0
\$\begingroup\$

enter image description here

I am trying to design an operational amplifier with a gain of more than 200000. My first stage is a differential amplifier with an active load. I am getting a gain of about 1900 from it. Second stage is a common collector amplifier with unity gain just to minimize the loading effect on the first stage. Until the output of the 2nd stage (common collector amplifier), it all works well.

Problem arise when I try to connect a common emitter amplifier to the output of second stage to increase the overall gain. From what I can see, the BC junction of transistor 10 is forward biased so the CE amplifier doesn't work as an amplifier. I have tried many things like changing resistance values and biasing current. Nothing works.

Would really appreciate if anyone can help. I am stuck.

\$\endgroup\$
1
  • \$\begingroup\$ Please ask a specific question \$\endgroup\$
    – Voltage Spike
    May 18 at 21:25
5
\$\begingroup\$

The node between Q2's and Q6's collector puts out a current that is proportional to the input differential voltage; that is, the differential stage is a so-called transconductance amplifier. It doesn't produce a usable voltage on its own. To get a voltage from this, you'd typically use a PNP miller integerator (the one in the linked picture uses an NPN transistor, just flip it around for PNP). This integrator circuit will then put out a voltage that you can buffer with another NPN emitter follower. By choosing the integrator's capacitor appropriately, you can also stabilize the OpAmp (because it won't be stable if the gain doesn't roll off at high frequencies). This of course means that you'll have a very high gain at DC but the gain will drop off as the frequency increases. Building OpAmps with a constant high gain for all frequencies is sadly not physically possible, which is why at least one capacitor is required in every OpAmp - even in OpAmp chips. An unstable OpAmp will oscillate and not produce any usable output signal.

Here is an article about a very simple discrete OpAmp. You can clearly see the miller integrator formed by Q3 and C1. This circuit is the most basic OpAmp you can build and it's the basis for more complex circuits. You can improve it by adding current mirror loading to the differential pair, a current load for the miller transistor, darlington transistors at various places in the circuit and much more.

If you need more gain than this simple configuration can provide, you might need to look into cascode differential amplifiers, Wilson current mirrors and/or nested feedback loops. You might also need higher-bandwidth transistors so you can get more gain without your OpAmp becoming unstable. You sadly can't just add another common-emitter amplifier stage at the output of the OpAmp because that'll mess with the DC bias and also make the OpAmp unstable due to too much gain at high frequencies.

The output of the miller integrator in any OpAmp will have to be buffered, for example with an NPN emitter follower, because any loading on the miller integrator's output node will drastically reduce your overall open-loop gain.

Finally, in order to properly measure a discrete OpAmp, whether in reality or in a simulation, you'll need to provide external feedback to it so its operating point is properly stabilized. This will of course drop the gain of the complete circuit dramatically. To still measure the open-loop gain of the OpAmp alone in a simulation, you can apply an AC signal to the circuit and do an AC sweep. The expression "V(out)/V(in+,in-)" will show you the gain of the OpAmp itself (over frequency), where V(in+,in-) is the voltage difference between the OpAmp's non-inverting and inverting inputs. Alternatively, you can calculate the OpAmp's gain by multiplying the differential stage's transconductance with the miller integrator's DC transresistance.

\$\endgroup\$
1
  • \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. \$\endgroup\$
    – Voltage Spike
    May 18 at 15:07
0
\$\begingroup\$

The 2nd stage Darlington is current starved expecting 100 ohm load to be driven by the same current as the front end.
This is your main issue.

  • you could move the 8k from collector of Q7 to the emitter to boost the Q8 mirror , just from my cursory glance.
  • But it IS MUCH better to use a separate current mirror with the output going into CE with a diode multiplier cascode to a differential Darlington Push-pull output.
  • then the high side of the diode multiplier is a good place to add a 27 pF Miller cap back to the base of Q9 as that will be negative feedback into high impedance.(if you raise R1 by an order of magnitude which also increases gain)
\$\endgroup\$
2
  • \$\begingroup\$ Thanks again Jonathan...So you are saying that I should increase the biasing current of the CE amplifier? \$\endgroup\$
    – Blackbeard
    May 17 at 17:27
  • \$\begingroup\$ Yes you must pull up the collector with more current to match + 10% the expected Ve/Re current + 3rd stage current you need. \$\endgroup\$ May 17 at 17:50
0
\$\begingroup\$

Transistor Q10 is deeply saturated-- collector voltage ~ emitter voltage. This transistor is essentially a wire. Reduce this branch current until you get Vce positive with some margin.

\$\endgroup\$

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.