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the first value of Q1='0' (off). how can I calculate VA voltage, the output of op amp ? I am not sure if is it like a buffer cause I've never seen some buffer like this. thanks ahead for any help

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this is what i get after i try to simulate how is possible the current flow with same voltage?? besided R1?

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    \$\begingroup\$ Hint, what is the effect of Q1 when the latch output is 0 and when it is 1? \$\endgroup\$ – The Photon Jun 13 at 20:49
  • \$\begingroup\$ I think when Q1 is 1 the VA voltage will be -Vin,(va=-vin), but I can't prove myself what is the value of VA in this case of Q1=0 @ThePhoton \$\endgroup\$ – Knowledge Jun 13 at 20:52
  • \$\begingroup\$ Okay, good so far. How does the BJT act when the base voltage is 0? \$\endgroup\$ – The Photon Jun 13 at 20:53
  • \$\begingroup\$ cut off mode, yea i know that collector of bjt not connect the V+,(V+ free) that what makes me problem in my mind. @ThePhoton \$\endgroup\$ – Knowledge Jun 13 at 20:55
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    \$\begingroup\$ BTW, please add designators to your schematic so we can talk about "R1" or "R3" instead of "the 10 k resistor connected to the non-inverting input of the op-amp". \$\endgroup\$ – The Photon Jun 13 at 21:14
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schematic

simulate this circuit – Schematic created using CircuitLab

Vin+ is set to a fixed voltage Vout changes to null the input difference "error".

Iin to both Op Amp inputs is close to zero so we neglect this current. Same for Q1 when OFF. Thus R1 has (almost pA~uA?) no current flowing thru it when Q1 is OFF.

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  • \$\begingroup\$ First -Thanks, it is understood why V+=Vin and the current will not flow thru it when Q1 is off, BUT why the current in R2,R3 neglected(i know that in op amp the current input is closed to zero but there is another path.. \$\endgroup\$ – Knowledge Jun 13 at 21:37
  • \$\begingroup\$ Only current input OA is neglected. I(R2)=I(R3) always so if Vbat is 2 Vout must be -2 to get 0V on Vin- \$\endgroup\$ – Sunnyskyguy EE75 Jun 13 at 21:39
  • \$\begingroup\$ but the voltage Va is equaled to Vin when Q1 is off.... \$\endgroup\$ – Knowledge Jun 13 at 21:50
  • \$\begingroup\$ Yes as said in RED text \$\endgroup\$ – Sunnyskyguy EE75 Jun 13 at 21:54
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    \$\begingroup\$ This circuit looks like an over-complicated 1 shot on power up, but demonstrates the ability to invert voltages with a transistor. \$\endgroup\$ – Sunnyskyguy EE75 Jun 13 at 21:55

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