amplifierdiodes
I can't figure out how the DC current through the diodes is calculated in eq. 8.22 on this PDF from University of North Carolina.
$$
I_D = \frac{V_{CC}/2-V_{BE}}{R_f+R_2}
$$
\$I_D\$ should be equal to the current through the resistor \$R_2\$ minus the current going through the base of the transistor \$Q_1\$. I am able to obtain this equation only supposing \$I_B = 0\$, which is not true, specially because they are matching the diode to the transistor to avoid thermal problems.
The 1N4001 diode is considered a general purpose rectifier. It's in a family along which includes the 1N4001 through 1N4007 diodes.
I have never heard of (and cannot find) a 1N4008.
You should look at the datasheet for these diodes to determine if you can substitute them safely.
The main difference is in the maximum ratings:
You will have to make the determination of what constitutes an acceptable replacement because we do not know the characteristics of your device.
If you rearrange your formula, you get $$ (U_m - 2U_T) = I_m (2R_d + R) $$ which looks like $$ V = IR $$ where \$R\$ is the sum of all of the resistors in the loop and \$V\$ is the voltage across that summed resistor.
Kirchoff tells you that the voltage across this resistor is the same as the voltage looking at the rest of the circuit, which is \$U_m - 2U_T\$, because there must be a voltage drop across each of the diodes for current to flow.
Best Answer
Yes, that equation assumes Ib=0. We all know that can't be true IRL, but if the transistors aren't in saturation, it's likely that Ib is far lower than the bias currents used in the resistor/diode network. So the estimate isn't a perfect solution in the real world, but it's close enough...and engineering is never about perfection, it's all about being close enough.