General design theory help

To start with you'll need to convert that circuit to it's AC small signal equivalent circuit.  Then you can calculate the input impedance, among other things.

How's your knowledge of complex numbers and Laplace transforms?

edit: great idea for a thread.  I'm a bit rusty with small signal analysis so this could help.

So, according to the Thomas L. Floyd electronic's book I own, input impedance of the amp is (1+Abo*(R3/(R3+R4)))*Zen. Abo is the open loop gain of the amp, Zen its input impedance (found in the datasheet). Then I guees you have to parallel it with R2. And you add the (small) value of R1.

I guess, since the input impedance is really high, the only value which matters is R2 (and R1).

As for AC analysis, how can it help here ? Th only thing I could do is shortcut V1 and V2.

Please someone correct me if I'm wrong ! I might come back, with questions this time.

cant be sure, but here goes. correct me if i'm wrong.

input impedance is dominated by R1 and R2. The input impedance of the direct input is usually VERY large, depending on the types of input transistors. In an ideal case, it is assumed to be infinity. You don't really have to look at this if it's:

1) large enough
2) you know how to deal with the base currents involved
3) you know that it's an monolithic opamp and your external circuits will dominate your Zi.

As for the gain of the circuit, it's R4/R3 + 1. You can actually derive this by using Av=VO/VI and assuming that the input impedance is infinity, zero base current, and some basic thevenin/norton equivalent circuit on the voltage dividers.

R2 of 20K to ground is VERY small, it should generally be 10x the POT (if you have one), if not, then we can use just the 20K as input impedance.

The formula to calculate the RC filter:

f = 1/(2 x Pi x R x C)

so,

5Hz = 1/(2 x Pi x 20,000 x C)

can be rewritten as 

C = 1/5Hz/(2 x Pi x 20,000)

C = 1.59 uF

5Hz is really low. I'm not sure you can as down as 40Hz with headphones. Calculating it for 20Hz allows to use 0.47µF.

My question is, with a design like the AMB Beta22, how do you calculate the current that goes through R9 and R5 (which the same, right ?) It is much easier with the constant current sources the Dynalo uses. :)

Quote:

Originally Posted by bidoux 

5Hz is really low. I'm not sure you can as down as 40Hz with headphones. Calculating it for 20Hz allows to use 0.47µF.

Capacitor effects stick around for a few octaves above their -3db point in a high pass filter like this. I have seen statements elsewhere (from listening reports, take them for what they are worth) that more than one roll off in the audible band causes funky bass smears. Many people actually shoot for ~2hz as a -3db point. The cap does not need to be too big if you select R2=100Kohm

Quote:

Originally Posted by bidoux 

My question is, with a design like the AMB Beta22, how do you calculate the current that goes through R9 and R5 (which the same, right ?) It is much easier with the constant current sources the Dynalo uses. :)

I think your supposed to measure the voltage across R9 and calculate back from that, and yes the current is the same in both of those resistors: they are perfectly in series (also in series with the BJT&fet).

I have no idea how you would do it if you wanted to design your own input stage or amp this complex though. Spice? 

Im kind of curious why R5&R6 =! R7&R8. Is it to account for differences in the VGS of the P & N channel fets?

Quote:

Originally Posted by bidoux 

5Hz is really low. I'm not sure you can as down as 40Hz with headphones. Calculating it for 20Hz allows to use 0.47µF.

 

My question is, with a design like the AMB Beta22, how do you calculate the current that goes through R9 and R5 (which the same, right ?) It is much easier with the constant current sources the Dynalo uses. :)

ACTUALLY, the calculation is the same.

The input differential stage is self biased. Meaning that input stage current depends on how much voltage you drop across the source resistors and the IDSS curve of the device itself (this varies between devices). The VR1 is there to set the current for the input stage. Current that goes through R9 and R5 can be easily determined by V=IR, I=V/R by measuring the voltage drop across both resistors and applying the formula mentioned.

Quote:

Originally Posted by nikongod 

 

Capacitor effects stick around for a few octaves above their -3db point in a high pass filter like this. I have seen statements elsewhere (from listening reports, take them for what they are worth) that more than one roll off in the audible band causes funky bass smears. Many people actually shoot for ~2hz as a -3db point. The cap does not need to be too big if you select R2=100Kohm

 

I think your supposed to measure the voltage across R9 and calculate back from that, and yes the current is the same in both of those resistors: they are perfectly in series (also in series with the BJT&fet).

 

I have no idea how you would do it if you wanted to design your own input stage or amp this complex though. Spice? 

 

Im kind of curious why R5&R6 =! R7&R8. Is it to account for differences in the VGS of the P & N channel fets?

Because they have different transconductance curve between P and N devices. Amb wanted it to be balanced symmetrically.