Can anyone help me solve the equation for I-reference? The final answer is given, but I don't know how it is obtained. Thanks enter image description here

  • \$\begingroup\$ Homework. What were you able to figure out yourself so far? \$\endgroup\$ – Nick Alexeev Nov 3 '15 at 20:48
  • \$\begingroup\$ it is from the lecture notes, and I am trying to understand the concept \$\endgroup\$ – user65652 Nov 3 '15 at 20:49
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    \$\begingroup\$ yes I did try but I got the polarity of VEE and VEB2 reversed \$\endgroup\$ – user65652 Nov 3 '15 at 20:50
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    \$\begingroup\$ Eh, I'll give him credit for just not seeing this right. Sometimes you need a slap to uncross your eyes. \$\endgroup\$ – Daniel Nov 3 '15 at 20:51

The base and collector of Q1 and Q2 are tied together.


is simply the voltage across R.

Edit to include comment discussion: The voltage across R is just the total voltage drop in the current path, (Vcc + Vee) minus the other drops (Vbe1 + Vbe2). You are left with the drop across R, or

$$(V_{CC} + V_{EE}) - (V_{BE1}+V_{BE2})$$

  • \$\begingroup\$ yes I can see that , but how did you add these up? \$\endgroup\$ – user65652 Nov 3 '15 at 20:52
  • \$\begingroup\$ Vee is given as "-Vee". I think that may be where you're confused. \$\endgroup\$ – Daniel Nov 3 '15 at 20:53
  • \$\begingroup\$ Vcc + Vee is potential difference (voltage) between Q1's emitter and Q2's emitter \$\endgroup\$ – Daniel Nov 3 '15 at 20:54
  • \$\begingroup\$ I can see the potential difference for Q1 which s VCC-VBE1 but I don't get the one for Q2 \$\endgroup\$ – user65652 Nov 3 '15 at 20:55
  • \$\begingroup\$ -Vee + Vbe2 right? \$\endgroup\$ – Daniel Nov 3 '15 at 20:57

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