Driver to drive?

[John Devereux]

Spehro Pefhany <speffSNIP@interlogDOTyou.knowwhat> writes:

On Sun, 7 Jun 2015 04:17:06 -0700 (PDT), the renowned
rev.11d.meow@gmail.com wrote:


FAKE ICs MUST DIE!

And FTDI changes the Dev ID for Fake Chips and that is that.

You can program it back in the chip and modify the inf to make it work, supposedly.

But who'd trust a Fake FTDI IC?


Maybe a real CH340 from WCH in Nanjing is better?

Nice!

<http://www.ebay.co.uk/itm/USB-To-RS232-TTL-CH340G-Converter-Module-Adapter-STC-replace-Pl2303-CP2102-HOT-/141665281093?pt=LH_DefaultDomain_3&hash=item20fbe8a445>


<http://www.ebay.co.uk/itm/XP-7-8-Linux-MAC-USB2-0-to-RS485-Serial-Converter-Adapter-CH340G-Support-Windows-/390958738648?pt=LH_DefaultDomain_3&hash=item5b06f4ecd8>

Bought some just for fun, see if they work with linux, might be useful
for development.


--

John Devereux

 

[Spehro Pefhany]

On Sun, 07 Jun 2015 14:07:41 +0100, the renowned John Devereux
<john@devereux.me.uk> wrote:

Spehro Pefhany <speffSNIP@interlogDOTyou.knowwhat> writes:

On Sun, 7 Jun 2015 04:17:06 -0700 (PDT), the renowned
rev.11d.meow@gmail.com wrote:


FAKE ICs MUST DIE!

And FTDI changes the Dev ID for Fake Chips and that is that.

You can program it back in the chip and modify the inf to make it work, supposedly.

But who'd trust a Fake FTDI IC?


Maybe a real CH340 from WCH in Nanjing is better?

Nice!

http://www.ebay.co.uk/itm/USB-To-RS232-TTL-CH340G-Converter-Module-Adapter-STC-replace-Pl2303-CP2102-HOT-/141665281093?pt=LH_DefaultDomain_3&hash=item20fbe8a445


http://www.ebay.co.uk/itm/XP-7-8-Linux-MAC-USB2-0-to-RS485-Serial-Converter-Adapter-CH340G-Support-Windows-/390958738648?pt=LH_DefaultDomain_3&hash=item5b06f4ecd8

Bought some just for fun, see if they work with linux, might be useful
for development.

The Linux driver readme says:

Note: 1.Please run followed executable programs as root privilege
2.Current Driver support versions of linux kernel range from
2.6.25 to 3.13.x
3.Current Driver support 32bits and 64bits linux systems

--
Best regards,
Spehro Pefhany
Amazon link for AoE 3rd Edition: http://tinyurl.com/ntrpwu8
Microchip link for 2015 Masters in Phoenix: http://tinyurl.com/l7g2k48

 

[rickman]

On 6/7/2015 8:43 AM, Spehro Pefhany wrote:

On Sun, 7 Jun 2015 04:17:06 -0700 (PDT), the renowned
rev.11d.meow@gmail.com wrote:


FAKE ICs MUST DIE!

And FTDI changes the Dev ID for Fake Chips and that is that.

You can program it back in the chip and modify the inf to make it work, supposedly.

But who'd trust a Fake FTDI IC?


Maybe a real CH340 from WCH in Nanjing is better?

That's what I'm thinking. I bought some lowest price "Prolific"
adapters on eBay and was surprised to find they didn't work with the
Prolific drivers. That was about the time the counterfeit thing caught
my attention with the issue that Prolific only supports their newer
chips (not yet cloned) under Win8 as a means of squeezing out the
clones. I complained to the vendor and got my money back, but the irony
is that the CH340 works everywhere and anywhere with no problem once you
have the right driver from WCH.

So at this point I'd say WCH is the go to guy for USB<>232 chips...
unless you have a problem with the drivers not being automagically
downloaded and installed. I think I had to search for them myself. If
they fixed that I would leave Prolific and FTDI behind.

--

Rick

 

[John Devereux]

Spehro Pefhany <speffSNIP@interlogDOTyou.knowwhat> writes:

On Sun, 07 Jun 2015 14:07:41 +0100, the renowned John Devereux
john@devereux.me.uk> wrote:

Spehro Pefhany <speffSNIP@interlogDOTyou.knowwhat> writes:

On Sun, 7 Jun 2015 04:17:06 -0700 (PDT), the renowned
rev.11d.meow@gmail.com wrote:


FAKE ICs MUST DIE!

And FTDI changes the Dev ID for Fake Chips and that is that.

You can program it back in the chip and modify the inf to make it work, supposedly.

But who'd trust a Fake FTDI IC?


Maybe a real CH340 from WCH in Nanjing is better?

Nice!

http://www.ebay.co.uk/itm/USB-To-RS232-TTL-CH340G-Converter-Module-Adapter-STC-replace-Pl2303-CP2102-HOT-/141665281093?pt=LH_DefaultDomain_3&hash=item20fbe8a445


http://www.ebay.co.uk/itm/XP-7-8-Linux-MAC-USB2-0-to-RS485-Serial-Converter-Adapter-CH340G-Support-Windows-/390958738648?pt=LH_DefaultDomain_3&hash=item5b06f4ecd8

Bought some just for fun, see if they work with linux, might be useful
for development.

The Linux driver readme says:

Note: 1.Please run followed executable programs as root privilege
2.Current Driver support versions of linux kernel range from
2.6.25 to 3.13.x
3.Current Driver support 32bits and 64bits linux systems

Thanks

I probably won't be installing vendor binaries in linux just for that
but it looks like the chip is becoming popular after the FTDI debacle. I
expect it will be in the mainline kernel shortly (if it is not already).

--

John Devereux

 

[N. Coesel]

John Devereux schreef op 06/07/2015 om 09:28 AM:

We would never be safe.

That is also my viewpoint. I'm designing out all FTDI devices. Not only
because of the driver problems but also due to stability issues.

 

[Jasen Betts]

On 2015-06-07, rickman <gnuarm@gmail.com> wrote:

On 6/7/2015 8:43 AM, Spehro Pefhany wrote:
On Sun, 7 Jun 2015 04:17:06 -0700 (PDT), the renowned
rev.11d.meow@gmail.com wrote:


FAKE ICs MUST DIE!

And FTDI changes the Dev ID for Fake Chips and that is that.

You can program it back in the chip and modify the inf to make it work, supposedly.

But who'd trust a Fake FTDI IC?


Maybe a real CH340 from WCH in Nanjing is better?

That's what I'm thinking. I bought some lowest price "Prolific"
adapters on eBay and was surprised to find they didn't work with the
Prolific drivers. That was about the time the counterfeit thing caught
my attention with the issue that Prolific only supports their newer
chips (not yet cloned) under Win8 as a means of squeezing out the
clones. I complained to the vendor and got my money back, but the irony
is that the CH340 works everywhere and anywhere with no problem once you
have the right driver from WCH.

So at this point I'd say WCH is the go to guy for USB<>232 chips...
unless you have a problem with the drivers not being automagically
downloaded and installed. I think I had to search for them myself. If
they fixed that I would leave Prolific and FTDI behind.

Why not SiLabs CP2101 or CP2102

AFAIK they have never messed with the drivers.


--
umop apisdn

 

[N. Coesel]

Jasen Betts schreef op 06/08/2015 om 09:18 AM:

On 2015-06-07, rickman <gnuarm@gmail.com> wrote:
On 6/7/2015 8:43 AM, Spehro Pefhany wrote:
On Sun, 7 Jun 2015 04:17:06 -0700 (PDT), the renowned
rev.11d.meow@gmail.com wrote:


FAKE ICs MUST DIE!

And FTDI changes the Dev ID for Fake Chips and that is that.

You can program it back in the chip and modify the inf to make it work, supposedly.

But who'd trust a Fake FTDI IC?


Maybe a real CH340 from WCH in Nanjing is better?

That's what I'm thinking. I bought some lowest price "Prolific"
adapters on eBay and was surprised to find they didn't work with the
Prolific drivers. That was about the time the counterfeit thing caught
my attention with the issue that Prolific only supports their newer
chips (not yet cloned) under Win8 as a means of squeezing out the
clones. I complained to the vendor and got my money back, but the irony
is that the CH340 works everywhere and anywhere with no problem once you
have the right driver from WCH.

So at this point I'd say WCH is the go to guy for USB<>232 chips...
unless you have a problem with the drivers not being automagically
downloaded and installed. I think I had to search for them myself. If
they fixed that I would leave Prolific and FTDI behind.

Why not SiLabs CP2101 or CP2102

AFAIK they have never messed with the drivers.

I agree. From my experience the CP210x devices are also more stable when
using them on the bench when connecting/disconnecting to a target. With
the FTDI chips I need to unplug/plug when I switch my bench PSU on or off.

 

[rickman]

On 7/4/2015 9:43 PM, Jeff Liebermann wrote:

On Sat, 04 Jul 2015 19:33:30 -0400, Jerry Stuckle
jstucklex@attglobal.net> wrote:

On 7/4/2015 7:22 PM, Jeff Liebermann wrote:
On Sat, 04 Jul 2015 19:04:01 -0400, Jerry Stuckle
jstucklex@attglobal.net> wrote:
Think of it this way, without the math. On the transmitter side of the
network, the match is 1:1, with nothing reflected back to the transmitter.

So you have a signal coming back from the antenna. You have a perfect
matching network, which means nothing is lost in the network. The
feedline is perfect, so there is no loss in it. The only place for the
signal to go is back to the antenna.

Wikipedia says that if the source is matched to the line, any
reflections that come back are absorbed, not reflected back to the
antenna:

https://en.wikipedia.org/wiki/Impedance_matching
"If the source impedance matches the line, reflections
from the load end will be absorbed at the source end.
If the transmission line is not matched at both ends
reflections from the load will be re-reflected at the
source and re-re-reflected at the load end ad infinitum,
losing energy on each transit of the transmission line."

And you believe everything Wikipedia says? ROFLMAO.
But that also explains your ignorance.

Let's see if I understand you correctly. You claim that with a power
amplifier (source) output impedance that is perfectly matched to the
coax cable, but not necessarily the load (antenna), any reflected
power from the load (antenna) is bounced back to the load (antenna) by
the perfectly matched source (power amp). Is that what you're saying?

Yet, when I have a perfectly matched load (antenna), all the power it
is fed is radiated and nothing is reflected. You can't have it both
ways because the reflected power from the load (antenna), becomes the
incident power going towards the source (power amp). Matched and
mismatched loads do NOT act differently depending on the direction of
travel. If you claim were true, then transmitting into a matched
antenna or dummy load would reflect all the power back towards the
transmitter.

I think this is one of those situations where a casual explanation won't
work. You can use a "casual" explanation when the various
qualifications for a simplification apply. But to do that, the
qualifiers have to be fully understood and no one here is showing what
the qualifiers are much less that they are met. So until we get a real
explanation I will stick with what I recall. In the end, to settle this
we may have to use the math.

I'm sure someone in s.e.d could explain this properly. Some of them may
be purely argumentative, but some really know their stuff. I believe
the description of a conjugate match is the mathematical inverse of the
complex impedance of the antenna "viewed" through the feed line, but I
have to admit I don't really know what that implies or if it is even an
accurate description.

--

Rick

 

[Clifford Heath]

On 06/07/15 07:59, makolber@yahoo.com wrote:

For sake of discussion, assume the tx line is 50 ohms.
If the tx output is an ideal 50 ohm match, any reflection from the ant will be absorbed by the tx.
If the tx however, instead present a conjugate match to whatever z the line present it,
Any reflection from the ant will be completely re- reflected by the tx back to the ant.
The conjugate matched case will provide more radiated power.

At the cost, as Tim said, of possibly excessive voltages or currents.

A friend melted the aluminium tube driven elements on his six-metre Yagi
antenna with a 1000W HF amp. The amp, cable and connectors were fine...
the solid tube *elements* melted from excessive current.

 

For sake of discussion, assume the tx line is 50 ohms.
If the tx output is an ideal 50 ohm match, any reflection from the ant will be absorbed by the tx.
If the tx however, instead present a conjugate match to whatever z the line present it,
Any reflection from the ant will be completely re- reflected by the tx back to the ant.
The conjugate matched case will provide more radiated power.
Mark

 

[Tauno Voipio]

On 5.7.15 18:56, rickman wrote:

On 7/4/2015 9:43 PM, Jeff Liebermann wrote:
On Sat, 04 Jul 2015 19:33:30 -0400, Jerry Stuckle
jstucklex@attglobal.net> wrote:

On 7/4/2015 7:22 PM, Jeff Liebermann wrote:
On Sat, 04 Jul 2015 19:04:01 -0400, Jerry Stuckle
jstucklex@attglobal.net> wrote:
Think of it this way, without the math. On the transmitter side of
the
network, the match is 1:1, with nothing reflected back to the
transmitter.

So you have a signal coming back from the antenna. You have a perfect
matching network, which means nothing is lost in the network. The
feedline is perfect, so there is no loss in it. The only place for
the
signal to go is back to the antenna.

Wikipedia says that if the source is matched to the line, any
reflections that come back are absorbed, not reflected back to the
antenna:

https://en.wikipedia.org/wiki/Impedance_matching
"If the source impedance matches the line, reflections
from the load end will be absorbed at the source end.
If the transmission line is not matched at both ends
reflections from the load will be re-reflected at the
source and re-re-reflected at the load end ad infinitum,
losing energy on each transit of the transmission line."

And you believe everything Wikipedia says? ROFLMAO.
But that also explains your ignorance.

Let's see if I understand you correctly. You claim that with a power
amplifier (source) output impedance that is perfectly matched to the
coax cable, but not necessarily the load (antenna), any reflected
power from the load (antenna) is bounced back to the load (antenna) by
the perfectly matched source (power amp). Is that what you're saying?

Yet, when I have a perfectly matched load (antenna), all the power it
is fed is radiated and nothing is reflected. You can't have it both
ways because the reflected power from the load (antenna), becomes the
incident power going towards the source (power amp). Matched and
mismatched loads do NOT act differently depending on the direction of
travel. If you claim were true, then transmitting into a matched
antenna or dummy load would reflect all the power back towards the
transmitter.

I think this is one of those situations where a casual explanation won't
work. You can use a "casual" explanation when the various
qualifications for a simplification apply. But to do that, the
qualifiers have to be fully understood and no one here is showing what
the qualifiers are much less that they are met. So until we get a real
explanation I will stick with what I recall. In the end, to settle this
we may have to use the math.

I'm sure someone in s.e.d could explain this properly. Some of them may
be purely argumentative, but some really know their stuff. I believe
the description of a conjugate match is the mathematical inverse of the
complex impedance of the antenna "viewed" through the feed line, but I
have to admit I don't really know what that implies or if it is even an
accurate description.

You're right. To get a 1:1 match for the piece of feedline between
the transmitter and the antenna tuning unit, the tuning unit has
to present a conjugate match to the feedline from the tuning unit
to the antenna. If there is a reflected wave from the antenna, it
will be re-reflected back toward the antenna from the tuning unit.
The ping-ponging signal will die out by antenna radiation or feedline
losses. The situation is quite OK with slow modulations (like voice),
but the ping-ponging is unacceptable for fast signals (like analog TV).

--

-TV

 

[Tim Williams]

If there is incident energy upon a correctly matched transmitter's output,
it need not absorb 100% of the energy.

If the transmitter is an ideal, linear, Thevenin/Norton source, it must, as
is consistent with linearity, and the power transfer theorem, and
transmission line theory, and all that.

But, it is quite common for transmitters to reflect incident power. This
characteristic is captured by the scattering parameters, namely s22, the
output reflectance. (Well, that would be Gamma_22, but they're equivalent
parameters in the end.)

This corresponds, in turn, with the dynamic output impedance, which you may
recall from audio amplifiers, need not equal the "best load" impedance. For
instance, an amplifier for 8 ohm loads might have a dynamic impedance of
0.03 ohms, a very good voltage source in comparison -- acting like a short
circuit to incident energy (and therefore reflecting energy, out of phase,
back towards the loudspeaker, thus giving it a high "damping factor").

The same is true of RF amplifiers, except rather than constant voltage
characteristics due to the use of voltage negative feedback, the
characteristic is usually constant current (high impedance), due to the lack
of negative feedback (or the use of current feedback) and the high intrinsic
impedance of the devices (i.e., the collector / drain / plate resistance).

As the energy "piles up on" and reflects off the transmitter, internal
voltages and currents may get into dangerous ranges, causing excessive
dissipation or breakdown; this is partly why transmitters shouldn't be
operated with high SWR (the other part being, efficiency and power capacity
suck).

The only amplifiers that have a relatively matched intrinsic output
imepdance are vacuum tube triode amplifiers. Though even these tend to have
a poor match, either being operated in class AB with R_L > Rp, or class B/C
with R_L < Rp (and lots of grid current to push plate current up there).
Because, again, nonlinear devices don't need to obey the linearity theorems,
and can have impedances different from the "best load" value.

Tim

--
Seven Transistor Labs
Electrical Engineering Consultation
Website: http://seventransistorlabs.com

"rickman" <gnuarm@gmail.com> wrote in message
news:mnbk18$hvf$1@dont-email.me...

On 7/4/2015 9:43 PM, Jeff Liebermann wrote:
On Sat, 04 Jul 2015 19:33:30 -0400, Jerry Stuckle
jstucklex@attglobal.net> wrote:

On 7/4/2015 7:22 PM, Jeff Liebermann wrote:
On Sat, 04 Jul 2015 19:04:01 -0400, Jerry Stuckle
jstucklex@attglobal.net> wrote:
Think of it this way, without the math. On the transmitter side of
the
network, the match is 1:1, with nothing reflected back to the
transmitter.

So you have a signal coming back from the antenna. You have a perfect
matching network, which means nothing is lost in the network. The
feedline is perfect, so there is no loss in it. The only place for
the
signal to go is back to the antenna.

Wikipedia says that if the source is matched to the line, any
reflections that come back are absorbed, not reflected back to the
antenna:

https://en.wikipedia.org/wiki/Impedance_matching
"If the source impedance matches the line, reflections
from the load end will be absorbed at the source end.
If the transmission line is not matched at both ends
reflections from the load will be re-reflected at the
source and re-re-reflected at the load end ad infinitum,
losing energy on each transit of the transmission line."

And you believe everything Wikipedia says? ROFLMAO.
But that also explains your ignorance.

Let's see if I understand you correctly. You claim that with a power
amplifier (source) output impedance that is perfectly matched to the
coax cable, but not necessarily the load (antenna), any reflected
power from the load (antenna) is bounced back to the load (antenna) by
the perfectly matched source (power amp). Is that what you're saying?

Yet, when I have a perfectly matched load (antenna), all the power it
is fed is radiated and nothing is reflected. You can't have it both
ways because the reflected power from the load (antenna), becomes the
incident power going towards the source (power amp). Matched and
mismatched loads do NOT act differently depending on the direction of
travel. If you claim were true, then transmitting into a matched
antenna or dummy load would reflect all the power back towards the
transmitter.

I think this is one of those situations where a casual explanation won't
work. You can use a "casual" explanation when the various qualifications
for a simplification apply. But to do that, the qualifiers have to be
fully understood and no one here is showing what the qualifiers are much
less that they are met. So until we get a real explanation I will stick
with what I recall. In the end, to settle this we may have to use the
math.

I'm sure someone in s.e.d could explain this properly. Some of them may
be purely argumentative, but some really know their stuff. I believe the
description of a conjugate match is the mathematical inverse of the
complex impedance of the antenna "viewed" through the feed line, but I
have to admit I don't really know what that implies or if it is even an
accurate description.

--

Rick

 

On Sun, 05 Jul 2015 21:37:58 -0500, John S <Sophi.2@invalid.org>
wrote:

On 7/5/2015 11:00 AM, Clifford Heath wrote:
On 06/07/15 07:59, makolber@yahoo.com wrote:
For sake of discussion, assume the tx line is 50 ohms.
If the tx output is an ideal 50 ohm match, any reflection from the ant
will be absorbed by the tx.
If the tx however, instead present a conjugate match to whatever z the
line present it,
Any reflection from the ant will be completely re- reflected by the tx
back to the ant.
The conjugate matched case will provide more radiated power.

At the cost, as Tim said, of possibly excessive voltages or currents.

A friend melted the aluminium tube driven elements on his six-metre Yagi
antenna with a 1000W HF amp. The amp, cable and connectors were fine...
the solid tube *elements* melted from excessive current.

How much current (power, temperature) does it take to melt a (solid?)
aluminium (sic) tube (rod?) driven element? An inquiring mind wants to know.

There is a current maximum (and voltage minimum) at the centre of the
dipole. There are voltage maximum (and current minimum) at the ends of
the dipole.

Some long yagis might have quite low feedpoint impedance and hence
even larger current.

In addition you would have to consider skin effect losses.

However, while the feed point might get hot, aluminum has a good heat
transfer and thus would dissipate the power at the ends of the dipole.
Thus, I think it is quite unlikely that the element would melt away.
More likely the element was already broken for some other reason and
by coincidence the element dropped while transmitting.

 

[John S]

On 7/5/2015 11:00 AM, Clifford Heath wrote:

On 06/07/15 07:59, makolber@yahoo.com wrote:
For sake of discussion, assume the tx line is 50 ohms.
If the tx output is an ideal 50 ohm match, any reflection from the ant
will be absorbed by the tx.
If the tx however, instead present a conjugate match to whatever z the
line present it,
Any reflection from the ant will be completely re- reflected by the tx
back to the ant.
The conjugate matched case will provide more radiated power.

At the cost, as Tim said, of possibly excessive voltages or currents.

A friend melted the aluminium tube driven elements on his six-metre Yagi
antenna with a 1000W HF amp. The amp, cable and connectors were fine...
the solid tube *elements* melted from excessive current.

How much current (power, temperature) does it take to melt a (solid?)
aluminium (sic) tube (rod?) driven element? An inquiring mind wants to know.

 

[Jeff Liebermann]

On Sun, 05 Jul 2015 21:37:58 -0500, John S <Sophi.2@invalid.org>
wrote:

On 7/5/2015 11:00 AM, Clifford Heath wrote:
On 06/07/15 07:59, makolber@yahoo.com wrote:
For sake of discussion, assume the tx line is 50 ohms.
If the tx output is an ideal 50 ohm match, any reflection from the ant
will be absorbed by the tx.
If the tx however, instead present a conjugate match to whatever z the
line present it,
Any reflection from the ant will be completely re- reflected by the tx
back to the ant.
The conjugate matched case will provide more radiated power.

At the cost, as Tim said, of possibly excessive voltages or currents.

A friend melted the aluminium tube driven elements on his six-metre Yagi
antenna with a 1000W HF amp. The amp, cable and connectors were fine...
the solid tube *elements* melted from excessive current.

How much current (power, temperature) does it take to melt a (solid?)
aluminium (sic) tube (rod?) driven element? An inquiring mind wants to know.

Ok, let's see if my thermodynamics is still intact. Corrections
welcome as my math usually sucks and I always make at least one big
mistake.


I'll assume a 3 element 6 meter beam. If it had more elements, this
would be a good time to disclose the details.
<http://www.cushcraftamateur.com/Product.php?productid=A50-3S>
3.2 kg for the entire antenna. I would guess(tm) that the boom is
about 1.0 Kg, leaving 2.2 Kg of aluminum for the elements.

6061-T6 melts at about 580C = 853K
I'll assume an ambient temp of 27C = 300K

Heat absorbed before melting:
Specific heat capacity of Al:
C = 0.897 J/(gm*K)
Heat_1 = 0.897 * m * change_in_temp
Heat_1 = 0.897 * 2,200gm * 553K = 1100*10^3 KiloJoules

Heat of fusion for Al:
Lf = 0.395 kJ/g
Heat_2 = m * Lf = 2,200gm * 0.395 = 869 KiloJoules

Total heat required = 1100*10^3 + 869*10^3 = 1969 kJ
I'll roundoff to 2000 KiloJoules = 2 kilowatt-seconds

So, if 100% of the 1000 watts of power was dissipated by the driven
elements of the antenna, it would melt in 2 seconds. However, since
it's unlikely all 1000 watts actually made it to the antenna, and that
it's likely that the antenna actually radiated much of the power, the
actual meltdown times will be much longer. Still, it looks quite
possible to melt the elements.




--
Jeff Liebermann jeffl@cruzio.com
150 Felker St #D http://www.LearnByDestroying.com
Santa Cruz CA 95060 http://802.11junk.com
Skype: JeffLiebermann AE6KS 831-336-2558

 

[John S]

On 7/5/2015 11:10 PM, Jeff Liebermann wrote:

On Sun, 05 Jul 2015 21:37:58 -0500, John S <Sophi.2@invalid.org
wrote:

On 7/5/2015 11:00 AM, Clifford Heath wrote:
On 06/07/15 07:59, makolber@yahoo.com wrote:
For sake of discussion, assume the tx line is 50 ohms.
If the tx output is an ideal 50 ohm match, any reflection from the ant
will be absorbed by the tx.
If the tx however, instead present a conjugate match to whatever z the
line present it,
Any reflection from the ant will be completely re- reflected by the tx
back to the ant.
The conjugate matched case will provide more radiated power.

At the cost, as Tim said, of possibly excessive voltages or currents.

A friend melted the aluminium tube driven elements on his six-metre Yagi
antenna with a 1000W HF amp. The amp, cable and connectors were fine...
the solid tube *elements* melted from excessive current.

How much current (power, temperature) does it take to melt a (solid?)
aluminium (sic) tube (rod?) driven element? An inquiring mind wants to know.

Ok, let's see if my thermodynamics is still intact. Corrections
welcome as my math usually sucks and I always make at least one big
mistake.


I'll assume a 3 element 6 meter beam. If it had more elements, this
would be a good time to disclose the details.
http://www.cushcraftamateur.com/Product.php?productid=A50-3S
3.2 kg for the entire antenna. I would guess(tm) that the boom is
about 1.0 Kg, leaving 2.2 Kg of aluminum for the elements.

6061-T6 melts at about 580C = 853K
I'll assume an ambient temp of 27C = 300K

Heat absorbed before melting:
Specific heat capacity of Al:
C = 0.897 J/(gm*K)
Heat_1 = 0.897 * m * change_in_temp
Heat_1 = 0.897 * 2,200gm * 553K = 1100*10^3 KiloJoules

Heat of fusion for Al:
Lf = 0.395 kJ/g
Heat_2 = m * Lf = 2,200gm * 0.395 = 869 KiloJoules

Total heat required = 1100*10^3 + 869*10^3 = 1969 kJ
I'll roundoff to 2000 KiloJoules = 2 kilowatt-seconds

I'm a bit confused. I thought one joule was one watt*second. If so,
wouldn't 2000 kilojoules be 2000 kilowatt*seconds or 2 million watt*seconds?

So, if 100% of the 1000 watts of power was dissipated by the driven
elements of the antenna, it would melt in 2 seconds. However, since
it's unlikely all 1000 watts actually made it to the antenna, and that
it's likely that the antenna actually radiated much of the power, the
actual meltdown times will be much longer. Still, it looks quite
possible to melt the elements.

 

[Phil Hobbs]

On 7/5/2015 10:37 PM, John S wrote:

On 7/5/2015 11:00 AM, Clifford Heath wrote:
On 06/07/15 07:59, makolber@yahoo.com wrote:
For sake of discussion, assume the tx line is 50 ohms.
If the tx output is an ideal 50 ohm match, any reflection from the ant
will be absorbed by the tx.
If the tx however, instead present a conjugate match to whatever z the
line present it,
Any reflection from the ant will be completely re- reflected by the tx
back to the ant.
The conjugate matched case will provide more radiated power.

At the cost, as Tim said, of possibly excessive voltages or currents.

A friend melted the aluminium tube driven elements on his six-metre Yagi
antenna with a 1000W HF amp. The amp, cable and connectors were fine...
the solid tube *elements* melted from excessive current.

How much current (power, temperature) does it take to melt a (solid?)
aluminium (sic) tube (rod?) driven element? An inquiring mind wants to
know.

Depends on how badly oxidised it is. If the contact area is small, you
can get a pretty large current density.

I'd be very surprised if you could melt an aluminum antenna element
without the feedline getting hot enough to melt the dielectric and short
out.

Of course, if the antenna is sufficiently badly mismatched, so that the
antenna + matching network has some absurd Q, you get a current
enhancement by a factor Q+1.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics

160 North State Road #203
Briarcliff Manor NY 10510

hobbs at electrooptical dot net
http://electrooptical.net

 

[John S]

On 7/6/2015 8:58 AM, John S wrote:

On 7/5/2015 11:10 PM, Jeff Liebermann wrote:
On Sun, 05 Jul 2015 21:37:58 -0500, John S <Sophi.2@invalid.org
wrote:

On 7/5/2015 11:00 AM, Clifford Heath wrote:
On 06/07/15 07:59, makolber@yahoo.com wrote:
For sake of discussion, assume the tx line is 50 ohms.
If the tx output is an ideal 50 ohm match, any reflection from the ant
will be absorbed by the tx.
If the tx however, instead present a conjugate match to whatever z the
line present it,
Any reflection from the ant will be completely re- reflected by the tx
back to the ant.
The conjugate matched case will provide more radiated power.

At the cost, as Tim said, of possibly excessive voltages or currents.

A friend melted the aluminium tube driven elements on his six-metre
Yagi
antenna with a 1000W HF amp. The amp, cable and connectors were fine...
the solid tube *elements* melted from excessive current.

How much current (power, temperature) does it take to melt a (solid?)
aluminium (sic) tube (rod?) driven element? An inquiring mind wants
to know.

Ok, let's see if my thermodynamics is still intact. Corrections
welcome as my math usually sucks and I always make at least one big
mistake.


I'll assume a 3 element 6 meter beam. If it had more elements, this
would be a good time to disclose the details.
http://www.cushcraftamateur.com/Product.php?productid=A50-3S
3.2 kg for the entire antenna. I would guess(tm) that the boom is
about 1.0 Kg, leaving 2.2 Kg of aluminum for the elements.

6061-T6 melts at about 580C = 853K
I'll assume an ambient temp of 27C = 300K

Heat absorbed before melting:
Specific heat capacity of Al:
C = 0.897 J/(gm*K)
Heat_1 = 0.897 * m * change_in_temp
Heat_1 = 0.897 * 2,200gm * 553K = 1100*10^3 KiloJoules

Heat of fusion for Al:
Lf = 0.395 kJ/g
Heat_2 = m * Lf = 2,200gm * 0.395 = 869 KiloJoules

Total heat required = 1100*10^3 + 869*10^3 = 1969 kJ
I'll roundoff to 2000 KiloJoules = 2 kilowatt-seconds


I'm a bit confused. I thought one joule was one watt*second. If so,
wouldn't 2000 kilojoules be 2000 kilowatt*seconds or 2 million
watt*seconds?

I forgot to add:

Also, the flaw in the ointment is that this method of estimating implies
that the aluminum will melt with one watt applied for 2 million seconds.
I don't think so.

 

[George Herold]

On Sunday, July 5, 2015 at 10:37:32 PM UTC-4, John S wrote:

On 7/5/2015 11:00 AM, Clifford Heath wrote:
On 06/07/15 07:59, makolber@yahoo.com wrote:
For sake of discussion, assume the tx line is 50 ohms.
If the tx output is an ideal 50 ohm match, any reflection from the ant
will be absorbed by the tx.
If the tx however, instead present a conjugate match to whatever z the
line present it,
Any reflection from the ant will be completely re- reflected by the tx
back to the ant.
The conjugate matched case will provide more radiated power.

At the cost, as Tim said, of possibly excessive voltages or currents.

A friend melted the aluminium tube driven elements on his six-metre Yagi
antenna with a 1000W HF amp. The amp, cable and connectors were fine...
the solid tube *elements* melted from excessive current.

How much current (power, temperature) does it take to melt a (solid?)
aluminium (sic) tube (rod?) driven element? An inquiring mind wants to know.

If it's DC you can look up fusing currents of wires.
For AC it get more complicated.
(skin effect and all that.)

George H.

 

[rickman]

On 7/6/2015 12:10 AM, Jeff Liebermann wrote:

On Sun, 05 Jul 2015 21:37:58 -0500, John S <Sophi.2@invalid.org
wrote:

On 7/5/2015 11:00 AM, Clifford Heath wrote:
On 06/07/15 07:59, makolber@yahoo.com wrote:
For sake of discussion, assume the tx line is 50 ohms.
If the tx output is an ideal 50 ohm match, any reflection from the ant
will be absorbed by the tx.
If the tx however, instead present a conjugate match to whatever z the
line present it,
Any reflection from the ant will be completely re- reflected by the tx
back to the ant.
The conjugate matched case will provide more radiated power.

At the cost, as Tim said, of possibly excessive voltages or currents.

A friend melted the aluminium tube driven elements on his six-metre Yagi
antenna with a 1000W HF amp. The amp, cable and connectors were fine...
the solid tube *elements* melted from excessive current.

How much current (power, temperature) does it take to melt a (solid?)
aluminium (sic) tube (rod?) driven element? An inquiring mind wants to know.

Ok, let's see if my thermodynamics is still intact. Corrections
welcome as my math usually sucks and I always make at least one big
mistake.


I'll assume a 3 element 6 meter beam. If it had more elements, this
would be a good time to disclose the details.
http://www.cushcraftamateur.com/Product.php?productid=A50-3S
3.2 kg for the entire antenna. I would guess(tm) that the boom is
about 1.0 Kg, leaving 2.2 Kg of aluminum for the elements.

6061-T6 melts at about 580C = 853K
I'll assume an ambient temp of 27C = 300K

Heat absorbed before melting:
Specific heat capacity of Al:
C = 0.897 J/(gm*K)
Heat_1 = 0.897 * m * change_in_temp
Heat_1 = 0.897 * 2,200gm * 553K = 1100*10^3 KiloJoules

Please check your math. I think you made an error of 10^3.

Heat of fusion for Al:
Lf = 0.395 kJ/g
Heat_2 = m * Lf = 2,200gm * 0.395 = 869 KiloJoules

Total heat required = 1100*10^3 + 869*10^3 = 1969 kJ
I'll roundoff to 2000 KiloJoules = 2 kilowatt-seconds

But I guess you corrected it here.

So, if 100% of the 1000 watts of power was dissipated by the driven
elements of the antenna, it would melt in 2 seconds. However, since
it's unlikely all 1000 watts actually made it to the antenna, and that
it's likely that the antenna actually radiated much of the power, the
actual meltdown times will be much longer. Still, it looks quite
possible to melt the elements.

--

Rick