I've had some trouble with my amp because it is non inverting. Is it possible to derive NFB using the supply's rails? If I put a 1M resistor between + and the signal and between the signal and - , would that work as NFB? When there's a positive signal on the input, the signal goes positive, pulling the positive rail down and therefore the voltage divider on the input would pull the input down. Would this work?
Feedback signals.
Hi Solid Snake ( how did you pick this name?),
Feedback refers to a signal and not static DC voltages like power supplies. So when you say applying feedback , we mean taking a "signal" at the output of the device and 'feeding' it to some part of the input stage ( or earlier stage ) .
So if you just connect a resistor from your input ( or output) to the power supply , you are not applying negative feedback.
Have I misunderstood what you were saying ?
Cheers
Hi Solid Snake ( how did you pick this name?),
Feedback refers to a signal and not static DC voltages like power supplies. So when you say applying feedback , we mean taking a "signal" at the output of the device and 'feeding' it to some part of the input stage ( or earlier stage ) .
So if you just connect a resistor from your input ( or output) to the power supply , you are not applying negative feedback.
Have I misunderstood what you were saying ?
Cheers
Feedback and signal to noise ratio.
I thought that a numeric example might help with the concept of increasing signal to noise.
Say you have an amp with a gain of 100 and an input noise of 0.001V. Now apply a signal of 0.1V at the input to get 10 volts at the output. Now noise and input signal are amplified the same amount , so you get a signal to noise ratio of 0.1x100/0.001x100 which is 100 ( 40db).
Now we apply negative feedback which drops the gain to 20.
Now noise at the output will be 0.001x20V
We will need to raise the input signal to 0.5 volts to get a 10 volt output. So signal to noise ratio is now 0.5x20/0.001x20
which is 500 ( about 54db).
So you can see that the input noise is the same . We are just increasing the input voltage to get a better S/N. S/N goes up and input sensitivity goes down.
In engineering if you get something somewhere you have to give something elsewhere. There is no free lunch !
Cheers.
I thought that a numeric example might help with the concept of increasing signal to noise.
Say you have an amp with a gain of 100 and an input noise of 0.001V. Now apply a signal of 0.1V at the input to get 10 volts at the output. Now noise and input signal are amplified the same amount , so you get a signal to noise ratio of 0.1x100/0.001x100 which is 100 ( 40db).
Now we apply negative feedback which drops the gain to 20.
Now noise at the output will be 0.001x20V
We will need to raise the input signal to 0.5 volts to get a 10 volt output. So signal to noise ratio is now 0.5x20/0.001x20
which is 500 ( about 54db).
So you can see that the input noise is the same . We are just increasing the input voltage to get a better S/N. S/N goes up and input sensitivity goes down.
In engineering if you get something somewhere you have to give something elsewhere. There is no free lunch !
Cheers.
What I meant was that if you take a 1M resistor from signal to + and signal to -, the power supply will reflect the output signal because the supply only runs one audio channel. Basically when there is no signal, both rails are at 18v. When the signal swings positive, the positive rail may go down to 17v while the negative rail stays at 18v. This will make the point between the two resistors go to -0.5 volts, negating the input signal.
Supplies don't vary
Hi Solid Snake,
I think you got it all wrong.
The supplies do not vary at all !
Only your signal point moves up and down relative to its rest point - no voltage condition.
So +18 is always at +18 even if the input is +5Volts. I am assuming the power supply is good.
cheers.
Hi Solid Snake,
I think you got it all wrong.
The supplies do not vary at all !
Only your signal point moves up and down relative to its rest point - no voltage condition.
So +18 is always at +18 even if the input is +5Volts. I am assuming the power supply is good.
cheers.
Supply drop
Hi Solid Snake,
It is difficult to say what is happening without looking at your circuit diagram. If your supply is dropping , it is because of the current drawn by the load ( 8 ohm , 4 ohm or whatever it is).
If you remove the load and still apply an input signal , you will find no change in the power supply voltage. I will be VERY surprised if it does. In any case you better put up the circuit diagram for any serious solutions.
Cheers.
Hi Solid Snake,
It is difficult to say what is happening without looking at your circuit diagram. If your supply is dropping , it is because of the current drawn by the load ( 8 ohm , 4 ohm or whatever it is).
If you remove the load and still apply an input signal , you will find no change in the power supply voltage. I will be VERY surprised if it does. In any case you better put up the circuit diagram for any serious solutions.
Cheers.
Traderbam suggested I put a resistor between the emitter and ground. I see how this would work for the first stage because it has it's emitters on ground. The input signal is common to ground also, so I could see that this would turn the input stage into a sort of emitter follower. I don't see how putting a resistor on the output stage (the + side has emitter to + and the - side has emitter to -) would introduce NFB. The signal fed to the second stage always comes from ground but varies in current. In other words, the signal fed to the output stage is 0 volts, which sets off the output transistors, the only thing that varies about this signal is the current. I don't see how a resistor on the output devices could act as NFB. Would I need some kind of resistor network to convert current into voltage?
Snake,
If I understand your circuit you have two challenges to deal with straight away. One is NFB and the other is to maintain a sensible bias current through your output transistors.
As I recall, your circuit is a mirror of an input npn in common emitter (emitter to 0V) that then feeds an output pnp in common emitter (emmitter to +V) and then the pnp collector feeds the speaker. This is duplicated with a pnp input and npn output device. Is this correct?
So the two output devices need to have a contant bias current flowing through them when the input signal is zero. Have you solved this?
Your idea to take the NFB from the power rails is interesting and creative. The trouble is that the voltage signal on the rails tends to be a poor representation of the output voltage signal. For example, the rails have a lot of mains frequency noise on them from the psu rectifier recharghing the capacitors. Also, the impedance of the psu may not be very uniform or linear. To work well the feedback signal needs to follow the output signal as closely as possible without any extra noise or distortion.
One way to get NFB is, as I suggested earlier, is to run a resistor between speaker output (the node where the output transistor collectors meet) and the input emitter node. The emitter node then needs to have a resistor to 0V or else the feedback signal gets grounded and is lost. For example, you might have a 100-ohm resistor between input emitter node and 0V and a 2.2k resistor between output node and emitter node. This would make the amps voltage gain tend towards 23 = (2200+100)/100
I say tend towards 23 because the accuracy will depend upon the voltage to current gain of the input transistors and the current gain of the output transistors. The higher these are the closer to a gain of 23 you'll get.
I hope I am helping here. I'm not quite sure what stage you are at. Aplogies if I'm stating the obvious or answering the wrong questions.
BAM
If I understand your circuit you have two challenges to deal with straight away. One is NFB and the other is to maintain a sensible bias current through your output transistors.
As I recall, your circuit is a mirror of an input npn in common emitter (emitter to 0V) that then feeds an output pnp in common emitter (emmitter to +V) and then the pnp collector feeds the speaker. This is duplicated with a pnp input and npn output device. Is this correct?
So the two output devices need to have a contant bias current flowing through them when the input signal is zero. Have you solved this?
Your idea to take the NFB from the power rails is interesting and creative. The trouble is that the voltage signal on the rails tends to be a poor representation of the output voltage signal. For example, the rails have a lot of mains frequency noise on them from the psu rectifier recharghing the capacitors. Also, the impedance of the psu may not be very uniform or linear. To work well the feedback signal needs to follow the output signal as closely as possible without any extra noise or distortion.
One way to get NFB is, as I suggested earlier, is to run a resistor between speaker output (the node where the output transistor collectors meet) and the input emitter node. The emitter node then needs to have a resistor to 0V or else the feedback signal gets grounded and is lost. For example, you might have a 100-ohm resistor between input emitter node and 0V and a 2.2k resistor between output node and emitter node. This would make the amps voltage gain tend towards 23 = (2200+100)/100
I say tend towards 23 because the accuracy will depend upon the voltage to current gain of the input transistors and the current gain of the output transistors. The higher these are the closer to a gain of 23 you'll get.
I hope I am helping here. I'm not quite sure what stage you are at. Aplogies if I'm stating the obvious or answering the wrong questions.
BAM
Thanks Bam, your method is very clever. I have the output transistors biased through the input transistors. The input transistors are biased with 500K ohm resistors and a pot to balance the bias between + and -. I can usually get the idle voltage across the speaker within a few mV of zero. Before the input stage, I have a class A input stage. Does the class A stage require any NFB?
Also, my rails are -18, 0, 18. I can put enough power through an 8 ohm speaker to make me want to cover my ears, but the sound begins to distort when the volume is turned up. I tried using a 2 ohm speaker instead to get more power from the amp, but the volume stayed pretty much the same and distorted at the same power level as the 8 ohm speaker. I have a TIP41/TIP42 on my output stage that can dissipate 65 watts and pass 6 amps. I'm confused at what is limiting my amplifier.
Also, my rails are -18, 0, 18. I can put enough power through an 8 ohm speaker to make me want to cover my ears, but the sound begins to distort when the volume is turned up. I tried using a 2 ohm speaker instead to get more power from the amp, but the volume stayed pretty much the same and distorted at the same power level as the 8 ohm speaker. I have a TIP41/TIP42 on my output stage that can dissipate 65 watts and pass 6 amps. I'm confused at what is limiting my amplifier.
Feedback of current...
If you put a resistor betwen the - of the loudspeaker and ground...say 1 Ohm and take the feedback from the union of this resistor and the minus of the loudspeaker...
Yes you have a current amplifier...aka a current source!
Yes ...because there be no damping at all...
regards
If you put a resistor betwen the - of the loudspeaker and ground...say 1 Ohm and take the feedback from the union of this resistor and the minus of the loudspeaker...
Yes you have a current amplifier...aka a current source!
Yes ...because there be no damping at all...
regards
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