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?

Phase Splitter

| improve this question | | | | |

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.

enter image description here

| improve this answer | | | | |
  • 1
    \$\begingroup\$ it all makes sense now. \$\endgroup\$ – Neamus Jan 8 '19 at 18:48
  • 1
    \$\begingroup\$ 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. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jan 8 '19 at 19:07
  • 1
    \$\begingroup\$ tinyurl.com/ydyepq6c optimized \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jan 8 '19 at 19:55
  • 1
    \$\begingroup\$ min Vce>=2 for high current and Vce>0.7 for low current. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jan 8 '19 at 20:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.