Can you explain this differential amp?

volker said:

Nothing unusual about it, other than it's drawn upside down voltage-wise. The three transistors together form a crude operational amplifier. The left one of the differential pair is the negative input, signal input through the 2,2k resistor. The output of the circuit is the collector of QE05/06, so with RE46/47 connected from negative input to output, it is an inverting amplifier. The positive input, the right transistor of the LTP is tied to ground through a RC. Notice that the base-collector of QE05 is shorted out, there should be a compensation capacitor as in the lower part of the schematic.

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samgraysound said:

what makes the left transistor the negative input and the right the positive?

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clintrubber said:

Defined by the 'internal hookup' of the opamp (the 3 BJTs): starting from the left input, output of the diff-pair has same signal polarity. Then comes the rightmost BJT (say QE05), which inverts the phase. We started from the left input, so that one is then the negative input.

The positive input of the opamp is connected to ground, via RE45.

The applied feedback network is 'in agreement': the total amplifier stage is inverting, establishes a virtual earth @ the neg opamp input. The input resistor RE39 goes from amplifier input to the virtual earth (=the node of the neg opamp input), and RE47 is the feedback resistor, as placed between the virtual earth and the opamp output (the collector of QE05).

Page 22 of this link:

https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&cad=rja&uact=8&ved=2ahUKEwjf37Ca8P_fAhXOb1AKHZgxASUQFjAAegQIBxAC&url=http%3A%2F%2Fwww.cypress.com%2Ffile%2F65366%2Fdownload&usg=AOvVaw0ZYKHsOeIL_eBaWNWUn8On

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