Unity Gain Phase Splitter- Why outputs have different voltages

I’m trying to calculate the values of the biasing resistors for this phase splitter circuit. (I know the actual values of the resistors, but I just wanted to manually work it out). When I made the circuit on a Multisim, I realised that Vc=11.25V and Ve=3.75V when. I’ve been looking around the web to see why this is the case, but I can’t seem to find anything. I understand the outputs are different polarity’s but why are these voltages like this? Thanks in advance. ~Neamus.

To be able to achieve the maximum voltage swing at the two outputs we need to set the DC Q point at the "middle". The middle point is equal to $$\\frac{V_{CC}}{2} = 7.5V\$$. So we have two equal "halves" upper one and the lower one.

The lower emitter half we additional divided by 2 to get the full symmetric swing at the emitter output. Hence:

$$\V_E = \frac{7.5V}{2} = \frac{V_{CC}}{4} = 3.75V\$$

We do exactly the same thing with the upper collector half and set the collector voltage at:

$$\V_C = 7.5V + 3.75V = 11.25V\$$

I hope you now see why this is the case.

If not I hope that this diagram explains it well.

• it all makes sense now. – Neamus Jan 8 at 18:48
• This in practice won't be very linear if Vce drops <2V so offset the Ve to maximize linearity and reduce swing by 2V. ALso Load R is included with Rc and Re load must be >>2Re. if AC coupled. – Sunnyskyguy EE75 Jan 8 at 19:07
• tinyurl.com/ydyepq6c optimized – Sunnyskyguy EE75 Jan 8 at 19:55
• min Vce>=2 for high current and Vce>0.7 for low current. – Sunnyskyguy EE75 Jan 8 at 20:10