# triode exercise

Discussion in 'Electronic Basics' started by Allan Adler, Sep 9, 2003.

I'm having trouble doing simple exercises involving triode tubes.
I'm doing exercises from Kloeffler's book Electron Tubes on my own.
I'd be glad to know of a better book. Meanwhile, here is one problem
I've looked at, which I'll rewrite in my own words.

Page 78, #20. You have a 6AV6 triode. Its grid is connected to ground by
an unknown resistor and is at ground potential. The cathode is connected
to ground by a parallel RC circuit with a 5K resistor and a 1 microf capacitor.
The plate is connected to 300 V by a 145K resistor and connected to ground
by a series RC circuit wherein the 1 microf capacitor is closer to the plate
and the resistor is 100K. No other connection between 300 V and ground is
actually drawn in their circuit. They include a spec sheet for the 6AV6
for use with the problem.

They don't actually say that the grid is at ground potential. They just
say there is no voltage drop across the unknown resistor.

I don't want to get into the specific things the question calls for,
since I'm more interested in the comprehensive analysis of the circuit,
but I'll mention them: static load line, cathode grid-bias line, coordinates
of Q point, grid bias voltage, potentials across the capacitors and what
they will become if the cathode should suddenly burn out.

Regarding the analysis of the circuit: I began by drawing it as follows:

145K
(3)-----^v^v^------(2)-----------------
| | |
| --- Plate
| --- 1mmf (TUBE)
----- | Cathode
--- (4) |
----- 300V | |
--- < |
| > 100K |
| < |
| > |
| | ---^v^v^--- |
| | | 5K | |
------------------(0)-- --(1)
| 1mmf |
----) |----

where I've put some numbers in parentheses to indicate reference points
and where the 1 microf capacitor below (2) is supposed to be electrolytic
like the other one. Reference point (0) is ground. The current from (1)
through the tube is i_b. The current in the same direction through the
5K resistor is i1, the current through the capacitor in parallel with it
is i2, so i_b=i1+i2. The current through the series RC circuit joining
(2) to (0) is i4 and the current from (3) to (2) along the 145 K resistor
is i3, so i_b+i3=i4. I'll let V(a,b) denote the voltage drop from reference
point (a) to reference point (b). The value of V(2,1) is the voltage denoted
e_b in the spec sheets for the tube.

Since the problem tells us no current flows between grid and ground, I simply
left the grid out of the circuit and treated the tube as a black box with
an input (cathode) and one output (plate).

Since the spec sheets focus on i_b, e_b and the grid voltage, it seems
desirable to express everything in the circuit in terms of them. I eventually
did so, but it was necessary to solve a 1st order differential equation for i1
and a 1st order differential equation for i4. This expressed them as
indefinite integrals and I would need to know two boundary conditions in order
to know the constants of integration. Presumably the spec sheet ought to
help with that but I'm a little confused as to how to proceed with it.

Assuming I didn't make any mistakes in the above procedures, I then get
an integro-differential equation for i_b and e_b when I say that V(3,0)
is 300 volts along the various routes from (3) to (0).

Another resource, and I don't know if I am supposed to use it, is Child's
equation, which relates i_b, e_b and the grid voltage e_c within certain
operating limits, namely i_b is proportional to the 3/2 power of (e_b+mu.e_c),
where mu is the amplification factor. If I differentiate Child's equation
and use the fact that e_c is at ground potential, I get a relation between
the derivative of i_b and the derivative of e_b, which should help to simplify
the integro-differential equation, but I haven't used it yet to simplify it,
partly because the integro-differential equation looks ugly and I don't
want to deal with it if I don't really have to, and partly because I believe
that the reader for whom this book was intended was not supposed to think

So, what I would like to know is how people on this newsgroup would analyze
a circuit like this one.

Ignorantly,

****************************************************************************
* *
* Disclaimer: I am a guest and *not* a member of the MIT Artificial *
* Intelligence Lab. My actions and comments do not reflect *
* in any way on MIT. Moreover, I am nowhere near the Boston *
* metropolitan area. *
* *
****************************************************************************

2. ### spam the other guy please not meGuest

: You have a 6AV6 triode....

[snip text]
: They don't actually say that the grid is at ground potential. They just
: say there is no voltage drop across the unknown resistor.

Which means the tube draws no grid current.

: since I'm more interested in the comprehensive analysis of the circuit,
: but I'll mention them: static load line, cathode grid-bias line, coordinates
: of Q point, grid bias voltage, potentials across the capacitors and what
: they will become if the cathode should suddenly burn out.

The AC and DC load lines are different. No mention is made of the output
load and it's location, Anode or Cathode?

With no grid current, the bias voltage should be the Voltage drop across
the cathode resistor.

A burnt out cathode is not a standard term. Do you mean open or shorted
elements/operation..?

: So, what I would like to know is how people on this newsgroup would analyze
: a circuit like this one.

We would answer it based on our education and actual experience.

The result would be the reverse engineered values, and/or the generic
operation based on knowledge of how a triode operates.

skipp
http://sonic.ucdavis.edu  