This might be a question for one or another of the other
radio amateur newsgroups, but I figured I kind of already
know of some great engineering minds residing in
antenna-land, so I would post the question here first.

When I was a kid back in the 50s I built my first novice
transmitter with a one tube circuit using a 117N7 which put
out a couple of watts on CW through my end fed random wire
tuned with a tiny electric bulb and a single loop attached
between the two terminals of the bulb.

Oddly enough this circuit only used the hot side of the AC
outlet and a cold water pipe ground to the chassis of the
transmitter. Our house was built in the early 40s if that
tells you anything about how they wired outlets in those
days.

I have since dug up a reference to this transmitter circuit
in "Ham Radio Projects for Novice and Technician by Bert
Simon -W2UUN". Even though this little book came out in 1968
I know that I found the circuit in some earlier publication
because I built this thing in 1953. And I am alive to tell
the story...eg. not electrocuted. It keyed to the cathode
which was also tied to the top grid....kind of like a beam
pentode(?). I think I owned one 80m xtal at the time and a
Hallicrafters S-53A. [I got distracted from ham radio for 50
years, but I am back....with alot of catching up to do].

My question:
If I were to take a volt (amp) meter and put one probe in the
hot side of an AC house outlet and the other probe to a metal
rod stuck in the ground out in the middle of a field
somewhere (presumably nowhere near a neutral leg), what would
my meter read and why?
Deep electro-philosophical answers welcome as long as it is
expressed in terms a child could understand. (It seems that
this little odd transmitter circuit avoided the neutral leg
altogether-- just used the hot side and a ground).



Bill K6TAJ

"zeno" wrote
Oddly enough this circuit only used the hot side of the AC
outlet and a cold water pipe ground to the chassis of the
transmitter. Our house was built in the early 40s if that
tells you anything about how they wired outlets in those
days.

My question:
If I were to take a volt (amp) meter and put one probe in the
hot side of an AC house outlet and the other probe to a metal
rod stuck in the ground out in the middle of a field
somewhere (presumably nowhere near a neutral leg), what would
my meter read and why?
Deep electro-philosophical answers welcome as long as it is
expressed in terms a child could understand. (It seems that
this little odd transmitter circuit avoided the neutral leg
altogether-- just used the hot side and a ground).
Bill K6TAJ

Hi Bill,

Ground is referenced at both the generating station and your home in order
to complete the circuit. In your home's service mains panel, ground and
neutral are bonded together. Before the days of the third wire, added purely
for safety in the event of an equipment fault, a faulted piece of equipment
still had a return path, just not the added measure we have today of a very
short ground path (no pun intended) G

With a properly resistive load you could still perform your field exercise,
but it would have nothing to do with avoiding a nearby neutral wire which
serves the same function anyway. Neutral was always and still is, the return
path of a parallel circuit, nothing more or less. It is cold when no load is
connected to it and hot (minus the resistance-consumed current of a load
applied) when a load is connected.

73's

Jack
Virginia Beach, VA

Hi Jack,

So could I light up a 100 watt light bulb with one terminal stuck in the earth
out in the middle of a field and the other to the hot side of an AC
circuit?...I guess so.

The Earth is the ultimate return path I guess. The neutral as I understand it
is the center tap of the high voltage step down transformer which takes two hot
leads from the high voltage line and steps it down to 240V with a center tap
being the "neutral". The neutral is then grounded for extra safety. The neutral
(center tap) does not need to be grounded to provide the return path, but what
my question is why the Earth per se is also a return path? I guess being such a
large mass it is theoretically a zero potential.

Would my theoretical light bulb get brighter as I drove the rod deeper into the
earth?


Bill K6TAJ

Jack Painter wrote:

"zeno" wrote
Oddly enough this circuit only used the hot side of the AC
outlet and a cold water pipe ground to the chassis of the
transmitter. Our house was built in the early 40s if that
tells you anything about how they wired outlets in those
days.

My question:
If I were to take a volt (amp) meter and put one probe in the
hot side of an AC house outlet and the other probe to a metal
rod stuck in the ground out in the middle of a field
somewhere (presumably nowhere near a neutral leg), what would
my meter read and why?
Deep electro-philosophical answers welcome as long as it is
expressed in terms a child could understand. (It seems that
this little odd transmitter circuit avoided the neutral leg
altogether-- just used the hot side and a ground).
Bill K6TAJ

Hi Bill,

Ground is referenced at both the generating station and your home in order
to complete the circuit. In your home's service mains panel, ground and
neutral are bonded together. Before the days of the third wire, added purely
for safety in the event of an equipment fault, a faulted piece of equipment
still had a return path, just not the added measure we have today of a very
short ground path (no pun intended) G

With a properly resistive load you could still perform your field exercise,
but it would have nothing to do with avoiding a nearby neutral wire which
serves the same function anyway. Neutral was always and still is, the return
path of a parallel circuit, nothing more or less. It is cold when no load is
connected to it and hot (minus the resistance-consumed current of a load
applied) when a load is connected.

73's

Jack
Virginia Beach, VA

"zeno" wrote
Hi Jack,

So could I light up a 100 watt light bulb with one terminal stuck in the
earth
out in the middle of a field and the other to the hot side of an AC
circuit?...I guess so.

Bill,
Just before it blew up, it would light up, yes. Remember I said "With a
properly resistive load" and that would be a very dangerous however possible
experiment to attempt. The circuit breakers are set to prevent your
haphazard determination of what current load is too much - the hot wire
supply in should be matched with "cooler" (neutral) return, accounting for
acceptable current usage by the load... up to the point where too much
current generated heat trips the breaker. Nothing forces you (in your
experiment) from keeping all of that available voltage (0v felt on neutral)
as long as the current did not exceed 15a or whatever your breaker allows.
Obviously a 100w light bulb shorted to ground would blow instantly, before
the breaker could protect it..


The Earth is the ultimate return path I guess. The neutral as I understand
it
is the center tap of the high voltage step down transformer which takes
two hot
leads from the high voltage line and steps it down to 240V with a center
tap
being the "neutral". The neutral is then grounded for extra safety. The
neutral
(center tap) does not need to be grounded to provide the return path, but
what
my question is why the Earth per se is also a return path? I guess being
such a
large mass it is theoretically a zero potential.

Would my theoretical light bulb get brighter as I drove the rod deeper
into the
earth?

Assuming a series of (18) 100w bulbs on a 15a circuit (120v), yes the ground
rod's resistance (5-15 ohm depending on soil) would allow less than the full
brightness of the bulbs, until or if you were able to reduce the impedance
by using a larger surface area rod and drive it deeper into wet soil. This
is the same principle of making the very best and lowest impedance ground
system you possibly can for lightning protection!

73's

Jack



Bill K6TAJ

Jack Painter wrote:

"zeno" wrote
Oddly enough this circuit only used the hot side of the AC
outlet and a cold water pipe ground to the chassis of the
transmitter. Our house was built in the early 40s if that
tells you anything about how they wired outlets in those
days.

My question:
If I were to take a volt (amp) meter and put one probe in the
hot side of an AC house outlet and the other probe to a metal
rod stuck in the ground out in the middle of a field
somewhere (presumably nowhere near a neutral leg), what would
my meter read and why?
Deep electro-philosophical answers welcome as long as it is
expressed in terms a child could understand. (It seems that
this little odd transmitter circuit avoided the neutral leg
altogether-- just used the hot side and a ground).
Bill K6TAJ

Hi Bill,

Ground is referenced at both the generating station and your home in
order
to complete the circuit. In your home's service mains panel, ground and
neutral are bonded together. Before the days of the third wire, added
purely
for safety in the event of an equipment fault, a faulted piece of
equipment
still had a return path, just not the added measure we have today of a
very
short ground path (no pun intended) G

With a properly resistive load you could still perform your field
exercise,
but it would have nothing to do with avoiding a nearby neutral wire
which
serves the same function anyway. Neutral was always and still is, the
return
path of a parallel circuit, nothing more or less. It is cold when no
load is
connected to it and hot (minus the resistance-consumed current of a load
applied) when a load is connected.

73's

Jack
Virginia Beach, VA

On 8-May-2004, "Jack Painter" wrote:

experiment) from keeping all of that available voltage (0v felt on neutral)
as long as the current did not exceed 15a or whatever your breaker allows.
Obviously a 100w light bulb shorted to ground would blow instantly, before
the breaker could protect it..

Uhhhh, No! The most voltage from either wire of a 117 Volt household circuit to any other wire or
to any made ground is 117 Volts. The light bulb would be quite happy to glow at something up to its
normal brightness for as long as you wanted. Now if the grounded conductor was somewhere out in a
field instead of being the local house ground, then the light bulb would not receive the full 117
Volts, because of the resistance of the intervening earth, and would be unhappily dim.

As for the Original Poster's question about the 117N7 transmitter, they were inherently unsafe
unless built on a chassis insulated from the antenna and ground. The usual method was to use a
floating negative (not connected to the metal chassis) inside the transmitter. The antenna was
isolated from the 117 VAC circuit by the DC blocking capacitor to the pi network. The external
antenna ground connected to the chassis. The cathode of the tube and the negative side of the DC
Power Supply had to be bypassed to the chassis through a large capacitor for an RF ground. In no
case should the neutral conductor be left unconnected, even if a water pipe ground could carry the
neutral current. That would leave 117 VAC on the chassis (and the metal shafts of tuning
capacitors.. and the ON/OFF toggle switch) if the ground wire came loose.

I grew up in that era too and got many shocks from AC-DC radios.

Ken Fowler

Thanks Ken, you're right. I was thinking that without current-sensing
abilities on the neutral side of the circuit that it might allow the full
available current, but of course a light bulb would just work as a light
bulb, drawing no more power than it was designed for! I feel a little dim
myself right now ;-(

Jack

"Ken Fowler" wrote in message
...

On 8-May-2004, "Jack Painter" wrote:

experiment) from keeping all of that available voltage (0v felt on
neutral)
as long as the current did not exceed 15a or whatever your breaker
allows.
Obviously a 100w light bulb shorted to ground would blow instantly,
before
the breaker could protect it..

Uhhhh, No! The most voltage from either wire of a 117 Volt household
circuit to any other wire or
to any made ground is 117 Volts. The light bulb would be quite happy to
glow at something up to its
normal brightness for as long as you wanted. Now if the grounded
conductor was somewhere out in a
field instead of being the local house ground, then the light bulb would
not receive the full 117
Volts, because of the resistance of the intervening earth, and would be
unhappily dim.

Zeno:

I think you are confusing cause and effect. There are many safety
considerations in designing an electrical power distribution system. One
consideration is this: Suppose some fault occurs which increases the leakage
from the primary voltage (anywhere from 4 kV to 54 kV) of your distribution
transformer to the secondary which supplies your home, and the secondary is
ungrounded. Contrary to the way it would be portrayed on movie and TV
programs, your light bulbs would still light normally, and appliances would
work correctly (except grounded radios and TVs which actually would blow up)
but if you touched anything connected to the secondary service, you would
enjoy instant electrocution. So, the reference for the secondary, the
center tap of the US 240 VAC system is connected to a ground rod(s) of a
specified maximum resistance to earth (which tends to be the potential your
body assumes while walking around your house). The design criteria for this
fault condition is that the primary fuse will open before the secondary
system voltage rises above a safe(?) level. So, neutral is grounded, to a
relatively good extent.

Now, to further pursue this mind game, consider the earth to be a copper
ball. Then, all of it that is nearby will be at the potential of your
electric service neutral. And a sheet metal screw driven into it anywhere
within reason would provide an adequate return path for AC drawn from either
phase of the service. Real earth is more resistive, depending on soil
content and moisture level, so one can approach the copper ball condition,
but usually it is more like a bag of loose carbon particles. And, the more
surface area of a driven rod exposed to the earth, the lower the resistance
and the less voltage drop for a given current, or the more current for a
given voltage drop. It seems to all follow Ohm's law, for some unknown
reason.

A couple of caveats. One, modern wiring tends to include Ground Fault
devices which measure differential current to a wiring device. If more
current flows in the hot line then returns there to that neutral, it
disconnects the device. So the connection you used would not operate. Two,
none of this applies to RF, as in antenna grounds, because reactance must be
considered in addition to pure resistance which dominates at 60 Hz. And
another: Some to none of this applies in most other countries which use
other standards and methods.

--
Crazy George
Remove N O and S P A M imbedded in return address

Bill,
The voltage on the end of the wire will be something
between 120 vac and 0 vac. Depends on how long the wire
is and the resistance of your 'dirt'. I wouldn't expect
the light bulb to be as bright as it could be (back to the
resistance and length thingys again), but I would expect
it to have some 'glow' to it...
'Doc

PS - Rubber gloves and boots would be 'nice' if you try
that experiment.

On Sun, 09 May 2004 04:13:55 GMT, zeno wrote:

My question:
If I were to take a volt (amp) meter and put one probe in the
hot side of an AC house outlet and the other probe to a metal
rod stuck in the ground out in the middle of a field
somewhere (presumably nowhere near a neutral leg), what would
my meter read and why?
Deep electro-philosophical answers welcome as long as it is
expressed in terms a child could understand. (It seems that
this little odd transmitter circuit avoided the neutral leg
altogether-- just used the hot side and a ground).



Bill K6TAJ

Bill,

Assuming that the transformer on the pole supplying the power to your
house has a grounded neutral, an "infinitely" high impedance voltmeter
will read total applied voltage across an open circuit, and so your
meter will read 110/120 volts.

If your voltmeter does not have an infinitely high impedance, the
internal impedance of the meter will be in series with the line/meter
lead/ground impedance and will read a portion of the applied voltage
equal to the voltage drop across it's internal impedance.

Ron, W1WBV