Grid Resistors
 

"Keith G" wrote in message
...

"Iain Churches" wrote in message
i.fi...

"West" wrote


...AND better no pot at all, but if you must, use a step attenuator with
quality resistors.


Hi West.

Better still, no preamp at all:-) With the addition of a stepped
attenuator, one can feed the CD player straight to the power amp.
Morgan Jones states: "No preamplifier is better than any
preamplifier!" I agree with that sentiment.


S'what I did with my 2A3 SET which is effectively a power amp with a
volume control - my thinking was that the 'pre' section could be a
nightmare for a beginner to get right (quiet) and that there are/were any
number of 'ready-made' pre's that could be used if necessary, until such
time I built my own! It has worked so well without a pre I never got round
to building one!

Yes. That's a very good solution. What is ther sense of a preamp
with excess gain which you need to attenuate? I like to have the
gain control at about 2 oclock for monitoring at a "healthy" level



Many take this a step further and add an input selector switch
as well as the SA to the power amp. This makes sense too.


Except that it eliminates the option of a 'single source direct' and
condemns the amp to possibly 'inescapable' crosstalk with the extra
internal wiring....


Never found that to be a problem, unless of course you switch
to other unused and open inputs. Once they are terminated,
one can get a fig of 85dB. If you need, you can use relay
input switching, with the unused inputs just looking into a dummy
load.

Cheers
Iain

Grid Resistors
 

On Thu, 15 Nov 2007 17:35:27 +0000, Nick Gorham
wrote:

Don Pearce wrote:


As a quick experiment I've just tried disconnecting the grid of the second
half of the ECC83 in a Quad QC2 pre-amp that happens to be on my bench at
the moment. With 280V HT the anode voltage is normally 235V (50K anode load,
2.2K cathode bias resistor), with the grid disconnected it drops to 160V.
Trying the same trick with the first half (470K anode load, 4.7K cathode
resistor) only results in the anode voltage dropping from 114V to 105V.

David.



OK - that is about what I would expect. The ECC83 has an internal
resistance of 80kohm at 100V, so that is about as close to saturated
as you could expect it to get. I suspect that if you could actually
remove the grid wire, the voltages would not have come out far
different.

d


I think you are confusing the rp with Rp, looking at the diode line

http://www.mif.pg.gda.pl/homepages/f...37/1/12AX7.pdf

I think its more like 46k, but the numbers still make sense,

Quite so. Something about 40k looks like the absolute lower limit.

160v across the diode line would expect about 3.4ma, you are getting a
bit less (280-160)/50000 = 3.2ma, so again near enough, maybe its
biasing itself off a little.

So any charge collected by the grid is doing next to nothing by way of
bias.

d

--
Pearce Consulting
http://www.pearce.uk.com

Grid Resistors
 

"Iain Churches" wrote


Yes. That's a very good solution. What is ther sense of a preamp
with excess gain which you need to attenuate? I like to have the
gain control at about 2 oclock for monitoring at a "healthy" level


12 o' clock is my psychological best for 'starting to get loud'!! :-)




Many take this a step further and add an input selector switch
as well as the SA to the power amp. This makes sense too.


Except that it eliminates the option of a 'single source direct' and
condemns the amp to possibly 'inescapable' crosstalk with the extra
internal wiring....


Never found that to be a problem, unless of course you switch
to other unused and open inputs.


More of a problem when more than one source is running, of course.


Once they are terminated,
one can get a fig of 85dB. If you need, you can use relay
input switching, with the unused inputs just looking into a dummy
load.


Or simple toggle switches, in the case of valve amps - very satisfying
to use!

Grid Resistors
 

Keith G wrote:
"Iain Churches" wrote in message
i.fi...
"West" wrote


...AND better no pot at all, but if you must, use a step attenuator
with
quality resistors.

Hi West.

Better still, no preamp at all:-) With the addition of a stepped
attenuator, one can feed the CD player straight to the power amp.
Morgan Jones states: "No preamplifier is better than any
preamplifier!" I agree with that sentiment.


S'what I did with my 2A3 SET which is effectively a power amp with a
volume control - my thinking was that the 'pre' section could be a
nightmare for a beginner to get right (quiet) and that there are/were
any number of 'ready-made' pre's that could be used if necessary, until
such time I built my own! It has worked so well without a pre I never
got round to building one!

Keith,

Exactly what I do as well.


Many take this a step further and add an input selector switch
as well as the SA to the power amp. This makes sense too.


Except that it eliminates the option of a 'single source direct' and
condemns the amp to possibly 'inescapable' crosstalk with the extra
internal wiring....

I do this with a separate shielded cable from each input jack to the
selector switch terminal, all grounds connected at the input jack end
only. I have had no crosstalk issues.

Raymond

Grid Resistors
 

"Nick Gorham" wrote in message
...

Yes, except, you don't actually need a cap, as long as grid current
generated across the grid resistor generates the required voltage.

Sorry no, the cap is an important part of the circuit. Firstly it charges to
the signal peak, meaning that grid current only flows on signal peaks, not
all the time. Secondly it connects the signal to the grid without providing
a DC shunt path for the high-value grid resistor.

David.

Grid Resistors
 

David Looser wrote:
"Nick Gorham" wrote in message
...

Yes, except, you don't actually need a cap, as long as grid current
generated across the grid resistor generates the required voltage.


Sorry no, the cap is an important part of the circuit. Firstly it charges to
the signal peak, meaning that grid current only flows on signal peaks, not
all the time. Secondly it connects the signal to the grid without providing
a DC shunt path for the high-value grid resistor.

David.



Ok, fair enough, but what do you think would happen without an input
signal or any other connection other than the high resistance connection
between the grid and cathode circuit.

I agree that practically it needs to be there to prevent the grid being
pulled to ground, but I think you will find the idea of charging it up
to signal "peaks" is just a bit strange. Whats going to charge up, the
grid side of the cap is referenced to ground, and will it all go wrong
after a period of silence?

We could play, hunt the hidden cap, can you say miller?

--
Nick

Grid Resistors
 

Flipper wrote:

Oops. My mistake. I meant "least negative," not most
negative. In
particular, it'll bias to the zero grid current point,
since there's
no grid current.

Right, seems a good point. I would expect it to be
determined by the thermionic relationship between grid and
cathode, more or less regardless of the anode current.

The reason I asked was that, considering the amp will stop
working, it seems unlikely that any damage would be done as
a result of the pot failing. Unless it's a power stage, in
most cases the valve would tolerate the extra current for a
long time, and it should be unlikely that resistors will get
too hot. The main thing to worry about if the pot fails is
the silence.

cheers, Ian

Grid Resistors
 

Ian Iveson wrote:
Flipper wrote:


Oops. My mistake. I meant "least negative," not most
negative. In
particular, it'll bias to the zero grid current point,
since there's
no grid current.


Right, seems a good point. I would expect it to be
determined by the thermionic relationship between grid and
cathode, more or less regardless of the anode current.

The reason I asked was that, considering the amp will stop
working, it seems unlikely that any damage would be done as
a result of the pot failing. Unless it's a power stage, in
most cases the valve would tolerate the extra current for a
long time, and it should be unlikely that resistors will get
too hot. The main thing to worry about if the pot fails is
the silence.

cheers, Ian



Just FYI, I made a similar mistake breadboarding a 2a3 output stage, I
forgot to reference the grid to the cathode circuit, the current was
double what I expected, so that would have left me with a glowing anode
if I had not noticed. There isn't the anode load in that case to help
out with the DC current.

I agree though, in normal use, the pot will go to a small signal valve,
and the sky won't fall.

--
Nick

Grid Resistors
 

"Nick Gorham" wrote in message
...

I agree that practically it needs to be there to prevent the grid being
pulled to ground, but I think you will find the idea of charging it up to
signal "peaks" is just a bit strange.

I don't find it remotely strange.

Whats going to charge up,

the capacitor.

the grid side of the cap is referenced to ground, and will it all go wrong
after a period of silence?

During silence the charge will leak away via the grid resistor.

With no-signal the grid is at chassis potential and the valve is operating
under zero-bias conditions. The capacitor is charged to the whatever voltage
is on the other end of it which maybe quite high if it comes from the anode
of a previous valve, or it might be zero if direct from a volume control;
let's assume the later.

Now apply a signal, as the signal swings positive the grid-cathode diode is
forward biased thus dramatically reducing the input resistance of the valve,
the capacitor now starts to charge up, aiming for the signal voltage and
with a time constant set by the value of the capacitor and the total circuit
resistance (source resistance plus valve grid-cathode resistance). Once the
signal has passed it's peak value the grid voltage starts to fall, turning
off the grid-cathode diode, thus significantly increasing the time constant.
The charge on the capacitor that it acquired during the positive half cycle
is thus retained creating a small amount of negative bias for the valve.
This charge will leak away slowly via the high value grid leak so that the
charge on the capacitor, and thus the negative bias, roughly follows the
envelope of the signal

The thing is analogous to the valve TV sync separator where positive-going
sync pulses drive the valve into grid-current forcing the coupling capacitor
to charge to the instantaneous anode voltage of the preceding video
amplifier. The difference here is that the signal amplitude is high enough
to cause cut-off of the sync separator valve during the subsequent picture
portion of the video waveform.

David.

Grid Resistors
 

On Fri, 16 Nov 2007 10:18:13 -0000, "David Looser"
wrote:

I agree that practically it needs to be there to prevent the grid being
pulled to ground, but I think you will find the idea of charging it up to
signal "peaks" is just a bit strange.

I don't find it remotely strange.

Whats going to charge up,

the capacitor.

It is actually a very standard technique called clamping. It was used
in televisions to keep black levels constant. It isn't the best way to
do things, because the level will drift around a bit with signal
amplitude, but when component count is important and volts are scarce
it is certainly good enough.

d
--
Pearce Consulting
http://www.pearce.uk.com