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Today's topics:

* Riding an air bag - 1 messages, 1 author
http://groups.google.com/group/sci.electronics.repair/t/0d8fff4e5e1ab554?hl=en
* OT: Is this question too challenging for a BSEE graduate? - 4 messages, 4
authors
http://groups.google.com/group/sci.electronics.repair/t/31ba2b6a402a3720?hl=en

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TOPIC: Riding an air bag
http://groups.google.com/group/sci.electronics.repair/t/0d8fff4e5e1ab554?hl=en
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== 1 of 1 ==
Date: Tues, Oct 26 2010 7:25 pm
From: RoadRunner

If you think that sitting on an air bag in an attempt to get airborne
might be a bad idea, you would be correct!

http://www.craigboyce.com/w/2010/10/riding-an-air-bag/

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TOPIC: OT: Is this question too challenging for a BSEE graduate?
http://groups.google.com/group/sci.electronics.repair/t/31ba2b6a402a3720?hl=en
==============================================================================

== 1 of 4 ==
Date: Tues, Oct 26 2010 8:06 pm
From: ehsjr

David Nebenzahl wrote:
> On 10/25/2010 10:37 PM ehsjr spake thus:
>
>> I think you might be missing the fact that the circuit has feedback,
>> from the output to the inverting input. Take a look at the schematic
>> again.
>>
>> The op amp "sees" a voltage on the + input and does whatever it can
>> to make the - input the same voltage. The output of the op amp is
>> connected back to the - input, so when the op amp raises or lowers
>> the voltage on the output pin, that voltage appears on the - input.
>> Thus, if you put X volts on the non-inverting (+) input, you'll get
>> X volts on the inverting (-) input.
>
>
> Now that I understand things a little better, yes, I do get the feedback
> here, and in other op amp circuits.
>
> But just a small quibble with the way you and others have described
> what's going on here. You say "the op amp ... does whatever it can to
> make the - input the same voltage" (as the + input). In fact, it does no
> such thing: the input is, after all, just an input.

In fact, _it most certainly does_. Without external connection - the
feedback - it will fail to make the inputs equal. But it will do all
it can do until those voltages are equal.

You seem to have missed the fact that my entire post was discussing
feedback. Sigh.

Ed

>
> What you might ought have said is that the *circuit*, including the
> feedback loop, forces the inverting input to (virtually) the same
> voltage as the noninverting input, right? The op amp, in and of itself,
> doesn't "do" anything to (that is, out of) either input. It's only by
> virtue of the feedback that this action occurs.
>
> Maybe just a semantic quibble. Other than that I'm with you here. Thanks
> to all for explaining.
>
>

== 2 of 4 ==
Date: Tues, Oct 26 2010 8:49 pm
From: David Nebenzahl

On 10/26/2010 8:06 PM ehsjr spake thus:

> David Nebenzahl wrote:
>
>> But just a small quibble with the way you and others have described
>> what's going on here. You say "the op amp ... does whatever it can to
>> make the - input the same voltage" (as the + input). In fact, it does no
>> such thing: the input is, after all, just an input.
>
> In fact, _it most certainly does_. Without external connection - the
> feedback - it will fail to make the inputs equal. But it will do all
> it can do until those voltages are equal.

You mean the little op amp will huff and puff and turn blue? Not sure
what you're getting at here. Without feedback, the inputs will simply be
whatever they are.

> You seem to have missed the fact that my entire post was discussing
> feedback. Sigh.

See the remainder of my reply that you posted below but didn't reply to.

>> What you might ought have said is that the *circuit*, including the
>> feedback loop, forces the inverting input to (virtually) the same
>> voltage as the noninverting input, right? The op amp, in and of itself,
>> doesn't "do" anything to (that is, out of) either input. It's only by
>> virtue of the feedback that this action occurs.

--
The fashion in killing has an insouciant, flirty style this spring,
with the flaunting of well-defined muscle, wrapped in flags.

- Comment from an article on Antiwar.com (http://antiwar.com)

== 3 of 4 ==
Date: Tues, Oct 26 2010 9:18 pm
From: "Michael A. Terrell"

tm wrote:
>
> "Michael A. Terrell" <mike.terrell@earthlink.net> wrote in message
> news:3YSdndK57NrYGFrRnZ2dnUVZ_qWdnZ2d@earthlink.com...
> >
> > PlainBill47@yawho.com wrote:
> >>
> >> On Sun, 24 Oct 2010 21:14:00 -0400, "Michael A. Terrell"
> >> <mike.terrell@earthlink.net> wrote:
> >>
> >> >
> >> >PlainBill47@yawho.com wrote:
> >> >>
> >> >> On Sat, 23 Oct 2010 13:05:23 -0700, "William Sommerwerck"
> >> >> <grizzledgeezer@comcast.net> wrote:
> >> >>
> >> >> >> Any electronics technician should be able to solve this
> >> >> >> by inspection; no calculator necessary. 3K/1k = x/40,
> >> >> >> therefore x= 120 ohms. Come up with more difficult ones
> >> >> >> next time.
> >> >> >
> >> >> >What do you mean by "inspection"? Are you applying a formula you
> >> >> >memorized?
> >> >> >Or do you /understand/ what's involved?
> >> >> >Probably better than you do. The voltage across R1 is 1/4 of +VDC.
> >> >> >An op-amp tries to force both inputs to the same voltage. Since it
> >> >> >was stipulated the op-amp is a 'classic, ideal' op amp, we can assume
> >> >> >it has none of the defects found in the real world. As a result the
> >> >> >voltage across R3 will also be 1/4 of VDC. The only way that can
> >> >> >happen is if the effective resistance of Q1 is 3 times the resistance
> >> >> >of R3, or 120 ohms.
> >> >>
> >> >> NOW, what is less certain is the proper answer to the problem
> >> >> "Calculate the equivalent resistance of this programmable load."
> >> >> Given that R1, R2, and R3 are all part of the load, the proper answer
> >> >> to the original diagram is 153.846 ohms. Except that circuit does not
> >> >> show any evidence of being 'programmable'.
> >> >
> >> >
> >> > You 'program' it by changing reistors.
> >> No kidding? You replace a simple variable power resistor, which only
> >> requires a screw driver to change the resistance with three resistors,
> >> an op-amp (which requires a separate power supply), and a mosfet. To
> >> change the value of the virtual resistor you have to change a
> >> resistor?
> >>
> >> It would seem to me a potentiometer would improve usability greatly.
> >
> >
> > It would seem to me you aren't familiar with a customized BOM or
> > SIT. I worked with both in manufacturing.
> >
>
> Ok, I get it now. It was for a cost plus government job.

No, you don't get it. Customized BOM allows lower costs by reusing
most of a design for different customer needs. SIT (Select In Test)
allows equipment to be built that exceeds the normal specifications of
the current state of the art. HP and Tektronix used to do it in a lot
of their products, as well. There was no 'cost plus' jobs for anyone.
We would get a request for bid, or inquiry to see if we could supply
what a customer needed. We would bid it, based on a current base model
and all required customization to meet their needs. If they agreed to
the price we built it and shipped it on schedule and on budget. It
didn't matter who the customer was. NASA, NOAA, the ESA or numerous
other customers in the Aerospace industry. Some of that equipment was
in use 24/7 for over 30 years with no repairs.

If you think that you can build state of the art microwave receivers
with off the shelf parts, you need help.

--
Politicians should only get paid if the budget is balanced, and there is
enough left over to pay them.

== 4 of 4 ==
Date: Tues, Oct 26 2010 9:48 pm
From: "tm"

"Michael A. Terrell" <mike.terrell@earthlink.net> wrote in message
news:3sqdnbd75qeRNVrRnZ2dnUVZ_qWdnZ2d@earthlink.com...
>
> tm wrote:
>>
>> "Michael A. Terrell" <mike.terrell@earthlink.net> wrote in message
>> news:3YSdndK57NrYGFrRnZ2dnUVZ_qWdnZ2d@earthlink.com...
>> >
>> > PlainBill47@yawho.com wrote:
>> >>
>> >> On Sun, 24 Oct 2010 21:14:00 -0400, "Michael A. Terrell"
>> >> <mike.terrell@earthlink.net> wrote:
>> >>
>> >> >
>> >> >PlainBill47@yawho.com wrote:
>> >> >>
>> >> >> On Sat, 23 Oct 2010 13:05:23 -0700, "William Sommerwerck"
>> >> >> <grizzledgeezer@comcast.net> wrote:
>> >> >>
>> >> >> >> Any electronics technician should be able to solve this
>> >> >> >> by inspection; no calculator necessary. 3K/1k = x/40,
>> >> >> >> therefore x= 120 ohms. Come up with more difficult ones
>> >> >> >> next time.
>> >> >> >
>> >> >> >What do you mean by "inspection"? Are you applying a formula you
>> >> >> >memorized?
>> >> >> >Or do you /understand/ what's involved?
>> >> >> >Probably better than you do. The voltage across R1 is 1/4 of
>> >> >> >+VDC.
>> >> >> >An op-amp tries to force both inputs to the same voltage. Since it
>> >> >> >was stipulated the op-amp is a 'classic, ideal' op amp, we can
>> >> >> >assume
>> >> >> >it has none of the defects found in the real world. As a result
>> >> >> >the
>> >> >> >voltage across R3 will also be 1/4 of VDC. The only way that can
>> >> >> >happen is if the effective resistance of Q1 is 3 times the
>> >> >> >resistance
>> >> >> >of R3, or 120 ohms.
>> >> >>
>> >> >> NOW, what is less certain is the proper answer to the problem
>> >> >> "Calculate the equivalent resistance of this programmable load."
>> >> >> Given that R1, R2, and R3 are all part of the load, the proper
>> >> >> answer
>> >> >> to the original diagram is 153.846 ohms. Except that circuit does
>> >> >> not
>> >> >> show any evidence of being 'programmable'.
>> >> >
>> >> >
>> >> > You 'program' it by changing reistors.
>> >> No kidding? You replace a simple variable power resistor, which only
>> >> requires a screw driver to change the resistance with three resistors,
>> >> an op-amp (which requires a separate power supply), and a mosfet. To
>> >> change the value of the virtual resistor you have to change a
>> >> resistor?
>> >>
>> >> It would seem to me a potentiometer would improve usability greatly.
>> >
>> >
>> > It would seem to me you aren't familiar with a customized BOM or
>> > SIT. I worked with both in manufacturing.
>> >
>>
>> Ok, I get it now. It was for a cost plus government job.
>
>
> No, you don't get it. Customized BOM allows lower costs by reusing
> most of a design for different customer needs. SIT (Select In Test)
> allows equipment to be built that exceeds the normal specifications of
> the current state of the art. HP and Tektronix used to do it in a lot
> of their products, as well. There was no 'cost plus' jobs for anyone.
> We would get a request for bid, or inquiry to see if we could supply
> what a customer needed. We would bid it, based on a current base model
> and all required customization to meet their needs. If they agreed to
> the price we built it and shipped it on schedule and on budget. It
> didn't matter who the customer was. NASA, NOAA, the ESA or numerous
> other customers in the Aerospace industry. Some of that equipment was
> in use 24/7 for over 30 years with no repairs.
>
> If you think that you can build state of the art microwave receivers
> with off the shelf parts, you need help.
>

I really do get it. I see you never worked on any "black" projects.

BOM and SIT are EE-101 so don't try to be so impressive with what you
think you know.

And by the way, I am not trying to piss you off. I was just trying to be a
bit sarcastic. By making a simple "resistor" using five parts and calling
it a "programmable" load sure sounds like someone was selling a design
at cost plus.

Regards (really),
tm

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