6
\$\begingroup\$

I'm unable to understand how the biasing of the 12AX7 Tube in the schematic below works. In particular, I can't understand how the LM334 can switch on, where does it find the required voltage to switch on? enter image description here

\$\endgroup\$
  • 2
    \$\begingroup\$ You should mention that the LM334 is a voltage regulator. The LM334 has 3 pins so it can be connected in many ways. There is only one way where it will work as a current source (that's also a hint to the answer to your question). \$\endgroup\$ – Bimpelrekkie Dec 21 '17 at 13:38
  • 3
    \$\begingroup\$ The voltage/current comes through the tube, of course! \$\endgroup\$ – Dave Tweed Dec 21 '17 at 14:35
9
\$\begingroup\$

The 12AX7 double triode is connected as a differential amplifier (long-tailed pair). In the Wikipedia article you see a similar circuit implemented with BJTs, but the principle is the same: consider the cathodes to be the emitters, etc.

For optimum performance the AC resistance on the common cathode should be high. This is achieved using an LM334 connected as a current source (in this case it acts as a current sink actually).

Due to the symmetry of the circuit when no signal is applied, the current sinked by the LM334 is shared in equal parts by the two triodes.

The shared bias current is set by the 34Ω resistor, as explained in the datasheet:

enter image description here

Note that the set current is temperature dependent, and will increase linearly with temperature (the LM334 is also used as a temperature sensor). At 25°C, i.e. 298K, from the formulas above you get:

$$ I_{SET}=\frac {227 \mu V / K} {34 \Omega} \times 298K \approx 1.98mA $$

Therefore each triode will have a quiescent cathode current of about 1mA.

...where does it find the required voltage to switch on?

The LM334 contains a complete feedback amplifier connected as a shunt regulator. It doesn't need a "power supply": it takes the power to function directly by sinking a current and regulating that current using negative feedback. Of course you have to design the circuit so that it sinks a minimum amount of current, which translates to a minimum voltage across the \$V^{+}\$ and \$V^{-}\$ terminals.

For the range of currents we are interested in, the datasheet declares this minimum voltage to be about 1V (emphasis mine):

enter image description here

Figure 11 tells you more about the voltage you can expect to find across the current source (enhanced by me):

enter image description here

Interpolating the curve for \$R_{SET} = 34 \Omega\$ (in blue) and setting the current to about 2mA, you see that you'll have about 1V across that source, which is no problem, since the whole triode pair is powered by 240V.

In other words, the circuits inside the LM334 automatically adjust the voltage across its terminals in order to maintain that 2mA shared current constant.

|improve this answer|||||
\$\endgroup\$
  • \$\begingroup\$ Thanks for the great explanation. So If I understand well, at "time 0" the LM334 sees 240V across V+ and V-, and it takes this voltage down to 0.9V thanks to its regulating action? The datasheet states that the max operating voltage is 40V. How does this cope with the 240V? \$\endgroup\$ – Enrico Dec 21 '17 at 14:39
  • 1
    \$\begingroup\$ There will be a transient, which is not accounted for by the analysis I carried out above. Initially both the triodes and the LM344 could be OFF, but they have different internal resistances when OFF. Chances are that the higher resistance is across anode-cathode of the tube, therefore initially, as a voltage divider, the majority of the voltage will drop across the tube, hopefully (if the circuit has been designed well, this is what will happen, until the LM334 starts regulating). \$\endgroup\$ – Lorenzo Donati -- Codidact.org Dec 21 '17 at 14:50
3
\$\begingroup\$

I'm sure I remember John Brooskie covered this @ tubecad.com in one of his articles. Which is where this schematic came from. Over the years, I've noticed how hard it is to search for something on his site, google is touch and go with the results.

To answer the question, the voltage source is from the cathode. But to further answer why it doesn't explode seeing the 240V at the ccs when you first apply power:

at start up time, when the tubes are cold, the tube is in a near non conductive state, its internal resistance is very high, thus protecting the ccs from the massive current flow that would be there otherwise.

|improve this answer|||||
\$\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.