# Audio Amplifier Transistor with Diodes

I'm trying to understand this audio amplifier. Whats the purpose of these transistors with the diodes in series?

• If the diodes were linked-out, do you then know what those transistors do? You also must provide a link to where you got the picture from i.e. credit the owner of the photo. Commented Oct 29, 2023 at 16:41
• It is an over-current protection circuit. The diodes are there to protect Q9, and Q10 from negative or positive voltages that appear at the amplifier output during normal operation. For example, the Q9 does not like negative Vce voltage. Are you sure this arrangement is correct?
– G36
Commented Oct 29, 2023 at 16:51
• @G36 Yes, it's correct, and i've seen the amplifier working. It is a circuit we're looking at in university. I think it is quite similar to the NAD S-300. What's the purpose of Q9 and Q10 then? Higher frequencies? Commented Oct 29, 2023 at 17:12
• Q9 is a overcurrent and SOA protection transistor for NPN output stage. And Q10 do the same think for the PNP output stage. youtube.com/watch?v=7bAghP_ARrc
– G36
Commented Oct 29, 2023 at 17:18
• sound-au.com/soa.htm#s5.0
– G36
Commented Oct 29, 2023 at 17:24

To understand the circuit better first try to understand how simple the current protection circuit works.

Where I simplified the circuit and $$\T_1\$$ represents the Darlington output stage with the β = 1000. And $$\T_2\$$ as an overcurrent protection circuit. The driver (VAS - voltage amplifier stage) and the bias stages were not shown.

In normal conditions, $$\T_2\$$ is OFF the amplifier works as if there was no $$\T_2\$$ in a circuit. To force $$\T_2\$$ into conduction the voltage across $$\R_E\$$ resistor should be larger than around $$\0.6V\$$. Therefore as soon as the load current increases to the value so that the voltage drop across $$\R_E\$$ becomes larger than $$\0.6V\$$ the $$\T_2\$$ will start to Turn-ON. When $$\T_2\$$ turns ON, $$\T_2\$$ starts to steal the base current from $$\T_1\$$ (start pulling I1 current away from the T1 base) thereby reducing the collector current of $$\T_1\$$. Hence the output current is limited to:

$$I_{max} ≈ \frac{V_{BE}}{R_E}$$

Also notice that the crucial thing for this circuit to work properly is that the driver stage (I1 = 5mA in my diagram) can only provide a limited amount of current. If this condition is not met the circuit will not work.

Now adding by a voltage divider at $$\T_2\$$ base ($$\R_{35}\$$, $$\R_{24}\$$) we have increased the output current protection circuit tripping point.

$$IL_{max} \approx \frac{V_{BE}}{R_3} \times \left(1 + \frac{R_{35}}{R_{24}} \right)$$

Now we can add another resistor from base $$\T_2\$$ to $$\V_{CC}\$$ so that now the $$\IL_{max}\$$ will no longer be fixed, but will vary together with output transistor ($$\T_1\$$) $$\V_{ce}\$$ voltage. But the main principle of operation stays the same.

For example, if we want to reduce the $$\IL_{max}\$$ when shorting to GND to 3A. We can achieve this if we use $$\R_{23}\$$.

$$\V_{R_E} = 3A * 0.22\Omega = 0.66V\$$ and $$\R_{24}\$$ current need to be equal to:

$$\I_{R_{24}} = \frac{0.7V - 0.66V}{R_{35}} + \frac{0.7V}{R_{24}} = 7.4mA\$$

Therefore

$$\R_{23} = \frac{12V - 0.7V}{7.4mA} = 1.5k\Omega\$$

This is the key. Just by adding $$\R_{23}\$$ we made the protection circuit whose maximum output current is not constant but will depend on Vce voltage value as well, it will reduce the current limit threshold as Vce increases, so we have a transistor SOA protection as well. You can try to derive the expression for this case yourself.

As for the purpose of a $$\D_3\$$ and $$\D_4\$$ diodes. They are here to protect the "overcurrent detection" transistors ($$\Q_9\$$ and $$\Q_{10}\$$). Notice that during normal operation the voltage at the output can be positive and negative as well. For example, without the $$\D_3\$$ diode negative voltage would "force" $$\Q_9\$$ base-collector junction into conduction. So the $$\D_3\$$ dioda job is to prevent forward-biased the $$\Q_9\$$ base-collector junction. $$\D_4\$$ do the same thing for $$\Q_{10}\$$.

The diode is there to protect the small-signal voltage amplifier stage from damage due to energy back-feeding from the amplifier output, in the event of any of the following:
(a) an output transistor failing short-circuit (C to B, or C to E).
(b) inadvertent connection of a power source (eg: battery) or unexpected load (such as back-to-back zener diodes) at the amplifier output, that can cause the amplifier output node voltage go beyond the voltage of the internal VAS (voltage amplifier stage), ie: the base of Q12 and the base of Q13.
(c) reaction of a highly inductive load to a rapid input signal.

In all cases, what is happening is that the B-C junction of Q9 (Q10 for negative signals) is being forward-biased (either permanently or momentarily), causing a low impedance path from its base to its collector. D3 (D4) prevents this path from destroying both Q9 (Q10), and the sensitive components connected to the base of Q12 (Q13).

The sequence of events for cases (b) and (c) can be deduced by considering case (a), as follows:

Case (a):
Consider the circuit with D3 was shorted out. During normal operation, Q17 fails short-circuit from C to E, causing +35V to appear its emitter. Since the voltage at Q12 base is really just the music signal plus 3 Vbe drops, the voltage at the base of Q9 is now much higher than the voltage at its collector. D3 prevents this voltage from reaching the VAS stage (base of Q12).