Driver to drive?

[John Larkin]

On Thu, 17 Jul 2014 04:12:48 +0000 (UTC), mroberds@att.net wrote:

Crossposted and followup-to sci.electronics.design .

cassiano@gmail.com wrote:
I work at a physics lab and we have some custom hardware that have
been developed by researches along the years. We are trying to fix one
of them, but it uses a component that we are having a hard time to
find: It is a N-channel dual gate GaAs MES FET by Sony called 3SK166A.

Just from Googling, one broker claims to have stock on a 3SK166,
3SK166-1, and 3SK166A. I have never dealt with this broker before.

http://www.utsource.net/3SK166.html

Brokers like this one buy up inventories of old parts from other
suppliers, and/or send out random parts with false part numbers on them.
Sometimes you will get the genuine part at a bargain price, and
sometimes you will get a part that does not work at all.

We need help finding a compatible replacement for this component.

I have cross-posted this reply to sci.electronics.design, which is much
more active than sci.electronics.components . It also has people
smarter than I am who might be able to suggest a replacement.

What claims to be the Sony datasheet is available at:

http://pdf.datasheetcatalog.com/datasheets/90/206535_DS.pdf
http://cs.utsource.net/goods_files/pdf/46/46558_SONY_3SK166A.pdf

Some of the interesting parameters seem to be: Vds 8 V, Vgs -6 V,
Id 80 mA, Pd 150 mW, gain about 20 dB. The datasheet gives S
parameters and noise figures for a range of 200 to 2,000 MHz. The
package is a Sony M-254; it doesn't list an EIA or JEDEC package code.

Standard disclaimers apply: I don't get money or other consideration
from any companies mentioned.

Matt Roberds

Looks a lot like an NE25139, also obsolete.


--

John Larkin Highland Technology Inc
www.highlandtechnology.com jlarkin at highlandtechnology dot com

Precision electronic instrumentation

 

[JW]

On Thu, 17 Jul 2014 04:12:48 +0000 (UTC) mroberds@att.net wrote in Message
id: <lq7ifu$5gc$1@dont-email.me>:

Crossposted and followup-to sci.electronics.design .

cassiano@gmail.com wrote:
I work at a physics lab and we have some custom hardware that have
been developed by researches along the years. We are trying to fix one
of them, but it uses a component that we are having a hard time to
find: It is a N-channel dual gate GaAs MES FET by Sony called 3SK166A.

Just from Googling, one broker claims to have stock on a 3SK166,
3SK166-1, and 3SK166A. I have never dealt with this broker before.

http://www.utsource.net/3SK166.html

Brokers like this one buy up inventories of old parts from other
suppliers, and/or send out random parts with false part numbers on them.
Sometimes you will get the genuine part at a bargain price, and
sometimes you will get a part that does not work at all.

They seem to have an Ebay presence as well, selling this part. It may be
easier getting it from there.

http://www.ebay.com/sch/i.html?_trksid=p2050601.m570.l1313.TR0.TRC0.H0&_nkw=3SK166A&_sacat=0&_from=R40

 

[Jeff Layman]

On 15/07/2014 14:33, rramlal@lakeheadu.ca wrote:

On Monday, November 25, 1996 3:00:00 AM UTC-5, Thomas N. Lockyer wrote:
In article <57bf57$7rd@mercury.galstar.com> tpappano@galstar.com (Thomas Pappano) writes:
From: tpappano@galstar.com (Thomas Pappano)
Subject: Re: klystron vs. magnetron
Date: 25 Nov 1996 06:42:47 GMT

(snip)

> Yes you can still get those tube's, Toshiba sells them for about quarter mill.

Has the price gone up or down in the 18 years since the post you replied
to was made?

--

Jeff

 

[Jan Panteltje]

On a sunny day (Fri, 18 Jul 2014 18:28:51 +0100) it happened Jeff Layman
<JMLayman@invalid.invalid> wrote in <lqblgi$c6v$1@news.albasani.net>:

On 15/07/2014 14:33, rramlal@lakeheadu.ca wrote:
On Monday, November 25, 1996 3:00:00 AM UTC-5, Thomas N. Lockyer wrote:
In article <57bf57$7rd@mercury.galstar.com> tpappano@galstar.com (Thomas Pappano) writes:
From: tpappano@galstar.com (Thomas Pappano)
Subject: Re: klystron vs. magnetron
Date: 25 Nov 1996 06:42:47 GMT

(snip)

Yes you can still get those tube's, Toshiba sells them for about quarter mill.

Has the price gone up or down in the 18 years since the post you replied
to was made?

--

Jeff

http://www.ebay.com/sch/i.html?_trksid=p2050601.m570.l1313.TR0.TRC0.H0.Xklystron&_nkw=klystron&_sacat=0&_from=R40

208 hits, cheapest on top of page 1:
QKK753 Reflex Klystron tube NOS $18.95

 

On Monday, 25 November 1996 08:00:00 UTC, Thomas Pappano wrote:

Magnetrons are a specialized diode that operates in the influence of
a magnetic field and usually have integral resonant circuitry that
defines the frequency band they run in. They
are used in pulsed applications such as
pulse radar and also in continuous wave applications like microwave
ovens.

Every microwave oven I have looked at (with an antenna and diode detector) is pulsed. I think each pulse lasts around 100us with a repetition rate of 100Hz (in the UK) at full power. In low-power modes the pulse repetition rate is reduced but the peak power is unchanged.

The frequency is unstable, so as the turntable rotates with a slightly asymmetric load the frequency drifts up and down (within the ISM band)

John

 

[Tim Wescott]

On Mon, 21 Jul 2014 02:04:24 -0700, jrwalliker wrote:

On Monday, 25 November 1996 08:00:00 UTC, Thomas Pappano wrote:

Magnetrons are a specialized diode that operates in the influence of a
magnetic field and usually have integral resonant circuitry that
defines the frequency band they run in. They are used in pulsed
applications such as pulse radar and also in continuous wave
applications like microwave ovens.

Every microwave oven I have looked at (with an antenna and diode
detector) is pulsed. I think each pulse lasts around 100us with a
repetition rate of 100Hz (in the UK) at full power. In low-power modes
the pulse repetition rate is reduced but the peak power is unchanged.

The frequency is unstable, so as the turntable rotates with a slightly
asymmetric load the frequency drifts up and down (within the ISM band)

Can we hand out awards for most years elapsed between question and
answer? Nearly 18 years may not be a record, but it's pretty good, all
in all.

--

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

 

On Tuesday, July 29, 2014 3:07:08 AM UTC+1, Michael Covington wrote:

> Hello everyone. I am an electronics student, and I am having trouble with one the problems in my text. The problem is a simple circuit with a 6V supply voltage, and two resistors in series. One resistor (R1) is 500kOhms and the other resistor (R2) doesn't have a value, but there is supposed to be a voltage of 1V at R2. I know that the current stays the same in a series circuit, but I tried using the current I calculated just using the info I have...6V and a 500k Ohm, but still cannot figure out how to get the resistance of R2!! I tried Kirchoff's loop law and simply have a difference of 5 volts. I would really appreciate it if someone could give me a hand with this, so I will know what i am doing in the future.

V=IR so for a given i, V is proportional to R. So in a chain where you have 5v across 500k, you have 1v across 100k.


NT

 

[Adrian Tuddenham]

Michael Covington <michaelcovington670@gmail.com> wrote:

...The problem is a simple circuit with a 6V
supply voltage, and two resistors in series. One resistor (R1) is
500kOhms and the other resistor (R2) doesn't have a value, but there is
supposed to be a voltage of 1V at R2.

Be careful with the terminology. The voltage appears *across* R2, not
"at" it.

Imagine the circuit is already set up and running with the correct value
for R2 (whatever it may be). You have a 6v supply and 1v is being lost
across R2, this leaves 5v across R1.

The value of R1 is 500kOhms, so the current through it (by Ohms Law)
must be 5/500000 = 10 microamps (1 x 10^-5 amps)

To develop 1 volt across R2 with a 10 microamp current through it, Ohms
Law says it would have to be 1/ (1 x 10^-5) Ohms = 100kOhms


As a quick check, you now have a potential divider of a 500k resistor in
series with a 100k resistor, the 500k drops 5v, the 100k drops 1v and
the total voltage is 6v. This looks correct


--
~ Adrian Tuddenham ~
(Remove the ".invalid"s and add ".co.uk" to reply)
www.poppyrecords.co.uk

 

[John Fields]

On Mon, 28 Jul 2014 19:07:08 -0700 (PDT), Michael Covington
<michaelcovington670@gmail.com> wrote:

Hello everyone. I am an electronics student, and I am having trouble with one the problems in my text. The problem is a simple circuit with a 6V supply voltage, and two resistors in series. One resistor (R1) is 500kOhms and the other resistor (R2) doesn't have a value, but there is supposed to be a voltage of 1V at R2. I know that the current stays the same in a series circuit, but I tried using the current I calculated just using the info I have...6V and a 500k Ohm, but still cannot figure out how to get the resistance of R2!! I tried Kirchoff's loop law and simply have a difference of 5 volts. I would really appreciate it if someone could give me a hand with this, so I will know what i am doing in the future.

Thanks,
Michael

---
1. For the circuit you describe: (View using a fixed-pitch font)

.. E1
.. |
.. [R1]
.. |
.. +--E2
.. |
.. [R2]
.. |
.. GND

what you know is:

.. 6V E1
.. |
.. [500k] R1
.. |
.. +-->1V E2
.. |
.. [R2]
.. |
.. GND

Then, since the voltage across R1 is E1 - E2, the current through it
must be:

.. E1 - E2 5V
.. I = --------- = -------- = 10ľA
.. R 500kR

Since current in a series circuit is everywhere the same, that 10ľA
must also go through R2, so its resistance must be:

.. E 1V
.. R = --- = ------ = 100k ohms
.. I 10ľA


2. For the circuit you describe, the voltage divider formula can be
used to good advantage by eliminating the a priori need to know the
current or as a way to check your work.

In its simplest form, it's:

.. E1 R2
.. E2 = ---------.
.. R1 + R2

Its other permutations are:


.. E2 (R1 + R2)
.. E1 = --------------,
.. R2

.. R2 (E1 - E2)
.. R1 = --------------, and
.. E2

.. E2 R1
.. R2 = ---------,
.. E1 - E2

which, in your case, you could use to solve your problem:

.. E2 R1 1V * 500kR 500kRV
.. R2 = --------- = ------------ = -------- = 100k ohms
.. E1 - E2 6V - 1V 5V


John Fields

 

On Monday, July 28, 2014 11:41:11 PM UTC-4, John Larkin wrote:

On Mon, 28 Jul 2014 19:07:08 -0700 (PDT), Michael Covington

michaelcovington670@gmail.com> wrote:



Hello everyone. I am an electronics student, and I am having trouble with one the problems in my text. The problem is a simple circuit with a 6V supply voltage, and two resistors in series. One resistor (R1) is 500kOhms and the other resistor (R2) doesn't have a value, but there is supposed to be a voltage of 1V at R2. I know that the current stays the same in a series circuit, but I tried using the current I calculated just using the info I have...6V and a 500k Ohm, but still cannot figure out how to get the resistance of R2!! I tried Kirchoff's loop law and simply have a difference of 5 volts. I would really appreciate it if someone could give me a hand with this, so I will know what i am doing in the future.



Thanks,

Michael



The current does NOT stay the same! That wouldn't make sense.

Maybe in your inebriated state...a series circuit by definition means "the same" current passes through all the components. If this "same" current has magnitude I, then I=V1/R1=V2/R2 from which R2/R1=V2/V1 or the resistance ratios are equal to the corresponding voltage drop ratios. Since the voltage drop ratio of V2/V1=1V/5V= 1/5 then that's the resistor ratio, making R2= 1/5 R1= 100K.

The current will be I = 6/(R1+R2), because R1+R2 is the resistance that the

battery sees.



So you can do a couple of simultaneous equations, one the thing above and

another one for R2. But that's a nuisance, and doesn't train your intuition.



Imagine a long skinny strip of resistive material, out in the open. Let's make

it 6 inches long for fun.



Apply 6 volts on one end, and ground the other.



Now get a voltmeter, ground the - probe, and slide the + probe along the strip.

Obviously you'll have 0 volts at one end and 6 volts at the other end.



How far along the strip will you see 3 volts? And where will you see 1 volt?



You could actually do this. A piece of wet paper would pretty much work.









--



John Larkin Highland Technology Inc

www.highlandtechnology.com jlarkin at highlandtechnology dot com



Precision electronic instrumentation

 

On Tuesday, July 29, 2014 12:41:05 PM UTC-4, John Larkin wrote:

On Tue, 29 Jul 2014 09:30:21 -0700 (PDT), bloggs.fredbloggs.fred@gmail.com

wrote:



On Monday, July 28, 2014 11:41:11 PM UTC-4, John Larkin wrote:

On Mon, 28 Jul 2014 19:07:08 -0700 (PDT), Michael Covington



michaelcovington670@gmail.com> wrote:







Hello everyone. I am an electronics student, and I am having trouble with one the problems in my text. The problem is a simple circuit with a 6V supply voltage, and two resistors in series. One resistor (R1) is 500kOhms and the other resistor (R2) doesn't have a value, but there is supposed to be a voltage of 1V at R2. I know that the current stays the same in a series circuit, but I tried using the current I calculated just using the info I have...6V and a 500k Ohm, but still cannot figure out how to get the resistance of R2!! I tried Kirchoff's loop law and simply have a difference of 5 volts. I would really appreciate it if someone could give me a hand with this, so I will know what i am doing in the future.







Thanks,



Michael







The current does NOT stay the same! That wouldn't make sense.



Maybe in your inebriated state...a series circuit by definition means "the same" current passes through all the components.



It looked to me like the OP calculated the current with just 500K in the

circuit, and then added R2 and assumed that the current wouldn't change. Which

is why his math didn't work.



Haven't had any alcohol in days. Two days, in fact. Half a beer on Sunday..



Another simple way to do it:



Assume the result, 1 volt across R2. That leaves 5 volts across R1. NOW

calculate the current. Then compute R2.

It's an educational problem, the student is supposed to walk away from it with something more than just a number. The problem leads the student into discovering voltage drops are in the same ratio as resistor values in a series circuit.

--



John Larkin Highland Technology Inc

www.highlandtechnology.com jlarkin at highlandtechnology dot com



Precision electronic instrumentation

 

[Michael Covington]

On Tuesday, July 29, 2014 2:52:15 AM UTC-5, Adrian Tuddenham wrote:

Michael Covington <michaelcovington670@gmail.com> wrote:



...The problem is a simple circuit with a 6V

supply voltage, and two resistors in series. One resistor (R1) is

500kOhms and the other resistor (R2) doesn't have a value, but there is

supposed to be a voltage of 1V at R2.



Be careful with the terminology. The voltage appears *across* R2, not

"at" it.



Imagine the circuit is already set up and running with the correct value

for R2 (whatever it may be). You have a 6v supply and 1v is being lost

across R2, this leaves 5v across R1.



The value of R1 is 500kOhms, so the current through it (by Ohms Law)

must be 5/500000 = 10 microamps (1 x 10^-5 amps)



To develop 1 volt across R2 with a 10 microamp current through it, Ohms

Law says it would have to be 1/ (1 x 10^-5) Ohms = 100kOhms





As a quick check, you now have a potential divider of a 500k resistor in

series with a 100k resistor, the 500k drops 5v, the 100k drops 1v and

the total voltage is 6v. This looks correct





--

~ Adrian Tuddenham ~

(Remove the ".invalid"s and add ".co.uk" to reply)

www.poppyrecords.co.uk

That really helped, and makes sense to me now!! Thank you very much.

 

[John Larkin]

On Tue, 29 Jul 2014 09:30:21 -0700 (PDT), bloggs.fredbloggs.fred@gmail.com
wrote:

On Monday, July 28, 2014 11:41:11 PM UTC-4, John Larkin wrote:
On Mon, 28 Jul 2014 19:07:08 -0700 (PDT), Michael Covington

michaelcovington670@gmail.com> wrote:



Hello everyone. I am an electronics student, and I am having trouble with one the problems in my text. The problem is a simple circuit with a 6V supply voltage, and two resistors in series. One resistor (R1) is 500kOhms and the other resistor (R2) doesn't have a value, but there is supposed to be a voltage of 1V at R2. I know that the current stays the same in a series circuit, but I tried using the current I calculated just using the info I have...6V and a 500k Ohm, but still cannot figure out how to get the resistance of R2!! I tried Kirchoff's loop law and simply have a difference of 5 volts. I would really appreciate it if someone could give me a hand with this, so I will know what i am doing in the future.



Thanks,

Michael



The current does NOT stay the same! That wouldn't make sense.

Maybe in your inebriated state...a series circuit by definition means "the same" current passes through all the components.

It looked to me like the OP calculated the current with just 500K in the
circuit, and then added R2 and assumed that the current wouldn't change. Which
is why his math didn't work.

Haven't had any alcohol in days. Two days, in fact. Half a beer on Sunday.

Another simple way to do it:

Assume the result, 1 volt across R2. That leaves 5 volts across R1. NOW
calculate the current. Then compute R2.


--

John Larkin Highland Technology Inc
www.highlandtechnology.com jlarkin at highlandtechnology dot com

Precision electronic instrumentation

 

[Maynard A. Philbrook Jr.]

In article <74ket91vg2jojoukasetutjq5ok6jlrbgs@4ax.com>,
jfields@austininstruments.com says...

On Mon, 28 Jul 2014 19:07:08 -0700 (PDT), Michael Covington
michaelcovington670@gmail.com> wrote:

Hello everyone. I am an electronics student, and I am having trouble with one the problems in my text. The problem is a simple circuit with a 6V supply voltage, and two resistors in series. One resistor (R1) is 500kOhms and the other resistor (R2) doesn't have a value, but there is supposed to be a voltage of 1V at R2. I know that the current stays the same in a series circuit, but I tried using the current I calculated just using the info I have...6V and a 500k Ohm,
but still cannot figure out how to get the resistance of R2!! I tried Kirchoff's loop law and simply have a difference of 5 volts. I would really appreciate it if someone could give me a hand with this, so I will know what i am doing in the future.

Thanks,
Michael

---
1. For the circuit you describe: (View using a fixed-pitch font)

. E1
. |
. [R1]
. |
. +--E2
. |
. [R2]
. |
. GND

what you know is:

. 6V E1
. |
. [500k] R1
. |
. +-->1V E2
. |
. [R2]
. |
. GND

Then, since the voltage across R1 is E1 - E2, the current through it
must be:

. E1 - E2 5V
. I = --------- = -------- = 10ľA
. R 500kR

Since current in a series circuit is everywhere the same, that 10ľA
must also go through R2, so its resistance must be:

. E 1V
. R = --- = ------ = 100k ohms
. I 10ľA


2. For the circuit you describe, the voltage divider formula can be

Thanks for spoon feeding the future techs that won't be worth anything
at job interview.

I suppose however if ever asked to do a simple little math quiz he
could always pop open his smart phone and ring your bell here!

Personally I think you got more out of it trying to prove to the
audience that you could do it. Those types usually are lacking and know
it.


Jamie

 

[Tim Wescott]

On Mon, 28 Jul 2014 19:07:08 -0700, Michael Covington wrote:

Hello everyone. I am an electronics student, and I am having trouble
with one the problems in my text. The problem is a simple circuit with a
6V supply voltage, and two resistors in series. One resistor (R1) is
500kOhms and the other resistor (R2) doesn't have a value, but there is
supposed to be a voltage of 1V at R2. I know that the current stays the
same in a series circuit, but I tried using the current I calculated
just using the info I have...6V and a 500k Ohm, but still cannot figure
out how to get the resistance of R2!! I tried Kirchoff's loop law and
simply have a difference of 5 volts. I would really appreciate it if
someone could give me a hand with this, so I will know what i am doing
in the future.

You are being given a story problem, and are expected to turn it into a
problem in pure math, and then to solve that.

You don't have sufficient information to calculate just the current,
because while it is the same in R1 and R2, it depends on the value of
R2. You don't have sufficient information to calculate just the value of
R2 -- but you would if you knew the current.

Try writing out the equation for the voltage at (I think you mean across)
R2 as a function of the current (which you do not know -- use a symbol).
Then write out the equation for the current as a function of the supply
voltage and the two resistances. Then see if you have enough information
to correctly solve the problem.

I _highly suggest_ that you retain your units everywhere -- ohms, amps,
volts, etc. If you end up finding that the value of R2 is in volts and
the value of the current is in ohms squared, then you know you've messed
up and you can go try again (dimensional analysis has saved my butt so
many times I couldn't even begin to count the number of times -- but the
count would be in butt-saves).

--

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

 

[John Fields]

On Tue, 29 Jul 2014 14:25:18 -0400, "Maynard A. Philbrook Jr."
<jamie_ka1lpa@charter.net> wrote:

In article <74ket91vg2jojoukasetutjq5ok6jlrbgs@4ax.com>,
jfields@austininstruments.com says...

On Mon, 28 Jul 2014 19:07:08 -0700 (PDT), Michael Covington
michaelcovington670@gmail.com> wrote:

Hello everyone. I am an electronics student, and I am having trouble with one the problems in my text. The problem is a simple circuit with a 6V supply voltage, and two resistors in series. One resistor (R1) is 500kOhms and the other resistor (R2) doesn't have a value, but there is supposed to be a voltage of 1V at R2. I know that the current stays the same in a series circuit, but I tried using the current I calculated just using the info I have...6V and a 500k Ohm,
but still cannot figure out how to get the resistance of R2!! I tried Kirchoff's loop law and simply have a difference of 5 volts. I would really appreciate it if someone could give me a hand with this, so I will know what i am doing in the future.

Thanks,
Michael

---
1. For the circuit you describe: (View using a fixed-pitch font)

. E1
. |
. [R1]
. |
. +--E2
. |
. [R2]
. |
. GND

what you know is:

. 6V E1
. |
. [500k] R1
. |
. +-->1V E2
. |
. [R2]
. |
. GND

Then, since the voltage across R1 is E1 - E2, the current through it
must be:

. E1 - E2 5V
. I = --------- = -------- = 10ľA
. R 500kR

Since current in a series circuit is everywhere the same, that 10ľA
must also go through R2, so its resistance must be:

. E 1V
. R = --- = ------ = 100k ohms
. I 10ľA


2. For the circuit you describe, the voltage divider formula can be

Thanks for spoon feeding the future techs that won't be worth anything
at job interview.

---
Well, now you'll be able to claim you know how to design a voltage
divider, so for the next job interview you go to maybe how to scoop
fries won't be on the quiz...
---

I suppose however if ever asked to do a simple little math quiz he
could always pop open his smart phone and ring your bell here!

Personally I think you got more out of it trying to prove to the
audience that you could do it. Those types usually are lacking and know
it.

---
What an unkind thing to say about the audience.

John Fields

 

On Monday, July 28, 2014 11:58:28 PM UTC-4, Robert Baer wrote:

Jim Thompson wrote:

https://pbs.twimg.com/media/BtU8by1CIAENQZ3.jpg:large

I think it may be worse than that.

For a while, it was no secret that the (newer immigrant) Mexicans
were plotting to forcibly overtake the US.

Don't forget libertarians. It's rumored they want to take over the
government and leave everyone alone.

Cheers,
James Arthur

 

[nuny@bid.nes]

On Monday, July 28, 2014 8:58:28 PM UTC-7, Robert Baer wrote:

Jim Thompson wrote:

https://pbs.twimg.com/media/BtU8by1CIAENQZ3.jpg:large
...Jim Thompson

I think it may be worse than that.

For a while, it was no secret that the (newer immigrant) Mexicans
were plotting to forcibly overtake the US.

Mexicans aren't a threat; they've pretty much given up the idea of Reconquista and establishing Aztlan.

I do have two points of concern though:

1) Have you noticed that most of them are not from Mexico, but instead from places like Nicaragua and Guatemala? It used to be Mexico's policy to stop all those who wanted to pass through from *their* southern border into the U. S., but apparently they've decided to just shuttle them through.

2) Did you know that in most (if not all) South American countries, if you don't live in a city, the government pretends you don't exist (until they want your land for something)? This attitude predictably has not generated a lot of loyalty to traditions followed in those cities, like membership in the Catholic Church. Islam is growing rapidly in most of rural South America.

Now, I may well be paranoid, but I'd like to know how many of those "children" know how to make IEDs.


Mark L. Fergerson

 

[John Larkin]

On Mon, 28 Jul 2014 20:58:28 -0700, Robert Baer
<robertbaer@localnet.com> wrote:

Jim Thompson wrote:
https://pbs.twimg.com/media/BtU8by1CIAENQZ3.jpg:large

...Jim Thompson
I think it may be worse than that.
For a while, it was no secret that the (newer immigrant) Mexicans
were plotting to forcibly overtake the US.

Mexicans are, in my experience, mostly decent, hard-working, family
people. Their kids speak standard English and assimilate just fine.

Our immigration policies should be selective, however, and keep out
the minority bad ones. Every country has the right to manage
immigration.

Immigrants are clobbering working-class jobs, by creating a pool of
cheap labor. Illegal immigrants are even cheaper, since many work off
the books, for cash, without benefits.

Immigration is, at least transiently, increasing income disparity in
the US. The guys who own the painting companies are getting rich using
Chinese and Mexican labor, and the native US guys who used to be
painters are unemployed.



--

John Larkin Highland Technology, Inc

jlarkin att highlandtechnology dott com
http://www.highlandtechnology.com

 

[John Larkin]

On Mon, 28 Jul 2014 12:52:02 -0700, Jim Thompson
<To-Email-Use-The-Envelope-Icon@On-My-Web-Site.com> wrote:

https://pbs.twimg.com/media/BtU8by1CIAENQZ3.jpg:large

...Jim Thompson

About an hour ago, I hired a new EE. Young girl, right out of a
Mexican college, BSEE, and she has really unusual (ie, really good)
electrical instincts.

I've met very few Mexican, or female, circuit designers. Should be
interesting.


--

John Larkin Highland Technology, Inc

jlarkin att highlandtechnology dott com
http://www.highlandtechnology.com