cathode follower

[Meat Plow]

I know some of you are old enough to answer this

http://www.drtube.com/schematics/ampeg/v4pre-jp.gif

V3 - 12DW7 1/2 pins 1,2,3.

Plate 390vdc
Grid 58vdc
Cath 210vdc

The grid should be 139 vdc.

The input cap (.01) was removed and tested, cathode circuit resistors
checked, output cap (.47) removed and tested. Grid voltage remains the
same with the .01 cap in or out of circuit.

The amp was brought to me with a 12AX7 substituted for the 12DW7.
Is this circuit going to depend on the electrical characteristics of a
12DW7 to get the right voltages? I don't have that tube but have since
ordered one since I can't see any reason for the low grid voltage
other than just the wrong tube in that circuit although a 12AX7 is a
sub.

 

[Geo]

On Thu, 08 Jan 2009 12:39:59 -0500, Meat Plow <meat@petitmorte.net> wrote:

I know some of you are old enough to answer this

http://www.drtube.com/schematics/ampeg/v4pre-jp.gif

V3 - 12DW7 1/2 pins 1,2,3.

Plate 390vdc
Grid 58vdc
Cath 210vdc

The grid should be 139 vdc.

The input cap (.01) was removed and tested, cathode circuit resistors
checked, output cap (.47) removed and tested. Grid voltage remains the
same with the .01 cap in or out of circuit.

What is the voltage at the junction of the 1M, 1k and 47k resistors?
I would have said with 219v across the 1k + 47k that you should get about 206
volts at the junction.
Either your meter is taking current through the 1M or the valve is...
If your meter has input resistance of 1M then you could expect to see 1/2 the
actual value.
--
Geo

 

[bg]

Meat Plow wrote in message <2g23aj.qks.17.2@news.alt.net>...

I know some of you are old enough to answer this

http://www.drtube.com/schematics/ampeg/v4pre-jp.gif

V3 - 12DW7 1/2 pins 1,2,3.

Plate 390vdc
Grid 58vdc
Cath 210vdc

The grid should be 139 vdc.

The input cap (.01) was removed and tested, cathode circuit resistors
checked, output cap (.47) removed and tested. Grid voltage remains the
same with the .01 cap in or out of circuit.

The amp was brought to me with a 12AX7 substituted for the 12DW7.
Is this circuit going to depend on the electrical characteristics of a
12DW7 to get the right voltages? I don't have that tube but have since
ordered one since I can't see any reason for the low grid voltage
other than just the wrong tube in that circuit although a 12AX7 is a
sub.

Remove the .01 and the .047 caps. This will isolate the tube from any
external influence. The bias for the grid is developed across the series 1k
and 47k resistors. Seeing as the grid does not draw current, the 1 meg
resistor drops no voltage. So the grid sits at whatever voltage is at the
junction of 1k and 47k. The cathode draws current and will be positve with
respect to ground. If there needs to be 139 volts from the junction of
1k/47k, then there needs to be about 3ma of current flowing thru the 47k .
The actual cathode to grid bias should be about 3 volts. If the part values
check out correct, then the tube is not drawing the correct current.
bg

 

[bg]

bg wrote in message ...

Meat Plow wrote in message <2g23aj.qks.17.2@news.alt.net>...
I know some of you are old enough to answer this

http://www.drtube.com/schematics/ampeg/v4pre-jp.gif

V3 - 12DW7 1/2 pins 1,2,3.

Plate 390vdc
Grid 58vdc
Cath 210vdc

The grid should be 139 vdc.

The input cap (.01) was removed and tested, cathode circuit resistors
checked, output cap (.47) removed and tested. Grid voltage remains the
same with the .01 cap in or out of circuit.

The amp was brought to me with a 12AX7 substituted for the 12DW7.
Is this circuit going to depend on the electrical characteristics of a
12DW7 to get the right voltages? I don't have that tube but have since
ordered one since I can't see any reason for the low grid voltage
other than just the wrong tube in that circuit although a 12AX7 is a
sub.

Remove the .01 and the .047 caps. This will isolate the tube from any
external influence. The bias for the grid is developed across the series 1k
and 47k resistors. Seeing as the grid does not draw current, the 1 meg
resistor drops no voltage. So the grid sits at whatever voltage is at the
junction of 1k and 47k. The cathode draws current and will be positve with
respect to ground. If there needs to be 139 volts from the junction of
1k/47k, then there needs to be about 3ma of current flowing thru the 47k .
The actual cathode to grid bias should be about 3 volts. If the part values
check out correct, then the tube is not drawing the correct current.
bg

Your figures would indicate that the tube is drawing 4.375ma - I=210v/48k
The voltage at the junction of 1k/47k should measure 205.625v . If there is

no current flowing in the 1 meg resistor, and there shouldn't be, then the
grid would measure 205.625 as well. Obviuosly 58 volts at the grid indicates
that the grid or something is drawing current thru the 1meg resistor(see
below). Perhaps the socket has carbon arcs or it is the voltmeter??
Using a simulator program with a 12ax7a and 390 volts on the plate gives me
these values -
cathode current = 2.18ma
Voltage at 1k/47k junction = grid voltage = 102.5v
Cathode voltage = 104.7v (so the actual grid bias is neg 2.2v with respect
to cathode)
Bear in mind that a 10 meg voltmeter connected to the grid and ground would
severly upset the grid voltage. It would measure about 33 volts. The only
way you can measure grid voltage here is to measure it at the junction of
the 1k/47k resistors.
I would assume that if you measure the voltage across the 47k resistor, use
ohms law, and come up with about 3ma, the circuit is working correctly.
Measure the resistance from grid to ground and it should be 1meg + 47k.
Measure the resistance from grid to plate and it should be infinite.

 

[hr(bob) hofmann@att.net]

On Jan 8, 11:39 am, Meat Plow <m...@petitmorte.net> wrote:

I know some of you are old enough to answer this

http://www.drtube.com/schematics/ampeg/v4pre-jp.gif

V3 - 12DW7 1/2 pins 1,2,3.

Plate 390vdc
Grid   58vdc
Cath  210vdc

The grid should be 139 vdc.

The input cap (.01) was removed and tested, cathode circuit resistors
checked, output cap (.47) removed and tested. Grid voltage remains the
same with the .01 cap in or out of circuit.

The amp was brought to me with a 12AX7 substituted for the 12DW7.
Is this circuit going to depend on the electrical characteristics of a
12DW7 to get the right voltages? I don't have that tube but have since
ordered one since I can't see any reason for the low grid voltage
other than just the wrong tube in that circuit although a 12AX7 is a
sub.

I take it you do not know vacuum tube functions at all, that's what I
grew up on. No knocks, that's just the way things happen.

Any tube will conduct current until the grid is at a somewhat lower
voltage than the cathode, ie the grid is negative with respect to the
cathode. Any meter measuring a high resistance/impedance circuit will
load the circuit and distort the readings. The meter should have an
input resistance of at least 10 times the highest resistance in the
circuit. SInce you circuit shows resistances of 1 megohm, your meter
should have an input resistance of at least 10 megohms on anay scale
youuse to measure the voltage. 20x would be even better. Pin 2
should have a somewhat lower voltage than pin 3, by the ratio of
47/48. The voltage at the junction of the 1K and 47 K resistors in
the cathode should be the same as the voltage measured at Pin 2 unless
the tube is defective or more likely, the voltmeter impedance is
changing the grid voltage.

 

[David]

"hr(bob) hofmann@att.net" <hrhofmann@att.net> wrote in
message
news:cebe5096-e128-4d8c-8644-88af85a82747@p36g2000prp.googlegroups.com...
On Jan 8, 11:39 am, Meat Plow <m...@petitmorte.net> wrote:

I know some of you are old enough to answer this

http://www.drtube.com/schematics/ampeg/v4pre-jp.gif

V3 - 12DW7 1/2 pins 1,2,3.

Plate 390vdc
Grid 58vdc
Cath 210vdc

The grid should be 139 vdc.

The input cap (.01) was removed and tested, cathode
circuit resistors
checked, output cap (.47) removed and tested. Grid voltage
remains the
same with the .01 cap in or out of circuit.

The amp was brought to me with a 12AX7 substituted for the
12DW7.
Is this circuit going to depend on the electrical
characteristics of a
12DW7 to get the right voltages? I don't have that tube
but have since
ordered one since I can't see any reason for the low grid
voltage
other than just the wrong tube in that circuit although a
12AX7 is a
sub.

I take it you do not know vacuum tube functions at all,
that's what I
grew up on. No knocks, that's just the way things happen.

Any tube will conduct current until the grid is at a
somewhat lower
voltage than the cathode, ie the grid is negative with
respect to the
cathode. Any meter measuring a high resistance/impedance
circuit will
load the circuit and distort the readings. The meter should
have an
input resistance of at least 10 times the highest resistance
in the
circuit. SInce you circuit shows resistances of 1 megohm,
your meter
should have an input resistance of at least 10 megohms on
anay scale
youuse to measure the voltage. 20x would be even better.
Pin 2
should have a somewhat lower voltage than pin 3, by the
ratio of
47/48. The voltage at the junction of the 1K and 47 K
resistors in
the cathode should be the same as the voltage measured at
Pin 2 unless
the tube is defective or more likely, the voltmeter
impedance is
changing the grid voltage.

I agree with the above, but these old high value resistors
have a tendency to go way up in value and that would just
exasperate the measurement problem. It may be worth
measuring the 1 meg resistor to see if it still is anything
close to 1 meg.

David

 

[Meat Plow]

On Thu, 8 Jan 2009 20:42:55 -0800 (PST), "hr(bob) hofmann@att.net"
<hrhofmann@att.net>wrote:

On Jan 8, 11:39?am, Meat Plow <m...@petitmorte.net> wrote:
I know some of you are old enough to answer this

http://www.drtube.com/schematics/ampeg/v4pre-jp.gif

V3 - 12DW7 1/2 pins 1,2,3.

Plate 390vdc
Grid ? 58vdc
Cath ?210vdc

The grid should be 139 vdc.

The input cap (.01) was removed and tested, cathode circuit resistors
checked, output cap (.47) removed and tested. Grid voltage remains the
same with the .01 cap in or out of circuit.

The amp was brought to me with a 12AX7 substituted for the 12DW7.
Is this circuit going to depend on the electrical characteristics of a
12DW7 to get the right voltages? I don't have that tube but have since
ordered one since I can't see any reason for the low grid voltage
other than just the wrong tube in that circuit although a 12AX7 is a
sub.

I take it you do not know vacuum tube functions at all, that's what I
grew up on. No knocks, that's just the way things happen.

Uh well yes but just not with a cathode follower/splitter.

Any tube will conduct current until the grid is at a somewhat lower
voltage than the cathode, ie the grid is negative with respect to the
cathode. Any meter measuring a high resistance/impedance circuit will
load the circuit and distort the readings. The meter should have an
input resistance of at least 10 times the highest resistance in the
circuit. SInce you circuit shows resistances of 1 megohm, your meter
should have an input resistance of at least 10 megohms on anay scale
youuse to measure the voltage. 20x would be even better. Pin 2
should have a somewhat lower voltage than pin 3, by the ratio of
47/48. The voltage at the junction of the 1K and 47 K resistors in
the cathode should be the same as the voltage measured at Pin 2 unless
the tube is defective or more likely, the voltmeter impedance is
changing the grid voltage.

The meter is a Fluke 77 so I would expect there to be at least 1 meg
on the probe without checking to be sure.

 

[Meat Plow]

On Fri, 9 Jan 2009 08:13:14 -0600, "David"
<someone@some-where.com>wrote:

I agree with the above, but these old high value resistors
have a tendency to go way up in value and that would just
exasperate the measurement problem. It may be worth
measuring the 1 meg resistor to see if it still is anything
close to 1 meg.

Yeah the resistors are all within +- 10%.

 

[Samuel M. Goldwasser]

"hr(bob) hofmann@att.net" <hrhofmann@att.net> writes:

On Jan 8, 11:39 am, Meat Plow <m...@petitmorte.net> wrote:
I know some of you are old enough to answer this

http://www.drtube.com/schematics/ampeg/v4pre-jp.gif

V3 - 12DW7 1/2 pins 1,2,3.

Plate 390vdc
Grid   58vdc
Cath  210vdc

The grid should be 139 vdc.

The input cap (.01) was removed and tested, cathode circuit resistors
checked, output cap (.47) removed and tested. Grid voltage remains the
same with the .01 cap in or out of circuit.

The amp was brought to me with a 12AX7 substituted for the 12DW7.
Is this circuit going to depend on the electrical characteristics of a
12DW7 to get the right voltages? I don't have that tube but have since
ordered one since I can't see any reason for the low grid voltage
other than just the wrong tube in that circuit although a 12AX7 is a
sub.

I take it you do not know vacuum tube functions at all, that's what I
grew up on. No knocks, that's just the way things happen.

Any tube will conduct current until the grid is at a somewhat lower
voltage than the cathode, ie the grid is negative with respect to the
cathode. Any meter measuring a high resistance/impedance circuit will
load the circuit and distort the readings. The meter should have an
input resistance of at least 10 times the highest resistance in the
circuit. SInce you circuit shows resistances of 1 megohm, your meter
should have an input resistance of at least 10 megohms on anay scale
youuse to measure the voltage. 20x would be even better. Pin 2
should have a somewhat lower voltage than pin 3, by the ratio of
47/48. The voltage at the junction of the 1K and 47 K resistors in
the cathode should be the same as the voltage measured at Pin 2 unless
the tube is defective or more likely, the voltmeter impedance is
changing the grid voltage.

Think: Depletion mode MOSFET.

--
sam | Sci.Electronics.Repair FAQ: http://www.repairfaq.org/
Repair | Main Table of Contents: http://www.repairfaq.org/REPAIR/
+Lasers | Sam's Laser FAQ: http://www.repairfaq.org/sam/lasersam.htm
| Mirror Sites: http://www.repairfaq.org/REPAIR/F_mirror.html

Important: Anything sent to the email address in the message header above is
ignored unless my full name AND either lasers or electronics is included in the
subject line. Or, you can contact me via the Feedback Form in the FAQs.

 

[hr(bob) hofmann@att.net]

On Jan 9, 6:57 pm, s...@repairfaq.org (Samuel M. Goldwasser) wrote:

"hr(bob) hofm...@att.net" <hrhofm...@att.net> writes:
On Jan 8, 11:39 am, Meat Plow <m...@petitmorte.net> wrote:
I know some of you are old enough to answer this

http://www.drtube.com/schematics/ampeg/v4pre-jp.gif

V3 - 12DW7 1/2 pins 1,2,3.

Plate 390vdc
Grid   58vdc
Cath  210vdc

The grid should be 139 vdc.

The input cap (.01) was removed and tested, cathode circuit resistors
checked, output cap (.47) removed and tested. Grid voltage remains the
same with the .01 cap in or out of circuit.

The amp was brought to me with a 12AX7 substituted for the 12DW7.
Is this circuit going to depend on the electrical characteristics of a
12DW7 to get the right voltages? I don't have that tube but have since
ordered one since I can't see any reason for the low grid voltage
other than just the wrong tube in that circuit although a 12AX7 is a
sub.

I take it you do not know vacuum tube functions at all, that's what I
grew up on.  No knocks, that's just the way things happen.

Any tube will conduct current until the grid is at a somewhat lower
voltage than the cathode, ie the grid is negative with respect to the
cathode.  Any meter measuring a high resistance/impedance circuit will
load the circuit and distort the readings.  The meter should have an
input resistance of at least 10 times the highest resistance in the
circuit.  SInce you circuit shows resistances of 1 megohm, your meter
should have an input resistance of at least 10 megohms on anay scale
youuse to measure the voltage.  20x would be even better.  Pin 2
should have a somewhat lower voltage than pin 3, by the ratio of
47/48.  The voltage at the junction of the 1K and 47 K resistors in
the cathode should be the same as the voltage measured at Pin 2 unless
the tube is defective or more likely, the voltmeter impedance is
changing the grid voltage.

Think: Depletion mode MOSFET.

--
    sam | Sci.Electronics.Repair FAQ:http://www.repairfaq.org/
 Repair | Main Table of Contents:http://www.repairfaq.org/REPAIR/
+Lasers | Sam's Laser FAQ:http://www.repairfaq.org/sam/lasersam.htm
        | Mirror Sites:http://www.repairfaq.org/REPAIR/F_mirror.html

Important: Anything sent to the email address in the message header above is
ignored unless my full name AND either lasers or electronics is included in the
subject line.  Or, you can contact me via the Feedback Form in the FAQs..- Hide quoted text -

- Show quoted text -

Now you have to explain that to me, my higher education on hardware
stuff stopped with PNPs and NPNs. Then I worked EMC issues for 40
years before retriring from Bell Labs.

 

[David]

"hr(bob) hofmann@att.net" <hrhofmann@att.net> wrote in
message
news:7fbe3a23-37ef-4bd1-99df-d916f57e5c65@i20g2000prf.googlegroups.com...
On Jan 9, 6:57 pm, s...@repairfaq.org (Samuel M. Goldwasser)
wrote:

"hr(bob) hofm...@att.net" <hrhofm...@att.net> writes:
On Jan 8, 11:39 am, Meat Plow <m...@petitmorte.net
wrote:
I know some of you are old enough to answer this

(snip)
Think: Depletion mode MOSFET.

--
sam | Sci.Electronics.Repair FAQ:http://www.repairfaq.org/
Repair | Main Table of
Contents:http://www.repairfaq.org/REPAIR/
+Lasers | Sam's Laser
FAQ:http://www.repairfaq.org/sam/lasersam.htm
| Mirror
Sites:http://www.repairfaq.org/REPAIR/F_mirror.html

Important: Anything sent to the email address in the
message header above is
ignored unless my full name AND either lasers or
electronics is included in the
subject line. Or, you can contact me via the Feedback Form
in the FAQs.- Hide quoted text -

- Show quoted text -

Now you have to explain that to me, my higher education on
hardware
stuff stopped with PNPs and NPNs. Then I worked EMC issues
for 40
years before retriring from Bell Labs.

Bob,

FET = Field Effect Transistor
JFET = Junction FET. These are depletion mode so a
Gate-Source voltage of the opposite polarity as the Drain
cuts off current just like a tube.
MOSFET = Metal Oxide Semiconductor FET. There come in
enhancement mode where a Gate-Source voltage toward the
drain turns it on. Depletion mode MOSFETs operate more like
a JFET.

David

 

[Jamie]

hr(bob) hofmann@att.net wrote:

On Jan 9, 6:57 pm, s...@repairfaq.org (Samuel M. Goldwasser) wrote:

"hr(bob) hofm...@att.net" <hrhofm...@att.net> writes:

On Jan 8, 11:39 am, Meat Plow <m...@petitmorte.net> wrote:

I know some of you are old enough to answer this

http://www.drtube.com/schematics/ampeg/v4pre-jp.gif

V3 - 12DW7 1/2 pins 1,2,3.

Plate 390vdc
Grid 58vdc
Cath 210vdc

The grid should be 139 vdc.

The input cap (.01) was removed and tested, cathode circuit resistors
checked, output cap (.47) removed and tested. Grid voltage remains the
same with the .01 cap in or out of circuit.

The amp was brought to me with a 12AX7 substituted for the 12DW7.
Is this circuit going to depend on the electrical characteristics of a
12DW7 to get the right voltages? I don't have that tube but have since
ordered one since I can't see any reason for the low grid voltage
other than just the wrong tube in that circuit although a 12AX7 is a
sub.

I take it you do not know vacuum tube functions at all, that's what I
grew up on. No knocks, that's just the way things happen.

Any tube will conduct current until the grid is at a somewhat lower
voltage than the cathode, ie the grid is negative with respect to the
cathode. Any meter measuring a high resistance/impedance circuit will
load the circuit and distort the readings. The meter should have an
input resistance of at least 10 times the highest resistance in the
circuit. SInce you circuit shows resistances of 1 megohm, your meter
should have an input resistance of at least 10 megohms on anay scale
youuse to measure the voltage. 20x would be even better. Pin 2
should have a somewhat lower voltage than pin 3, by the ratio of
47/48. The voltage at the junction of the 1K and 47 K resistors in
the cathode should be the same as the voltage measured at Pin 2 unless
the tube is defective or more likely, the voltmeter impedance is
changing the grid voltage.

Think: Depletion mode MOSFET.

--
sam | Sci.Electronics.Repair FAQ:http://www.repairfaq.org/
Repair | Main Table of Contents:http://www.repairfaq.org/REPAIR/
+Lasers | Sam's Laser FAQ:http://www.repairfaq.org/sam/lasersam.htm
| Mirror Sites:http://www.repairfaq.org/REPAIR/F_mirror.html

Important: Anything sent to the email address in the message header above is
ignored unless my full name AND either lasers or electronics is included in the
subject line. Or, you can contact me via the Feedback Form in the FAQs.- Hide quoted text -

- Show quoted text -


Now you have to explain that to me, my higher education on hardware
stuff stopped with PNPs and NPNs. Then I worked EMC issues for 40
years before retriring from Bell Labs.
A tube and JFET/Depletion mode Mos Fet, are naturally biased in the

on state.
These devices need to be turned off to control them instead of On
like you would logically think.

With a Tube and like the N channel fet, the voltage on the grid/gate
must be lower than that of the cathode/Source to enter the pinch off
state of the devices.

The pinch off voltage is most likely not the same spec as that of the
original tube.. The pinch off voltage = the voltage on the grid that is
lower than the voltage on the cathode, which turns off the tube.

In that circuit, it is most likely acceptable for that value to differ
from the original spec and have it still work with in reason.



http://webpages.charter.net/jamie_5"