Re: Question about grid behavior with triodes
- From: Patrick Turner <info@xxxxxxxxxxxxxxxxxx>
- Date: Tue, 02 Oct 2007 16:02:10 GMT
Wessel Dirksen wrote:
On 30 sep, 16:28, Patrick Turner <i...@xxxxxxxxxxxxxxxxxx> wrote:
Wessel Dirksen wrote:
On Sep 30, 5:07 am, Patrick Turner <i...@xxxxxxxxxxxxxxxxxx> wrote:
Eiron wrote:
Patrick Turner wrote:
Wessel Dirksen wrote:So putting 2V rms on the grid of a triode with a fixed bias of 2.4v
Hello RAT friends,
......snip...
may not be such a good idea. Here's a circuit to experiment with -
http://members.lycos.co.uk/fiultra/BatteryT68bis.jpg
--
Eiron.
2Vrms = +/-2.8pkV, and grid current would flow in a tube with fixed bias
= -2.4V.
However, as its done in the above fiultra preamp stage, is OK because
the gain attenuater is before the
417A and probably quite low level output would be wanted, and therefore
applied input voltage
probably will be below 0.2Vrms. The bias of 2.4V suits having a healthy
huge 20mA bias current in the 417A
which is supposed to sound very well.
Patrick Turner.
Thanks for the input everyone. I would like an opinion about the
interstage connection for my 3 stage amp using 6CG7's and KT88's all
in triode:
The first 6CG7 stage is a CCS unbypassed grounded cathode stage
running at about 7mA with -1.9V bias. With a 2VRMS max input. The grid
leak resistor is 1M. I calculate the Zin at 100K, and Zout at 12K. It
has a gain of around 17. This is DC coupled to the next stage at about
100V.
Therefore V1 Rk = 1.9V/7ma = 271 ohms.
The Zout is probably higher than 12k because the effective Ra with CCS
anode dc supply
= Ra + [(µ+1)xRk] = say 10k + 20x271 = 15.4k.
The second 6CG7 stage is a positive PS long tail phase splitter and
driver for the power tubes. The Rl's are adjusted for balance. The
total current through Rk is about 17.5mA. with about -6V bias. Zin is
300K. Zout is 5K.
Without a schematic you have us guessing, but lets assume a 300k R is
used from one LTP grid to the other,
with cap to 0V on one grid, and other taken to V1 anode directly.
The common Rk should of course be a CCS, not the 6k resistor you must
have there.
Then you'd get much easier balance and less thd.
Right. I want to do this! The next step in the learining curve.
If each anode RL average about 14k only, gain in each 1/2 should average
about 12, so
hence oa gain = 6.
Rout of each 1/2 of the LTP is much higher than you say when tested at
each anoide separately,
but anode to anode Rout is what is important, and = 2Ra plus 2RL in
parallel,
or 20k//28k = 12k.
The gain I calculate to be about 6. This is cap
coupled to the power tubes which are cathode biased now at about 70mA.
This sounds good to about 1/2 power.
What were all the actual voltage measurements and distortion
results?????
Then it sounds a wee bit thin,
not bad per se. So I'm thinking. The first stage would be +/- OK then.
What does "+/- OK then" mean? it isn't engineering language.
But would there be alot of grid current driving the second stage? 1V x
17= 17V RMS with a tube biased at -6V? Have I created a monster? If so
what are my options?
What have you measured?
I get a little fearful poking around inside at full output.
Fear is the promoter of safety precautions.
Use a well made 10:1 voltage divider of high input resistance to drop
measured voltages
to less fearful level.
But in
general, distortion rises gradually with power, but freq response
becomes "suddenly" more dependant on speaker load, leading me to
believe that somewhere in the pathway the impedance relationships
change. Thus the interest for grid behavior. I'll have to do more
woodsheding.
Indeed!
And all the voltages and onset of grid current can be well predicted
before you build it.
Perhaps rather than have us analyse your amp, you should do it instead.
Ok, too many newbie questions.
Newbies are capable as i was, they just need to focus and do more to
help themselves,
also like I did.
I'm doing just that most of the time, but sometimes the books don't
help completely. If I look at the curves, each individual stage looks
good enough. But I wondered if the first stage would provide too much
output voltage for the 2nd stage.
Draw the amp stages as a block diagram and with gain and maximum
clipping voltage in the output stage.
Then work backwards to the front of the amp to see what volts are
needed. Design all input stages to
produce more than the needed working signal voltages.
Soon you will see what stage is the first to saturate.
Tube craft is supposed to have been invented by an ageing God Of Triodes
to
prevent Old Guys getting Al Ziemer's problem :-)
Patrick Turner.
.
Thanks so much,
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