On to the exciting stuff !
Consider a transistor. If we neglect its base current, the thermal power it dissipates is :
p = Vce * Ie
If we wish to reduce thermal effects, we want to enforce a constant junction temperature, hence a constant dissipated power, in all critical transistors. This can be done in two ways :
They are simpler to explain and will make a good example before we start with the other, hairier stuff. Conventional wisdom has it that "current sources sound bad". Maybe this is because they add memory, as the measurements showed.
The critical transistor of our current source (Fig. 4-1) is Q1. Thermal Vbe variations will change the voltage on the emmiter resistor, hence the current.
The nice thing in a current source is that, obviously, the current is almost constant. Therefore, we will just put a cascode transistor (Q3) to enforce constant Vce on Q2, and then our constant Vce and Ie is done !
Fig. 4-1 : Current sources
Here, I activated the thermal model only on the current source transistors (and the cascodes) of both sources (input pair tail and VAS), but not in the rest of the amp. The measurements are self-explanatory :
|... and Re=1K||0.004|
It was known that cascodes improved the sound, but it was unclear wether or not it was because they give better frequency response. Doc Bottlehead sells cascode current sources to put in their highly famous single ended tube amps, so they must sound good. These results show that the effect of cascodes might be simply thermal.
From now we will keep the cascoded current sources, with Re=1K in the input stage and Re=650Ohms in the VAS. Voltage references will be taken from 8.2V Zeners, which is the voltage for which the Zeners are the less noisy.
Now that we have seen what tweaking an innocent-looking current source can do, let's tackle the input stage.
In order to reduce memory, we must ensure Constant thermal dissipation in critical transistors, regardless of :
Besides, this input stage should be insensitive to the drifts generated by the other stages, especially the VAS.
I will spare you the details, but I tried countless configurations, more or less inspired by the Lavardin patent [Download]. Their solution (read the patent !) is elegant but there is a hint of positive feedback, and it does not take into account the differential voltage. This is not very important as their amp is a very high feedback design, and error voltage is very small, but still, it leaves room for more brain-itching.
I finally arrived at the following configurations :
Fig. 4-2 : Input Stages
Configuration B is a Complimentary Feedback Pair. The current in Q1 and Q2 is simply the Vce of Q1B and Q2B divided by RIDiff. This Vce is quite constant, so the current is, too. However, Q1/Q2 Vce is not constant as it is simply one Vbe lower than the Vce in the standard case. Thermal power is not constant but should be lower, because the current is lower.
Configuration C is a simple cascode with JFETS. 2SK246 will suit this application well (go to Mr Borbely's site and read the articles on JFETs and cascodes). It will maintain its source around 2 volts above its gate, which is perfect here, as it will give us a constant Vce of around 2.6V. Mr. Borbely connects the gate of the JFET to the emitter of the transistor below in order to neutralize its parasitic gate-drain capacitance. It could work here, with one transistor.
Configuration D is the combination of B and C : a Complimentary pair to enforce constant current in Q1/2, and a Cascode to enforce constant Vce. Try to analyze the circuit, you will see it works. Q1 and Q2 are operated at constant power. Neat, eh ?
Of course, it looks complex and ugly, but it is the simplest way that I know of to achieve constant power. As an added bonus, the local feedback of the complimentary pair makes it much more linear than the standard configuration. And, the parasitic G-D cap on the JFETS prevents the whole mess from oscillating : it will not do dirty stuff at high frequencies.
Before measuring memory, let's make sure this thing is fit for the job by plotting the transfer function of Output (diff current) versus Input (Diff Voltage) (see Fig. 4-3).
The red curve is the new input stage, the black one is the simple, two-transistor version. Top curve shows Iout versus Vin, and the two curves at the bottom show the gain variations (derivative) of the previous curve. Same vertical scale is used on both, so we can see the new stage has a much flatter curve, meaning more linearity. The blue curve is the configuration B, the CFP, which has the same linearity as the new stage.
Fig. 4-3 : Input Stage Linearity
It will also be interesting to plot thermal power in the critical transistors Q1 and Q2 versus common mode (Fig. 4-4) and differential (Fig 4-5) input voltage :
Fig. 4-4 : Dissipated power in critical input transistor versus common mode input voltage
Fig. 4-5 : Delta of Dissipated power between critical input transistors versus differential input voltage
Numeric results are easier to handle (I added the cascode only case). The Common column is the variation in the dissipated power in one transistor for a 2 volts variation in common mode input voltage (-1 to +1 volts). The Diff column is the difference in power between the two transistors for a differential input voltage of 0.2V (-0.1V on one base, +0.1 on the other). The stages are loaded with 1k resistors int he collectors connected to a V+ of 40V. Powers are in milliwatt :
|A. Classic||8 mW||60 mW|
|B. CFP||< 0.001 mW||0.67 mW|
|C. Cascode||4.7 mW||4.7 mW|
|D. New||< 0.001 mW||0.04 mW|
This is very logical : the cascode and New stages are well protected against common mode. The CFP is well protected against differential voltage. The New circuit is the only one that can handle both with minimal thermal drift. Let's do a "real-life" test by incorporating it in the complete simulated amplifier. Here, the thermal models are active only in the input stage (left column), or in the whole amplifier (right column).
Results are conclusive : it works as advertised. We have to minimize the memory in the rest of the amp, though, it we want to benefit from it. Let's make a quick test to find who the culprit is : current mirror or VAS ? I will also keep the four input stages a for this one.
With a perfect VAS
With a perfect Current Mirror
We have interaction, or compensation effects, again between the current mirror and the VAS, as the perfect current mirror actually gives worse results than the "real" one. Anyway, it is pretty clear from these results, that :
That is precisely what we vill do in the next chapter !