electronrancher
Advanced Member level 1
get the 4th edition analog book by gray from bone hub, but i will give a small explanation here. and yes, it's definately possible to do DC (large-signal) analysis on a distributed amp, and is ALWAYS necessary to do this when designing one.. let's start from the beginning...
when biasing a BJT or mosfet (for audio amp), it first seems like a bad idea to have it halfway on, draining current through the device all the time - this is not true@! The reason we bias up in a certian spot (large signal) is so we have the most gain for our little tiny input signal (small signal).
If you know the simple common-emitter (or common source for MOS) amplifier, it is simply an inverter - it has a pullup resistor to Vcc. Turning the transistor on stronger makes the output (at collector/drain) voltage drop lower since the transistor can pull down more current. All of these amps act just like an inverter for LARGE SIGNAL applications! Vin=0 vout=5v, vin=5v, vout almost 0.
Now to change to small signal, you look at the region where the output is going from hi to low. This is usally a very sharp slope, dropping from 5v->0v within a few millivolts input. THIS IS THE SMALL SIGNAL REGION!
As we wiggle the "inverter" gate a few mV above and below the "balance point" of midrange, we can watch the output toggle high to low very strongly - now you are understanding the small signal operation.
Let's choose our Vin bias point so we get a nice 1/2way output voltage.. (Vout=2.5v with 0v & 5v rails). Now our input signal is on the razor-edge.. Move two millivolts up and the output goes from 2.5v down to 0. move two millivolts down and the output goes from 2.5v up to 5v, just like an inverter should. The small signal gain is 5V/4mV => 1,250. This razor edge is where you should place your bias points so all the stages of your audio amp have a large region to swing into.
Finding those bias points is trickier - you will have to do large-signal analysis on every stage to find the sharp slope on every amp, then choose inputs that are happy swinging around that particular midrange point (bias point).
Hints:
1) Start from the output and work your way in - mostly, you need a large output driver so it's bias point is not "optimal". If you choose the nicest input regions, you will over-amplify here and run into the top or bottom rail (clipping). Easier to decide on your power stage, and then make small adjustments in the earlier stages so they will combine to rest at output's midpoint....
2) Don't ask for gain of 10,000 from the first stage, and gain=1 from the power stage. Try 1000 and 10 to make it easier to line up those razor-edge toggle points (lower gain makes razor less sharp (flatter slope), so we have a WIDER place to place our input) If you misalign, or make the input way too big, you will run into the top or bottom rail, flattening the peaks of your output signal (clipping, or saturation/cutoff)
3) High gain amps will ALWAYS be unstable unless you put a small cap across one of the middle gain stages (assuming 3-stage amp). The good thing is that a small cap is multiplied by the gain of that middle stage, so not much cap is needed.
Example: 3-stage, gain=100,1000,10 => 1M! If we place a 10pF cap from the input-output of the middle stage, it appears to be a 10pF*1000 size cap, so it will easily stop oscillations from Rout*C down.
4) MOS is easier than BJT of course. BJT's have better gain at high frequency, but for audio you only need very low 40kHz response.
Assume B (beta) = 100 for a pretty hi-end output power BJT. Now a bumping system wil put 500W into an 8Ohm speaker, which gives us P=V*I^2 => I=6.5A on a 12V car battery. We would be dumping 6.5/100 => 65mA into the garbage just turning our NPN on! A few big fat power MOS (Si44xx) could easily put 100A into an 8-Ohm speaker with NO base current - just wiggle the gate voltage around the FET's razor-edge, and those little FETs will dump huge currents into your speaker without disturbing any of your input circuitry. (A hi-gain bjt stage is usally weak, and cannot drive 65mA into a power bjt output stage. On the other hand, a high-gain MOS stage does not need to drive ANY current into the next mos stage, it only has to have enough power to drive a small 100pF gate of the power mos - very easy to do at 0.5mA bias current)
post any questions on what is not easy to understand - audio stuff especially is really hard to work backwards from other people's schematic without understanding why they are putting a 50k on the base of something.
Also, the approach to small-signal where you ground sources is mostly for finding frequency response of a circuit. You guys are asking about SMALL SIGNAL DC GAIN - the linear slope of the sharp hi-to-low toggle(LARGE SIGNAL DC GAIN) When you ground a source, it just means that any wiggling at that node will NOT propagate backwards thru the device - it's resting against Vcc or something. Save AC until you need it. The little 2n3904 can amplify well over 100MHz so any audio stuff down at 10kHz is almost perfect with respect to the AC response.
Feel free to attack this opinion, but I know the math tricks and can reduce any AC model to that linear slope in about two lines. Go read grey!~
when biasing a BJT or mosfet (for audio amp), it first seems like a bad idea to have it halfway on, draining current through the device all the time - this is not true@! The reason we bias up in a certian spot (large signal) is so we have the most gain for our little tiny input signal (small signal).
If you know the simple common-emitter (or common source for MOS) amplifier, it is simply an inverter - it has a pullup resistor to Vcc. Turning the transistor on stronger makes the output (at collector/drain) voltage drop lower since the transistor can pull down more current. All of these amps act just like an inverter for LARGE SIGNAL applications! Vin=0 vout=5v, vin=5v, vout almost 0.
Now to change to small signal, you look at the region where the output is going from hi to low. This is usally a very sharp slope, dropping from 5v->0v within a few millivolts input. THIS IS THE SMALL SIGNAL REGION!
As we wiggle the "inverter" gate a few mV above and below the "balance point" of midrange, we can watch the output toggle high to low very strongly - now you are understanding the small signal operation.
Let's choose our Vin bias point so we get a nice 1/2way output voltage.. (Vout=2.5v with 0v & 5v rails). Now our input signal is on the razor-edge.. Move two millivolts up and the output goes from 2.5v down to 0. move two millivolts down and the output goes from 2.5v up to 5v, just like an inverter should. The small signal gain is 5V/4mV => 1,250. This razor edge is where you should place your bias points so all the stages of your audio amp have a large region to swing into.
Finding those bias points is trickier - you will have to do large-signal analysis on every stage to find the sharp slope on every amp, then choose inputs that are happy swinging around that particular midrange point (bias point).
Hints:
1) Start from the output and work your way in - mostly, you need a large output driver so it's bias point is not "optimal". If you choose the nicest input regions, you will over-amplify here and run into the top or bottom rail (clipping). Easier to decide on your power stage, and then make small adjustments in the earlier stages so they will combine to rest at output's midpoint....
2) Don't ask for gain of 10,000 from the first stage, and gain=1 from the power stage. Try 1000 and 10 to make it easier to line up those razor-edge toggle points (lower gain makes razor less sharp (flatter slope), so we have a WIDER place to place our input) If you misalign, or make the input way too big, you will run into the top or bottom rail, flattening the peaks of your output signal (clipping, or saturation/cutoff)
3) High gain amps will ALWAYS be unstable unless you put a small cap across one of the middle gain stages (assuming 3-stage amp). The good thing is that a small cap is multiplied by the gain of that middle stage, so not much cap is needed.
Example: 3-stage, gain=100,1000,10 => 1M! If we place a 10pF cap from the input-output of the middle stage, it appears to be a 10pF*1000 size cap, so it will easily stop oscillations from Rout*C down.
4) MOS is easier than BJT of course. BJT's have better gain at high frequency, but for audio you only need very low 40kHz response.
Assume B (beta) = 100 for a pretty hi-end output power BJT. Now a bumping system wil put 500W into an 8Ohm speaker, which gives us P=V*I^2 => I=6.5A on a 12V car battery. We would be dumping 6.5/100 => 65mA into the garbage just turning our NPN on! A few big fat power MOS (Si44xx) could easily put 100A into an 8-Ohm speaker with NO base current - just wiggle the gate voltage around the FET's razor-edge, and those little FETs will dump huge currents into your speaker without disturbing any of your input circuitry. (A hi-gain bjt stage is usally weak, and cannot drive 65mA into a power bjt output stage. On the other hand, a high-gain MOS stage does not need to drive ANY current into the next mos stage, it only has to have enough power to drive a small 100pF gate of the power mos - very easy to do at 0.5mA bias current)
post any questions on what is not easy to understand - audio stuff especially is really hard to work backwards from other people's schematic without understanding why they are putting a 50k on the base of something.
Also, the approach to small-signal where you ground sources is mostly for finding frequency response of a circuit. You guys are asking about SMALL SIGNAL DC GAIN - the linear slope of the sharp hi-to-low toggle(LARGE SIGNAL DC GAIN) When you ground a source, it just means that any wiggling at that node will NOT propagate backwards thru the device - it's resting against Vcc or something. Save AC until you need it. The little 2n3904 can amplify well over 100MHz so any audio stuff down at 10kHz is almost perfect with respect to the AC response.
Feel free to attack this opinion, but I know the math tricks and can reduce any AC model to that linear slope in about two lines. Go read grey!~
