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I came across this little circuit in an audio amplifier and I just wanted to ask if this is a certain configuration of VBE multiplier, if so what is the purpose of using the Q651 PNP BJT?

Schematic

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    \$\begingroup\$ It would help to see the rest of the circuit to provide some context, but it looks like a clamp (to around 3V) of some sort. \$\endgroup\$ Commented 2 days ago

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The transistors are configured as a Sziklai pair (AKA complementary feedback pair). This functions like an NPN transistor (the polarity is taken from the first transistor in the pair) with a high current gain similar to a Darlington pair. Unlike a Darlington, there is only a single diode drop between the 'base' and 'emitter' of the pair whereas in a Darlington the BE junctions are in series so you get a two diode drop.

So in this case, it's probably functioning as \$V_{BE}\$ multiplier, but with higher current gain than a single NPN would give have. This can give better linearity, but at the expense of possible instability. The first transistor acts as the temperature sensing component and may be mounted on the output transistor heatsink. Typically there is also a resistor from the collector of the first transistor to the emitter of the second (the 'collector' of the pair).

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Vbe multiplier reverse engineering

STEP 1: 2-transistor Vbe multiplier

The circuit behaves as a shunt voltage regulator or, as you said, as a kind of "Vbe multiplier".

schematic

simulate this circuit – Schematic created using CircuitLab

Here is a CircuitLab DC sweep simulation.

STEP 1

STEP 2: 1-transistor Vbe multiplier

For comparison, let's explore the classic 1-transistor Vbe amplifier.

schematic

simulate this circuit

STEP 2

STEP 3: 1-transistor Vbe follower

Finally, let's explore the version without a "multiplying voltage divider" (aka "diode-connected transistor").

schematic

simulate this circuit

STEP

STEP 4: Base-emitter junction

schematic

simulate this circuit

STEP 4

STEP 5: Diode

schematic

simulate this circuit

STEP 5

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    \$\begingroup\$ I got confused by the absence of any resistance in the loop to allow the \$V_{BE}\$ multiplier to develop its own 2.5V. I think a regular 0Ω ammeter with an explicit 1kΩ resistor outside would be clearer. \$\endgroup\$ Commented 2 days ago
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    \$\begingroup\$ @Simon Fitch, that's right... I usually explain it in the text and put the resistance in parentheses, but there's no text here. That's how I try not to clutter up the diagrams too much. \$\endgroup\$ Commented 2 days ago
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The circuit is designed to regulate the voltage (let's denoted as \$V_R\$) across the C657 capacitor. The presence of Q651 (PNP) is not primarily for multiplying the \$V_{BE}\$ but for enhancing the stability of the regulated voltage, resulting in better line regulation. If only Q652 (NPN) is used without Q651 (PNP) like below, the regulated voltage will exhibit more variation.

enter image description here

For instance, if the input voltage increases, the collector current (\$I_C\$) of NPN increases to maintain the \$V_R\$. This necessitates an increase in the base current (\$I_B\$), leading to higher current through \$I_{R1}\$, thus regulating at a higher \$V_R\$. However, with the PNP in place, the collector current (\$I_C\$) of NPN is significantly smaller, resulting in a smaller rise in \$V_R\$. Additionally, because a transistor's \$V_{BE}\$ changes with temperature, the regulated voltage would also change. If NPN is used alone, rising input voltage forces it to sink larger current, causing it to heat up and its \$V_{BE}\$ to drift, thus affecting the regulated voltage. Delegating current shunting to the PNP keeps the NPN's temperature relatively stable (as long as it is not placed too close to the PNP), thereby maintaining a stable regulated voltage.

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