Electronic – Bridging Discrete (BJT) Amp Channels

A buffer can serve many purposes in a circuit. There are other uses besides what I mention below, but these are some fairly common ones you might encounter.

One common use is as a time delay in data transmission. It can take a bit of time for a chip to recognize the change in a signal's voltage level and react to it (such as a digital HI to LO transmission). The amount of this slight propagation delay varies from one device to the next. Sometimes it is a few nano seconds, sometimes it is much more. Also, it does take some time for the signal to propagate through the copper from one part of the circuit board to another. When there are one or multiple data signals moving from one part of the circuit to another, it may be necessary to delay one so that they arrive in a specific order or at exactly the same time.

Buffers can be used as momentary data storage. Similar to the time delay, there are situations when you need a small amount of time to store a data bit such as pipelining, but this example can be a bit difficult to explain, depending upon your knowledge of computer architecture.

Another example is signal isolation. Because a buffer is essentially a signal repeater, it can be used to isolate the signal from two parts of the circuit. This is the specific use in your example, as noted by Ignacio in his answer. In this particular case, the DAC (digital to analog converter) is sensitive to loads. A load can be anything from a resistor, speaker, IC input, or even just a piece of wire. This "sensitivity to load" means the signal may change (distort) if too much or too little of a load is seen at the output of the DAC. To prevent this, a buffer chip is used to repeat the signal from the DAC output. The output of the buffer can be connected to whatever you want (within its own limitations) without affecting or distorting the output from the DAC in any way.

Lastly, a buffer can be used a signal amplifier. This is very similar to the previous example, but the reasons for using it can be different. While the output of some chip might not be extremely sensitive to loads, it may only be able to supply a few milliamps of current. In this case, you may want to use a buffer with a larger output current rating to amplifier the original signal so it can be used to drive some larger load.

1. A resistive load is not a bad model for headphones. If you wanted to be pedantic, you could add inductance for the speaker coils and parasitic L, C, R for the cabling, but a simple resistor is okay. I would look up the impedance your headphones and use the manufacturer supplied impedance.

2. I would consider R10 & R11 as a form of current limiting. Here's why. (I'm using the reference designators of the SGA-SOA-2 original schematic on sg-acoustics for this question, assume reference designators according to your schematic for the rest) The branch formed by Q6 & Q7 is the output stage of the op-amp. We know V_be6 + V_R10 = V_d5 + V_d6. V_d is related to the diode current in a logarithmic relation, and thus V_d5 + V_d6 will not change appreciably. Thus we can say V_be6 + V_R10 = constant. Thus at high output currents, R10 will take the voltage drop and decrease V_be6, thus limiting the amount of current that can flow through the output stage. This is good for protecting against short circuits. That's my guess anyways, it is a bit strange for short circuit protection, typically we see another transistor monitoring R10 and then shorting out V_be6. See (http://users.ece.gatech.edu/mleach/lowtim/prot.html, figure 3).

3. R8 implements Emitter Degeneration. There's a lot to be said on this topic, searching for common emitter with emitter degeneration will generate a lot of resources. The high level idea is we use R8 to set the bias current of the transistor because without it, the gain of the common emitter amplifier (Q7) is very sensitive to V_be7 and makes it difficult to break into large signal & small signal models. So R8 stabilizes our amplifier, but it also kills our gain. The voltage gain of the emitter degenerated common emitter amplifier is -R_C / R_E where R_C is the collector resistance and R_E is the emitter resistance. You can see that as R_E increases the gain decreases. This is bad as this amplifier stage provides the majority of the op-amp gain. In order to fix this, we use bypass capacitor C1. Capacitors have impedance 1 / (jwC) so at low frequency, it has very high impedance and at high frequency, there is low impedance. Let's just examine the DC (biasing case). At DC, C1 has high impedance and the parallel network C1 || R8 is approximately R8. So at DC, we still have all the benefits of emitter degeneration (basically helping us bias the transistor). At high frequency, we want to have large gain. In this situation, C1 || R8 is now dominated by C1 which has low impedance and C1 "bypasses" or shorts out R8. Now we are back to our standard common emitter amplifier which has much higher gain than the emitter degenerated version. Basically C1 makes the gain of the amplifier frequency dependent so we get both large gain at high frequency and nice biasing of the transistor.

4. Not really, you should expect comparable performance since you are using actual models and not ideal components. I've built discrete op-amps before and they typically match within 10% of simulation results.

5. Yes, certain transistors probably have better noise figures, but I'd be hesitant to say that the transistors themselves are the source of the harmonics instead of the design of the amplifier. I would carefully look at the biasing of the circuit before trying to use better transistors. Remember that if we break our small signal model assumption, we can't assume linearity over the amplifiers. Make sure each of your stages is linear over the range that you expect. Disconnect the output stage and ensure the diff pair + C-E is linear and remember that the input signal must be really small in order to meet your small signal assumption in open loop. Typical op-amp gain might be > 10^4. That means even a 1 mV signal would be amplified to 10 V swing. Test the diff pair & C-E open loop, verify linearity. Crossover distortion should be almost definitely eliminated by the op-amp feedback. If it isn't, then the op-amp gain isn't large enough.

6. Open loop measurements of op-amps are difficult because of the large gain. I would even try going lower to 100 uV. We can only expect linearity if our small signal models are valid, meaning the perturbations around the bias point is small enough that we can linearize. Reiterating what I said for question 5, check the linearity of each stage independently, the diff pair, C-E and output stage. The may mean creating bias circuitry to test each stage separately. If you are running into power rail saturation (running against 18, -18 V), probably check the biasing of the amplifiers to ensure there is enough voltage headroom at the output to allow for swing. If you see distortion, make sure small signal is still valid. Lastly, check the datasheets and make sure you're not running up against any current constraints. These could be in the BJT or part of our design as seen by the current limiting for R10, R11.