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Resistance of ammeter caused voltage drop

D

Dummy

Jan 1, 1970
0
I have a simple setup as below. The ammeter is used to measure the
transmit current of the radio. Power supply has an display of total
current as well.

Power supply (Vin=7.5V)-------ammeter--------- (Vout) radio

When connecting the ammeter, there's a 0.3V voltage drop across the
ammeter before transmitting. During transmit mode, total current as
displayed at ammeter is 1.80A, matched with the power supply current
display. But Vout is measured to be 6.14V only, thus causing the Tx
power to be lower.

Using direct cable connection without ammeter, the current measured is
almost similar, but the Tx power is much more higher. Current is
1.85A. Vout is 7.0V during transmission - a voltage drop of 0.5V.
I guess the ammeter is giving more resistance the the cable.

So I was wondering whether the ammeter should be used to measure any
high current in circuit level if it can cause some voltage drop. The
voltage drop might affect the circuit performance at the subsequent
stage.
 
T

Tim Wescott

Jan 1, 1970
0
Dummy said:
I have a simple setup as below. The ammeter is used to measure the
transmit current of the radio. Power supply has an display of total
current as well.

Power supply (Vin=7.5V)-------ammeter--------- (Vout) radio

When connecting the ammeter, there's a 0.3V voltage drop across the
ammeter before transmitting. During transmit mode, total current as
displayed at ammeter is 1.80A, matched with the power supply current
display. But Vout is measured to be 6.14V only, thus causing the Tx
power to be lower.

Using direct cable connection without ammeter, the current measured is
almost similar, but the Tx power is much more higher. Current is
1.85A. Vout is 7.0V during transmission - a voltage drop of 0.5V.
I guess the ammeter is giving more resistance the the cable.

So I was wondering whether the ammeter should be used to measure any
high current in circuit level if it can cause some voltage drop. The
voltage drop might affect the circuit performance at the subsequent
stage.

I'd look for a better ammeter. My old Heathkit electro-mechanical unit
only drops 1/4 volt at full scale -- I would imagine that a Fluke or
other serious DVM would be better (but I haven't checked, so I can't
guarantee it).
 
L

Larry Brasfield

Jan 1, 1970
0
Dummy said:
I have a simple setup as below. The ammeter is used to measure the
transmit current of the radio. Power supply has an display of total
current as well.

Power supply (Vin=7.5V)-------ammeter--------- (Vout) radio

When connecting the ammeter, there's a 0.3V voltage drop across the
ammeter before transmitting. During transmit mode, total current as
displayed at ammeter is 1.80A, matched with the power supply current
display. But Vout is measured to be 6.14V only, thus causing the Tx
power to be lower.

Using direct cable connection without ammeter, the current measured is
almost similar, but the Tx power is much more higher. Current is
1.85A. Vout is 7.0V during transmission - a voltage drop of 0.5V.
I guess the ammeter is giving more resistance the the cable.

It certainly looks that way.
So I was wondering whether the ammeter should be used to measure any
high current in circuit level if it can cause some voltage drop.

That would depend on how badly you want to measure
the current and what effect the voltage drop has. I am
going to guess about those things for now.

Assumption: The transmitter has a linear relationship
between supply voltage and supply current draw.

Assumption: The transmitter does not change its mode
of operation at the reduced voltage you see when the
ammeter is connected. (This is nearly but not quite
tautological with the linearity assumption. But it is
something you can verify independently when the
ammeter is not present.)

Assumption: You have a way, (such as a different cable,
a variable bench supply, or some low Ohm resistors), to
vary the supply with the ammeter connected.

Assumption: Your ammeter is a better instrument than
the meter built into the supply. (Otherwise I do not
know why we would be having this discussion.)

Measure the current with ammeter in at the ordinary
supply voltage. Call this Iao. Measure the voltage
at the transmitter with same lashup. Call this Vao.
Measure the drop across the ammeter, Vad.

Reduce the supply voltage by an amount similar
to what the ammeter drops, leaving the ammeter
in place. Measure current and voltage, to be
called Iar and Var respectively.

Calculate Rt = (Vao - Var) / (Iao - Iar)
This is the slope of the voltage versus current
characteristic for the transmitter.

Calculate Ina = Iao + Rt * Vad
This is an approximation of the current the
transmitter draws when you have no ammeter
to reduce the supply voltage it sees.
The voltage drop might affect the circuit performance at the subsequent
stage.

Yes, it probably does. It would help, when deciding
what to do about this, what you are trying to achieve
by measuring the current. Is this a one-time affair,
or will the ammeter become part of the setup? How
accurately do you need to know the current? (My
guess is that this does not matter much.)

There are current meters that impose no DC drop.
You could rent (or buy) one if you believe the
above procedure is too much trouble or not
sufficiently accurate. I do not advise this unless
you have more need for accuracy than I can see.
You could also put a lower shunt resistor across
your ammeter and calibrate the combination.
 
R

Robert Baer

Jan 1, 1970
0
Larry said:
It certainly looks that way.




That would depend on how badly you want to measure
the current and what effect the voltage drop has. I am
going to guess about those things for now.

Assumption: The transmitter has a linear relationship
between supply voltage and supply current draw.

Assumption: The transmitter does not change its mode
of operation at the reduced voltage you see when the
ammeter is connected. (This is nearly but not quite
tautological with the linearity assumption. But it is
something you can verify independently when the
ammeter is not present.)

Assumption: You have a way, (such as a different cable,
a variable bench supply, or some low Ohm resistors), to
vary the supply with the ammeter connected.

Assumption: Your ammeter is a better instrument than
the meter built into the supply. (Otherwise I do not
know why we would be having this discussion.)

Measure the current with ammeter in at the ordinary
supply voltage. Call this Iao. Measure the voltage
at the transmitter with same lashup. Call this Vao.
Measure the drop across the ammeter, Vad.

Reduce the supply voltage by an amount similar
to what the ammeter drops, leaving the ammeter
in place. Measure current and voltage, to be
called Iar and Var respectively.

Calculate Rt = (Vao - Var) / (Iao - Iar)
This is the slope of the voltage versus current
characteristic for the transmitter.

Calculate Ina = Iao + Rt * Vad
This is an approximation of the current the
transmitter draws when you have no ammeter
to reduce the supply voltage it sees.




Yes, it probably does. It would help, when deciding
what to do about this, what you are trying to achieve
by measuring the current. Is this a one-time affair,
or will the ammeter become part of the setup? How
accurately do you need to know the current? (My
guess is that this does not matter much.)

There are current meters that impose no DC drop.
You could rent (or buy) one if you believe the
above procedure is too much trouble or not
sufficiently accurate. I do not advise this unless
you have more need for accuracy than I can see.
You could also put a lower shunt resistor across
your ammeter and calibrate the combination.
The shunt scheme would seem to be a good idea.
Say one gets an 0.1 ohm 1% resistor and uses kelvin connections to
read the voltage drop across it with a DVM (lowest full scale range is
200mV).
Then at one amp, the voltage drop would be 100mV; 3 times better than
the presumed 300mV cited (which is not that much higher than "1/4 V).
BTW, i would put the resistor on the ground side...
 
J

Jamie

Jan 1, 1970
0
Dummy said:
I have a simple setup as below. The ammeter is used to measure the
transmit current of the radio. Power supply has an display of total
current as well.

Power supply (Vin=7.5V)-------ammeter--------- (Vout) radio

When connecting the ammeter, there's a 0.3V voltage drop across the
ammeter before transmitting. During transmit mode, total current as
displayed at ammeter is 1.80A, matched with the power supply current
display. But Vout is measured to be 6.14V only, thus causing the Tx
power to be lower.

Using direct cable connection without ammeter, the current measured is
almost similar, but the Tx power is much more higher. Current is
1.85A. Vout is 7.0V during transmission - a voltage drop of 0.5V.
I guess the ammeter is giving more resistance the the cable.

So I was wondering whether the ammeter should be used to measure any
high current in circuit level if it can cause some voltage drop. The
voltage drop might affect the circuit performance at the subsequent
stage.
if you want to experiment a bit, you could use an OP-Amp inputs with a
shunt ( very low value shunt). the Op-Amp can be an Amp for a simple
meter.
of course there is more that you need to do, this is just an idea for
you to ponder with.
 
J

JeffM

Jan 1, 1970
0
I was wondering whether the ammeter should be used to measure...
make sure your connections are all tight
Mark

Why are you responding to me?
Go up 3 levels in the thread and respond to the OP.

(Clueless Google posters are the bane of Usenet.)
 
K

Ken Taylor

Jan 1, 1970
0
Dummy said:
I have a simple setup as below. The ammeter is used to measure the
transmit current of the radio. Power supply has an display of total
current as well.

Power supply (Vin=7.5V)-------ammeter--------- (Vout) radio

When connecting the ammeter, there's a 0.3V voltage drop across the
ammeter before transmitting. During transmit mode, total current as
displayed at ammeter is 1.80A, matched with the power supply current
display. But Vout is measured to be 6.14V only, thus causing the Tx
power to be lower.

Using direct cable connection without ammeter, the current measured is
almost similar, but the Tx power is much more higher. Current is
1.85A. Vout is 7.0V during transmission - a voltage drop of 0.5V.
I guess the ammeter is giving more resistance the the cable.

So I was wondering whether the ammeter should be used to measure any
high current in circuit level if it can cause some voltage drop. The
voltage drop might affect the circuit performance at the subsequent
stage.

You could buy a shunt and use a voltmeter (on a suitably low range) to
measure the drop across it.

Ken
 
J

John Fields

Jan 1, 1970
0
I have a simple setup as below. The ammeter is used to measure the
transmit current of the radio. Power supply has an display of total
current as well.

Power supply (Vin=7.5V)-------ammeter--------- (Vout) radio

When connecting the ammeter, there's a 0.3V voltage drop across the
ammeter before transmitting. During transmit mode, total current as
displayed at ammeter is 1.80A, matched with the power supply current
display. But Vout is measured to be 6.14V only, thus causing the Tx
power to be lower.

Using direct cable connection without ammeter, the current measured is
almost similar, but the Tx power is much more higher. Current is
1.85A. Vout is 7.0V during transmission - a voltage drop of 0.5V.
I guess the ammeter is giving more resistance the the cable.

So I was wondering whether the ammeter should be used to measure any
high current in circuit level if it can cause some voltage drop. The
voltage drop might affect the circuit performance at the subsequent
stage.

---
so transmit, with the ammeter in the circuit, looks like this:

1.8A--->
Power supply (Vin=7.5V)-------ammeter--------- (Vout=6.14V) radio


Which makes the total (ammeter + cable) resistance look like:

Vin - Vout 1.36
Rt = ------------ = ------ ~ 0.756 ohms
I 1.8


With the ammeter out of the circuit, the cable resistance looks like:

Vin - Vout 0.5
Rc = ------------ = ------ ~ 0.270 ohms
I 1.85


which makes the ammeter resistance look like:


Ra = Rt - Rc = 0.756 - 0.270 = 0.476 ohms


I have a Fluke 8060A which measures 0.468 ohms on the 2 amp scale, so
0.476 ohms for Ra doesn't seem far-fetched if your ammeter is similar,
but 0.270 ohms seems awfully high for the cable, what with 100 feet of
#20 being about an ohm. Just to make sure that's not a problem I'd
check out all the wiring interfaces and make sure you don't have any
high-resistance connections anywhere.

Bottom line though, after you get everything cleaned up and sorted
out, if you know that what you need to make your measurements is 7.5V
at the radio, put your ammeter in the line if that's what you need to
do, then monitor the voltage at the radio's power input, then key the
transmitter and adjust the power supply until the voltage at the
radio's power input is 7.5V. If it goes up to 8V when you stop
transmitting, so what? You can always drop it back to 7.5 if you need
to work on the receiver, although I doubt that you'll need to, since
I'm sure the receiver's not running unregulated.
 
F

Fred Bloggs

Jan 1, 1970
0
Larry said:
Measure the current with ammeter in at the ordinary
supply voltage. Call this Iao. Measure the voltage
at the transmitter with same lashup. Call this Vao.
Measure the drop across the ammeter, Vad.

Reduce the supply voltage by an amount similar
to what the ammeter drops, leaving the ammeter
in place. Measure current and voltage, to be
called Iar and Var respectively.

Calculate Rt = (Vao - Var) / (Iao - Iar)
This is the slope of the voltage versus current
characteristic for the transmitter.

Calculate Ina = Iao + Rt * Vad
This is an approximation of the current the
transmitter draws when you have no ammeter
to reduce the supply voltage it sees.

Yeah? So the F___ what? And since when is resistance x voltage equal a
current?- and you don't even have the sign right. Here's the deal you
worthless, pretentious son-of-a-bitch-with-VD, you are a worthless
p.o.s.- we are wise to your dumb ass- go away.

You could also put a lower shunt resistor across
your ammeter and calibrate the combination.

Or buy/modify a p.s. with external sense compensation, damned worthless
idiot.
 
L

Larry Brasfield

Jan 1, 1970
0
Fred Bloggs said:
Yeah? So the F___ what? And since when is resistance x voltage equal a current?- and you don't even have the sign right.

(Finally, a positive contribution.)

That line should have read, of course, thusly:
Calculate Ina = Iao + Vad / Rt
And, contrary to what the esteemed Mr. Boggs
proclaims, the sign is correct. As defined, both
Vad and Rt are positive quantities (at least if
the transmitter draws more power at higher
voltages, which is already in evidence.) Since
Ina (pronounce as: I 'n'o 'a'mmeter) represents
the current predicted when no ammeter is present,
and that is already known to be higher, adding the
positive ratio Vad/Rt to the current measure with
the ammeter present is correct for getting such a
result. I would hope that this is now obvious
even to the most critical observer.
Here's the deal you worthless, pretentious son-of-a-bitch-with-VD, you are a worthless p.o.s.- we are wise to your dumb ass-

I perhaps should engage in some name-calling on
account of the above correction, but, as a low ranking
member of the excrement class, I am not up to it.

I already said 'No', Fred. Do you think repetition
is going to be effective? (It would appear so.)
Or buy/modify a p.s. with external sense compensation,

Spending money was an obvious option which
I mentioned in several of its many forms. The
OP's questions led me to believe he might be
interested in using the equipment he had. We
have seen no evidence to the contrary.
damned worthless idiot.

Fred, I appreciate your opinion. Honestly.
I tried to tell you that earlier, but I suspect
my meaning escaped your notice.
 
R

Robert Baer

Jan 1, 1970
0
Jamie said:
if you want to experiment a bit, you could use an OP-Amp inputs with a
shunt ( very low value shunt). the Op-Amp can be an Amp for a simple
meter.
of course there is more that you need to do, this is just an idea for
you to ponder with.

Again, pay attention and use kelvin connections.
 
R

Robert Baer

Jan 1, 1970
0
Ken said:
You could buy a shunt and use a voltmeter (on a suitably low range) to
measure the drop across it.

Ken

I think i said that, and added a cautionary note to use kelvin
connections.
 
J

John Fields

Jan 1, 1970
0
Spending money was an obvious option which
I mentioned in several of its many forms.

---
However, you never mentioned external sense compensation (using a
Kelvin connection at the load to supply feedback to the supply in
order to compensate for lead resistance) and had you known such a
thing existed you would surely have mentioned it as an "obvious
option".

Now that the cat's out of the bag, though, I suspect you'll soon
become the expert you'd like us to believe you already were.

Here, I'll save you a little time on Google:

On power supplies supplied with external sense compensation there are
two terminals, one usually marked "+ sense" or something like that,
and the other one marked "- sense" or something like that. In use, a
wire is connected from the "+ sense" terminal to the + input of the
load at the same point the supply lead is connected to the load, and
the "- sense" terminal is connected from the "- sense" terminal to the
- input of the load at the same point the supply lead is connected to
the load. That way, voltage variations _at the load_ are sensed and
fed back to the supply where the supply voltage is automatically
adjusted upward to compensate for the voltage dropped across the
supply leads. If sense compensation isn't needed, the sense terminals
are shorted to their respective supply outputs at the supply, and the
supply regulates the voltage at that point.

You're welcome.
 
L

Larry Brasfield

Jan 1, 1970
0
John Fields said:
---
However, you never mentioned external sense compensation (using a
Kelvin connection at the load to supply feedback to the supply in
order to compensate for lead resistance) and had you known such a
thing existed you would surely have mentioned it as an "obvious
option".

There are two silly assumptions you've made. As I have
stated elsewhere, I thought the OP would like a solution
utilizing what he mentioned he had on hand. I threw out
a few spending options without pretending to exhaust
them, merely to let him know he was not stuck with
just what he had. So I was not inclined to spend much
time trying to come up with a list that none of the smart
alecs around here would be able to "improve" upon.
So, assuming that my non-mention reflects ignorance
of remote sense power supplies is fatuous. The other
laughable assumption is that it should be an obvious
option to anybody who knew of such supplies. Why
should the OP go spend that kind of money when he
can simply use a shunt or measure the resistance of his
cable and use that and a voltmeter to measure current?
Now that the cat's out of the bag, though, I suspect you'll soon
become the expert you'd like us to believe you already were.

Here, I'll save you a little time on Google:

On power supplies supplied with external sense compensation there are
two terminals, one usually marked "+ sense" or something like that,
and the other one marked "- sense" or something like that. In use, a
wire is connected from the "+ sense" terminal to the + input of the
load at the same point the supply lead is connected to the load, and
the "- sense" terminal is connected from the "- sense" terminal to the
- input of the load at the same point the supply lead is connected to
the load. That way, voltage variations _at the load_ are sensed and
fed back to the supply where the supply voltage is automatically
adjusted upward to compensate for the voltage dropped across the
supply leads. If sense compensation isn't needed, the sense terminals
are shorted to their respective supply outputs at the supply, and the
supply regulates the voltage at that point.

Thanks, so much John.
You're welcome.

I note that your little description omits mention of the
1k or so resistors that normally obviate the need for
those jumpers when remote sensing is not used.

Not knowing how old you are, I may be actually wrong
about this, but there is a good chance that the incident
I relate below happened before you had any inkling of
what a circuit is or what 'electronics' means.

Before my job as an engineering tech which preceeded
my career as an electronics design engineer, I held a
job as a factory test tech. One day, in return for a
similar level of prank, I connected an RC network
between the so far unused sense terminals and the
output terminals on the power supply that my "pal"
would be using after lunch to continue troubleshooting
some equipment. (These machines were battery
operated but run off of a DC supply during most test
and troubleshooting.) After enjoying the spectacle of
him trying to figure out what was going on as his bench
supply was oscillating at a low level, I went and told a
few other techs so they could come and "help" (see).

As for the novelty of 4 wire connections for dealing
with cable and connection drops, you could pull up
one of my patents detailing a system that relies on that
very concept in order to operate effectively. (I do
not expect any such effort from you, attached as you
are to the notion of my ignorance. If not for that, it
ought to suffice to plug your "soon become" spew.)
 
J

John Fields

Jan 1, 1970
0
There are two silly assumptions you've made. As I have
stated elsewhere, I thought the OP would like a solution
utilizing what he mentioned he had on hand. I threw out
a few spending options without pretending to exhaust
them, merely to let him know he was not stuck with
just what he had. So I was not inclined to spend much
time trying to come up with a list that none of the smart
alecs around here would be able to "improve" upon.
So, assuming that my non-mention reflects ignorance
of remote sense power supplies is fatuous.

---
In light of the fact that his power supply might have been capable of
remote sensing and in view of your statement: "I thought the OP would
like a solution utilizing what he mentioned he had on hand." It seems
to me that your "familiarity" with remote sensing supplies would, at
the very least, brought forth the question of his power supply having
that capability and, if it did, a suggestion to use that capability.
---
The other
laughable assumption is that it should be an obvious
option to anybody who knew of such supplies.

---
Key phrase here is "knew of". If the OP wasn't aware that his supply
had that capability, or how to use it, then that option would hardly
have been obvious. You, though, knowing that such an option might
have been available didn't bring it up either.
---
Why should the OP go spend that kind of money when he
can simply use a shunt or measure the resistance of his
cable and use that and a voltmeter to measure current?

---
He probably shouldn't, but that's not what you advised him to do, you
sent him on some wild goose chase to measure this and that and
calculate this and that, when all you had to tell him was to monitor
the voltage at the input of the radio and crank the supply up to make
that voltage go to 7.5V when he was transmitting, LOL!

You also wrote:

"You could also put a lower shunt resistor across
your ammeter and calibrate the combination."

which is one of the stupidest things I've ever read. Do you know why
or would you like me to explain it to you while typing through fits of
laughter?
---
Thanks, so much John.

I note that your little description omits mention of the
1k or so resistors that normally obviate the need for
those jumpers when remote sensing is not used.

---
??? Funny... all of the stuff I've got around here _requires_
strapping the sense terminals to the output terminals if remote
sensing isn't used, but hey, that's only HP.
 
F

Fred Bloggs

Jan 1, 1970
0
Larry said:
(Finally, a positive contribution.)

That line should have read, of course, thusly:
Calculate Ina = Iao + Vad / Rt
And, contrary to what the esteemed Mr. Boggs
proclaims, the sign is correct. As defined, both
Vad and Rt are positive quantities (at least if
the transmitter draws more power at higher
voltages, which is already in evidence.) Since
Ina (pronounce as: I 'n'o 'a'mmeter) represents
the current predicted when no ammeter is present,
and that is already known to be higher, adding the
positive ratio Vad/Rt to the current measure with
the ammeter present is correct for getting such a
result. I would hope that this is now obvious
even to the most critical observer.




I perhaps should engage in some name-calling on
account of the above correction, but, as a low ranking
member of the excrement class, I am not up to it.




I already said 'No', Fred. Do you think repetition
is going to be effective? (It would appear so.)




Spending money was an obvious option which
I mentioned in several of its many forms. The
OP's questions led me to believe he might be
interested in using the equipment he had. We
have seen no evidence to the contrary.




Fred, I appreciate your opinion. Honestly.
I tried to tell you that earlier, but I suspect
my meaning escaped your notice.

What a simple-minded little fairy and brainless pussy you are- the OP
already said that he gets 1.85A at 7.0V without the ammeter- so who
needs your crap and pretentious pseudo- engineering formula? You want to
assume linearity for voltage drops less than 10%, then that there tells
you the effective power supply output resistance is 0.5V/1.85A=0.27
ohms. Then because the OP also states the load voltage is 6.14V at 1.80
with the ammeter, you have that 7.5V=1.8*0.27+1.8*Rammeter+6.14V or
Rammeter=0.486 ohms. The OP also states that without transmitting, the
drop due to the ammeter is 0.3V making I= 0.3V/(0.486+0.27)ohm=0.4 amps
at 7.2V load voltage- or a load power of 7.2*0.4=2.9 Watts. The load
power during transmit w/o ammeter is 7*1.85=13W and the load power w/
ammeter is 6.14*1.8=11W. The ammeter deprives the circuit of 2W of load
power,- so that the RF output will fall something like eta*2W where eta
is the transmit efficiency. Say for example eta is 33% more or less over
these power consumption ranges- then he must lose about 0.7W transmit
power- if he has a lot of Class A overhead stuff then eta may only be
10% and the power loss is 0.2W in transmit. Your little sissy model
conveys no usable information, and why would it- you don't what you're
doing- you are a pretentious fraud- you latch onto to same linearity
model for load Rt- you don't have a clue what's important and what's
not- and once again you can't understand what the OP is asking- like
mainly how much transmit power degradation can he expect. Get a clue and
take a hike- you are another "unwanted" NG p.o.s- just leave.
 
L

Larry Brasfield

Jan 1, 1970
0
Fred Bloggs said:
Larry said:
news:[email protected] [Some text and a load of crap cut.]
And, contrary to what the esteemed Mr. Boggs
proclaims, the sign is correct. As defined, both
Vad and Rt are positive quantities (at least if
the transmitter draws more power at higher
voltages, which is already in evidence.) Since
Ina (pronounce as: I 'n'o 'a'mmeter) represents
the current predicted when no ammeter is present,
and that is already known to be higher, adding the
positive ratio Vad/Rt to the current measure with
the ammeter present is correct for getting such a
result. I would hope that this is now obvious
even to the most critical observer.
Here's the deal you worthless, pretentious son-of-a-bitch-with-VD, you are a worthless p.o.s.- we are wise to your dumb ass-

So, you sincere, undiseased, wholesome paragon of virtue,
are you able to see and acknowledge your error? Why, if
you are such a valuable and respected contributor, do you
not help resolve the error you introduced? Is it that your
tenuous sense of superiority would be threatened? Or do
your invisible friends tell you to attempt some face saving?

........
[Much evidence of unfamiliarity with the paragraph cut.]

Nothing you said contravened the linear model for
the transmitter current consumption over a limited
supply voltage range. Have you managed to divine
some requirement of the OP's for accuracy that
would be compromised by using a linear model?
Get a clue and take a hike
[More crap into the bit bucket.]
just leave.

You must be daft, Fred. My answer is still No.
So maybe you better start your threatened action
to get me booted off without my cooperation.
Please amuse "us" with progress reports. <g>
 
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