Unity Gain Phase Splitter- Why outputs have different voltages

Question

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. 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 polarities, but why are these voltages like this?

score 3 · Accepted Answer · answered Jan 08 '19 at 18:40

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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.

answered Jan 08 '19 at 18:40

G36

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it all makes sense now. – Neamus Jan 08 '19 at 18:48
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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. – Tony Stewart EE75 Jan 08 '19 at 19:07
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http://tinyurl.com/ydyepq6c optimized – Tony Stewart EE75 Jan 08 '19 at 19:55
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min Vce>=2 for high current and Vce>0.7 for low current. – Tony Stewart EE75 Jan 08 '19 at 20:10

Unity Gain Phase Splitter- Why outputs have different voltages

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