# Current-Steering Circuit

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

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

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

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

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

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

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