
In a fairly basic design the open loop gain might be 3000-4000 and most of that would be used for nfb purposes.
In combination with a pre-amplifier line stage a closed loop gain of around 20 would be sufficient to generate full power.
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