Directly heated tube, cathode bias
 

I'm a little confused again.

Setting up a 1624 tube for cathode biasing and also need to create a
center tap for keying to ground.

The book says 610 ohms at my voltage. So would I use a pair of 1200 ohm
resistors in this case?

My logic is since there is not a separate cathode then the two R in
parallel would raise the filament 'cathode' 600 ohms above B-minus.

TIA and Merry Christmas,

Bill

Directly heated tube, cathode bias
 

Bill M wrote:
I'm a little confused again.

Setting up a 1624 tube for cathode biasing and also need to create a
center tap for keying to ground.

The book says 610 ohms at my voltage. So would I use a pair of 1200 ohm
resistors in this case?

My logic is since there is not a separate cathode then the two R in
parallel would raise the filament 'cathode' 600 ohms above B-minus.

TIA and Merry Christmas,

Bill

(Disclaimer -- I haven't done this. It's just knowledge gotten from
lots of books, and lots of solid-state circuit design experience).

No, because all the filament voltage will go to mildly heating up the
resistors, and none will go to wildly heating up the filament.

Use a center-tapped filament transformer, and put the 600 ohm resistor
between its center tap and ground. If necessary, use a separate
transformer just for the 1624s.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" gives you just what it says.
See details at http://www.wescottdesign.com/actfes/actfes.html

Directly heated tube, cathode bias
 

Bill M wrote:
I'm a little confused again.

Setting up a 1624 tube for cathode biasing and also need to create a
center tap for keying to ground.

The book says 610 ohms at my voltage. So would I use a pair of 1200 ohm
resistors in this case?

My logic is since there is not a separate cathode then the two R in
parallel would raise the filament 'cathode' 600 ohms above B-minus.

TIA and Merry Christmas,

Bill
If the tube filament is running on AC then a resistor from the
transformer winding center tap to ground would do the job. If the
filament winding isn't center tapped, then two resistors form the center
tap. Usually two low ohm resistors (say 22 ohms) are used with a third
resistor to ground for the cathode bias from the junction of the two
resistors.
You could just use two larger resistors as you suggest, but in the case
of an audio amplifier how would you bypass it with a capacitor? (use TWO
capacitors).

If the tube filament is running on AC then the problem is compounded
since you might not be able to isolate the filament supply from ground.
Guess battery operated radios used C batteries.

Directly heated tube, cathode bias
 

Tim Wescott wrote:
Bill M wrote:
I'm a little confused again.

Setting up a 1624 tube for cathode biasing and also need to create a
center tap for keying to ground.

The book says 610 ohms at my voltage. So would I use a pair of 1200
ohm resistors in this case?

My logic is since there is not a separate cathode then the two R in
parallel would raise the filament 'cathode' 600 ohms above B-minus.

TIA and Merry Christmas,

Bill

(Disclaimer -- I haven't done this. It's just knowledge gotten from
lots of books, and lots of solid-state circuit design experience).

No, because all the filament voltage will go to mildly heating up the
resistors, and none will go to wildly heating up the filament.

Use a center-tapped filament transformer, and put the 600 ohm resistor
between its center tap and ground. If necessary, use a separate
transformer just for the 1624s.

If his un-centertapped filament winding is ONLY supplying that one tube
and he has each leg going to ground through a 1200 ohm resistor that
would result in a 600 ohm connection to ground. So it should work.
Remember that's 2400 ohms across the filament, it won't suck up much
heater current at all.

Directly heated tube, cathode bias
 

"Bill M" wrote in message
...
I'm a little confused again.

Setting up a 1624 tube for cathode biasing and also need to create a
center tap for keying to ground.

The book says 610 ohms at my voltage. So would I use a pair of 1200 ohm
resistors in this case?



Correct, both resistors would essentially be in parallel, and from
the hot cathode to ground. Using two resistors would help
balance out the 60 cycle hum as an added benefit, which is
what I assume you're trying to accomplish.

Pete

Directly heated tube, cathode bias
 

On Dec 24, 8:58 pm, Bill M wrote:
I'm a little confused again.

Setting up a 1624 tube for cathode biasing and also need to create a
center tap for keying to ground.

The book says 610 ohms at my voltage. So would I use a pair of 1200 ohm
resistors in this case?

My logic is since there is not a separate cathode then the two R in
parallel would raise the filament 'cathode' 600 ohms above B-minus.

TIA and Merry Christmas,

Bill

hey OM:

Exactly and you will need at least 10 watts of resistors to be safe.
Because they also will have to handle plate, grid1, grid2, and the
current they draw across the filament. I suppose if they where wire
wound like 25 watt you could stick a ferrite rod in their hole and
make them your filament chokes also. But if you use carbon comp
resistors then part of your drive will go heating those resistors, so
not much gain going on there.

73 OM
n8zu

Directly heated tube, cathode bias
 

On Dec 25, 12:46 pm, raypsi wrote:
On Dec 24, 8:58 pm, Bill M wrote:

I'm a little confused again.

Setting up a 1624 tube for cathode biasing and also need to create a
center tap for keying to ground.

The book says 610 ohms at my voltage. So would I use a pair of 1200 ohm
resistors in this case?

My logic is since there is not a separate cathode then the two R in
parallel would raise the filament 'cathode' 600 ohms above B-minus.

TIA and Merry Christmas,

Bill

hey OM:

Exactly and you will need at least 10 watts of resistors to be safe.
Because they also will have to handle plate, grid1, grid2, and the
current they draw across the filament. I suppose if they where wire
wound like 25 watt you could stick a ferrite rod in their hole and
make them your filament chokes also. But if you use carbon comp
resistors then part of your drive will go heating those resistors, so
not much gain going on there.

73 OM
n8zu

Well they ain't filament chokes, but in the drive circuit the drive
signal won't go to heating them up if they are chokes. So they will
only be taking about 1 mil of current from the filament circuit.

73
n8zu

Directly heated tube, cathode bias
 

raypsi wrote:
On Dec 25, 12:46 pm, raypsi wrote:
On Dec 24, 8:58 pm, Bill M wrote:

I'm a little confused again.
Setting up a 1624 tube for cathode biasing and also need to create a
center tap for keying to ground.
The book says 610 ohms at my voltage. So would I use a pair of 1200 ohm
resistors in this case?
My logic is since there is not a separate cathode then the two R in
parallel would raise the filament 'cathode' 600 ohms above B-minus.
TIA and Merry Christmas,
Bill
hey OM:

Exactly and you will need at least 10 watts of resistors to be safe.
Because they also will have to handle plate, grid1, grid2, and the
current they draw across the filament. I suppose if they where wire
wound like 25 watt you could stick a ferrite rod in their hole and
make them your filament chokes also. But if you use carbon comp
resistors then part of your drive will go heating those resistors, so
not much gain going on there.

73 OM
n8zu

Well they ain't filament chokes, but in the drive circuit the drive
signal won't go to heating them up if they are chokes. So they will
only be taking about 1 mil of current from the filament circuit.

73
n8zu
One possible problem. If this is an RF amplifier circuit you probably
want to use NON-INDUCTIVE resistors in the circuit or they WILL act as
chokes!

Directly heated tube, cathode bias
 

On Thu, 25 Dec 2008 11:03:48 -0500, ken scharf wrote:

Tim Wescott wrote:
Bill M wrote:
I'm a little confused again.

Setting up a 1624 tube for cathode biasing and also need to create a
center tap for keying to ground.

The book says 610 ohms at my voltage. So would I use a pair of 1200
ohm resistors in this case?

My logic is since there is not a separate cathode then the two R in
parallel would raise the filament 'cathode' 600 ohms above B-minus.

TIA and Merry Christmas,

Bill

(Disclaimer -- I haven't done this. It's just knowledge gotten from
lots of books, and lots of solid-state circuit design experience).

No, because all the filament voltage will go to mildly heating up the
resistors, and none will go to wildly heating up the filament.

Use a center-tapped filament transformer, and put the 600 ohm resistor
between its center tap and ground. If necessary, use a separate
transformer just for the 1624s.

If his un-centertapped filament winding is ONLY supplying that one tube
and he has each leg going to ground through a 1200 ohm resistor that
would result in a 600 ohm connection to ground. So it should work.
Remember that's 2400 ohms across the filament, it won't suck up much
heater current at all.

Sorry. I was visualizing the 1200 ohms each in series with the
transformer, whose center tap is grounded.

Of course the alternate way makes perfect sense.

And here it is too late to ask Santa for more brains for Christmas.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" gives you just what it says.
See details at http://www.wescottdesign.com/actfes/actfes.html

Directly heated tube, cathode bias
 

"ken scharf" wrote in message
.. .
raypsi wrote:
On Dec 25, 12:46 pm, raypsi wrote:
On Dec 24, 8:58 pm, Bill M wrote:

I'm a little confused again.
Setting up a 1624 tube for cathode biasing and also need to create a
center tap for keying to ground.
The book says 610 ohms at my voltage. So would I use a pair of 1200
ohm
resistors in this case?
My logic is since there is not a separate cathode then the two R in
parallel would raise the filament 'cathode' 600 ohms above B-minus.
TIA and Merry Christmas,
Bill
hey OM:

Exactly and you will need at least 10 watts of resistors to be safe.
Because they also will have to handle plate, grid1, grid2, and the
current they draw across the filament. I suppose if they where wire
wound like 25 watt you could stick a ferrite rod in their hole and
make them your filament chokes also. But if you use carbon comp
resistors then part of your drive will go heating those resistors, so
not much gain going on there.

73 OM
n8zu

Well they ain't filament chokes, but in the drive circuit the drive
signal won't go to heating them up if they are chokes. So they will
only be taking about 1 mil of current from the filament circuit.

73
n8zu
One possible problem. If this is an RF amplifier circuit you probably
want to use NON-INDUCTIVE resistors in the circuit or they WILL act as
chokes!

Would that make a difference? I'd also assume there's two RF cathode
bypass caps in there; otherwise the circuit would be extremely
degenerative.

Pete

Directly heated tube, cathode bias
 

Tio Pedro wrote:
"ken scharf" wrote in message
.. .
raypsi wrote:
On Dec 25, 12:46 pm, raypsi wrote:
On Dec 24, 8:58 pm, Bill M wrote:

I'm a little confused again.
Setting up a 1624 tube for cathode biasing and also need to create a
center tap for keying to ground.
The book says 610 ohms at my voltage. So would I use a pair of 1200
ohm
resistors in this case?
My logic is since there is not a separate cathode then the two R in
parallel would raise the filament 'cathode' 600 ohms above B-minus.
TIA and Merry Christmas,
Bill
hey OM:

Exactly and you will need at least 10 watts of resistors to be safe.
Because they also will have to handle plate, grid1, grid2, and the
current they draw across the filament. I suppose if they where wire
wound like 25 watt you could stick a ferrite rod in their hole and
make them your filament chokes also. But if you use carbon comp
resistors then part of your drive will go heating those resistors, so
not much gain going on there.

73 OM
n8zu
Well they ain't filament chokes, but in the drive circuit the drive
signal won't go to heating them up if they are chokes. So they will
only be taking about 1 mil of current from the filament circuit.

73
n8zu
One possible problem. If this is an RF amplifier circuit you probably
want to use NON-INDUCTIVE resistors in the circuit or they WILL act as
chokes!

Would that make a difference? I'd also assume there's two RF cathode
bypass caps in there; otherwise the circuit would be extremely
degenerative.

Pete


So long as the inductance of the resistors doesn't resonate with the
bypass capacitors at some frequency of interest.

Directly heated tube, cathode bias
 

ken scharf wrote:

So long as the inductance of the resistors doesn't resonate with the
bypass capacitors at some frequency of interest.

Thanks for all the tips, guys.

-Bill

Directly heated tube, cathode bias
 

"ken scharf" wrote in message
. ..
Tio Pedro wrote:
"ken scharf" wrote in message
.. .
raypsi wrote:
On Dec 25, 12:46 pm, raypsi wrote:
On Dec 24, 8:58 pm, Bill M wrote:

I'm a little confused again.
Setting up a 1624 tube for cathode biasing and also need to create a
center tap for keying to ground.
The book says 610 ohms at my voltage. So would I use a pair of 1200
ohm
resistors in this case?
My logic is since there is not a separate cathode then the two R in
parallel would raise the filament 'cathode' 600 ohms above B-minus.
TIA and Merry Christmas,
Bill
hey OM:

Exactly and you will need at least 10 watts of resistors to be safe.
Because they also will have to handle plate, grid1, grid2, and the
current they draw across the filament. I suppose if they where wire
wound like 25 watt you could stick a ferrite rod in their hole and
make them your filament chokes also. But if you use carbon comp
resistors then part of your drive will go heating those resistors, so
not much gain going on there.

73 OM
n8zu
Well they ain't filament chokes, but in the drive circuit the drive
signal won't go to heating them up if they are chokes. So they will
only be taking about 1 mil of current from the filament circuit.

73
n8zu
One possible problem. If this is an RF amplifier circuit you probably
want to use NON-INDUCTIVE resistors in the circuit or they WILL act as
chokes!

Would that make a difference? I'd also assume there's two RF cathode
bypass caps in there; otherwise the circuit would be extremely
degenerative.

Pete
So long as the inductance of the resistors doesn't resonate with the
bypass capacitors at some frequency of interest.

But would the impedance ever be lower than the ohmic resistance
for the resistor?

Directly heated tube, cathode bias
 

Tio Pedro wrote:
"ken scharf" wrote in message
. ..
Tio Pedro wrote:
"ken scharf" wrote in message
.. .
raypsi wrote:
On Dec 25, 12:46 pm, raypsi wrote:
On Dec 24, 8:58 pm, Bill M wrote:

I'm a little confused again.
Setting up a 1624 tube for cathode biasing and also need to create a
center tap for keying to ground.
The book says 610 ohms at my voltage. So would I use a pair of 1200
ohm
resistors in this case?
My logic is since there is not a separate cathode then the two R in
parallel would raise the filament 'cathode' 600 ohms above B-minus.
TIA and Merry Christmas,
Bill
hey OM:

Exactly and you will need at least 10 watts of resistors to be safe.
Because they also will have to handle plate, grid1, grid2, and the
current they draw across the filament. I suppose if they where wire
wound like 25 watt you could stick a ferrite rod in their hole and
make them your filament chokes also. But if you use carbon comp
resistors then part of your drive will go heating those resistors, so
not much gain going on there.

73 OM
n8zu
Well they ain't filament chokes, but in the drive circuit the drive
signal won't go to heating them up if they are chokes. So they will
only be taking about 1 mil of current from the filament circuit.

73
n8zu
One possible problem. If this is an RF amplifier circuit you probably
want to use NON-INDUCTIVE resistors in the circuit or they WILL act as
chokes!
Would that make a difference? I'd also assume there's two RF cathode
bypass caps in there; otherwise the circuit would be extremely
degenerative.

Pete
So long as the inductance of the resistors doesn't resonate with the
bypass capacitors at some frequency of interest.

But would the impedance ever be lower than the ohmic resistance
for the resistor?


It would be an inductor with very low Q.

Directly heated tube, cathode bias
 

"ken scharf" wrote in message
.. .
Tio Pedro wrote:
"ken scharf" wrote in message
. ..
Tio Pedro wrote:
"ken scharf" wrote in message
.. .
raypsi wrote:
On Dec 25, 12:46 pm, raypsi wrote:
On Dec 24, 8:58 pm, Bill M wrote:

I'm a little confused again.
Setting up a 1624 tube for cathode biasing and also need to create
a
center tap for keying to ground.
The book says 610 ohms at my voltage. So would I use a pair of
1200 ohm
resistors in this case?
My logic is since there is not a separate cathode then the two R in
parallel would raise the filament 'cathode' 600 ohms above B-minus.
TIA and Merry Christmas,
Bill
hey OM:

Exactly and you will need at least 10 watts of resistors to be safe.
Because they also will have to handle plate, grid1, grid2, and the
current they draw across the filament. I suppose if they where wire
wound like 25 watt you could stick a ferrite rod in their hole and
make them your filament chokes also. But if you use carbon comp
resistors then part of your drive will go heating those resistors,
so
not much gain going on there.

73 OM
n8zu
Well they ain't filament chokes, but in the drive circuit the drive
signal won't go to heating them up if they are chokes. So they will
only be taking about 1 mil of current from the filament circuit.

73
n8zu
One possible problem. If this is an RF amplifier circuit you probably
want to use NON-INDUCTIVE resistors in the circuit or they WILL act as
chokes!
Would that make a difference? I'd also assume there's two RF cathode
bypass caps in there; otherwise the circuit would be extremely
degenerative.

Pete
So long as the inductance of the resistors doesn't resonate with the
bypass capacitors at some frequency of interest.

But would the impedance ever be lower than the ohmic resistance
for the resistor?


It would be an inductor with very low Q.

Yes, but it is still Hi Z, the resistive portion
is in series with the inductive reactance.

Pete