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I want to design a Class AB amplifier as seen below. But, I don't know how to find the values of the resistors and capacitors. I have found the KVL around the closed loop containing the diodes and resistors to find R; where R = (Vcc - 1.4V)/(2*I) but I dont know how to determinet I. Is there any way to find I, also my load resistor is 4 ohms. Thanksenter image description here

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  • \$\begingroup\$ bypassing the diodes with capacitors can improve the tolerance on the biassing current. Note that this schematic is more of a sketch for discussion, than a design for a high fidelity audio amplifier. \$\endgroup\$ – Neil_UK Apr 28 '17 at 7:30
  • \$\begingroup\$ I should be significantly larger than the base currents of TR1 and TR2 -like 5-10x larger. \$\endgroup\$ – Brian Drummond Apr 28 '17 at 8:50
  • \$\begingroup\$ As the transistors heat up, a positive feedback event will occur; the transistors will thermally run away, and destroy themselves unless enough lumpd Re is in the emitters. Play it safe, and insert 1_Ohm in each emitter. Or 10_Ohms for small transistors. \$\endgroup\$ – analogsystemsrf Apr 28 '17 at 11:25
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For basing diodes, 1-5mA range is quite acceptable.

But does not finish with this. You need some other info:

As you might know, output current passes through collectors. So each output transistor's base will need a current of \$I_B = I_C / \beta \$. Thus, R1 and R2 should also allow enough base current for output transistors.

  • Determine output power (\$P_O\$) and calculate output voltage (\$V_O\$) and current (\$I_{O}\$). Also, this circuit has a unity voltage gain:

$$V_{O-rms} = \sqrt{P_O \cdot 4\Omega} \ ,\ \ \ \ V_{O-pk} = V_{o-rms} \cdot 1.41$$ $$I_{O-rms} = \sqrt{P_O / 4\Omega} \ ,\ \ \ \ I_{O-pk} = I_{o-rms} \cdot 1.41$$

Select proper transistors having enough \$V_{CE}\$ for \$V_{O-pk}\$ and \$I_C\$ for \$I_{O-pk}\$. Select Vcc for enough voltage swing: \$V_{CC} = 2\cdot V_{o-pk} + 1V\$

  • From datasheets, calculate required base currents from \$I_B = I_C / \beta\$. Note that \$\beta\$ value can be quite low (e.g. between 10 and 25) for high output currents.
  • Calulate R1 and R4 for diode bias current and transistor base current.

Input coupling cap is calculated from minimum input frequency (\$f_L\$). If we assume input impedance is parallel equivalent of R1 and R2 (say \$R_i = R1 || R2\$) then input coupling cap is \$Ci = 1/(2\pi f_L R_i)\$. If you don't have any info, put a 10uF electrolytic and test.

Output coupling cap is also calculated from minimum input frequency: \$C_o = 1/(2\pi f_L \cdot 4\Omega)\$. If you don't have any info, put a 2200uF electrolytic (+ to common emitter, - to load resistance) and test.

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  • \$\begingroup\$ Thanks for the reply. Where did you get 1-5mA range as I would need to refer to it. \$\endgroup\$ – Swag96 Apr 30 '17 at 2:52
  • \$\begingroup\$ @Swag96 Any small-signal general purpose or fast switching silicon diodes serve good performance at that level of bias currents. For example, I recommend you to use 1N4148 (or LL4148 which is the SMD version) diodes because they have good performance at high temperatures and you can see from the datasheet that 1-10mA range is quite linear, so you can select any bias point on that range (e.g. for If = 3mA, Vf = 0.7V). \$\endgroup\$ – Rohat Kılıç Apr 30 '17 at 5:57

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