Detecting... load, current... um...

[John Fields]

On Sat, 04 Jul 2015 01:29:52 -0500, John Fields
<jfields@austininstruments.com> wrote:

On Sat, 04 Jul 2015 16:24:12 +1000, John G <john.g@green.com> wrote:

John G has brought this to us :
on 4/07/2015, John Larkin supposed :
On Wed, 1 Jul 2015 10:01:20 -0700 (PDT), George Herold
gherold@teachspin.com> wrote:

On Wednesday, July 1, 2015 at 1:17:18 AM UTC-4, Phil Allison wrote:
Phil Allison wrote:


** I would use a reed switch with a few dozen turns or so of wire wound
around it - eg:

http://www.arunet.co.uk/tkboyd/ec/ec1RehaReedP0331.jpg


** Probably I should explain that a bit more.

The coil wound around the reed switch *replaces* the previously mentioned
resistor in line with the 12V DC supply.

Adjust the number of turns ( which may be wound on a plastic tube for
convenience ) so the reed switch reliably closes when the current reaches
the higher level.

A 20mm long reed switch operates in less than a millisecond and needs
about 20 turns at 1 amp DC.


.... Phil

That's a neat idea Phil. Would it work with the wire just wrapped around
the body of the reed switch? It been decades since I use a reed
switch, but don't you want the magnetic field perpendicular to the
contact.

Axial, usually, a solenoid wound on the reed capsule. A hazard may be
the large pickup-dropoff (Peekup Andropov in CarTalk language) ratio.
A reed may energize at X amps and drop out at X/5.

Reed relays are rare these days, and deserve it.

Mouser has many thousands in stock from about from about $1.20

Actually he needs a reed SWITCH just the glas bit.
Not a reed RELAY which is a box with some windings as well. :-Z

---
It's generally considered to be a reed _switch_ when it's actuated
by a permanent magnet, and a reed _relay_ when it's actuated by an
electromagnet, just like a conventional non-reed relay.

---
Oops... you're right.

When he buys it it'll be a reed switch, but when he finishes it
it'll be a reed relay.

John Fields

 

[John G]

John Fields has brought this to us :

On Sat, 04 Jul 2015 01:29:52 -0500, John Fields
jfields@austininstruments.com> wrote:

On Sat, 04 Jul 2015 16:24:12 +1000, John G <john.g@green.com> wrote:

John G has brought this to us :
on 4/07/2015, John Larkin supposed :
On Wed, 1 Jul 2015 10:01:20 -0700 (PDT), George Herold
gherold@teachspin.com> wrote:

On Wednesday, July 1, 2015 at 1:17:18 AM UTC-4, Phil Allison wrote:
Phil Allison wrote:


** I would use a reed switch with a few dozen turns or so of wire
wound around it - eg:

http://www.arunet.co.uk/tkboyd/ec/ec1RehaReedP0331.jpg


** Probably I should explain that a bit more.

The coil wound around the reed switch *replaces* the previously
mentioned resistor in line with the 12V DC supply.

Adjust the number of turns ( which may be wound on a plastic tube for
convenience ) so the reed switch reliably closes when the current
reaches the higher level.

A 20mm long reed switch operates in less than a millisecond and needs
about 20 turns at 1 amp DC.


.... Phil

That's a neat idea Phil. Would it work with the wire just wrapped
around the body of the reed switch? It been decades since I use a reed
switch, but don't you want the magnetic field perpendicular to the
contact.

Axial, usually, a solenoid wound on the reed capsule. A hazard may be
the large pickup-dropoff (Peekup Andropov in CarTalk language) ratio.
A reed may energize at X amps and drop out at X/5.
Reed relays are rare these days, and deserve it.
Mouser has many thousands in stock from about from about $1.20

Actually he needs a reed SWITCH just the glas bit.
Not a reed RELAY which is a box with some windings as well. :-Z

---
It's generally considered to be a reed _switch_ when it's actuated
by a permanent magnet, and a reed _relay_ when it's actuated by an
electromagnet, just like a conventional non-reed relay.

---
Oops... you're right.

When he buys it it'll be a reed switch, but when he finishes it
it'll be a reed relay.

John Fields

My only intention on pointing out the difference was that they are in
different groups in the Mouser and other catalogues.
I wasn't meaning to be pedantic or anything.

--
John G Sydney.

 

[whit3rd]

On Saturday, July 4, 2015 at 7:34:27 AM UTC-7, Jasen Betts wrote:

On 2015-07-03, whit3rd <whit3rd@gmail.com> wrote:
On Friday, July 3, 2015 at 2:17:00 AM UTC-7, Phil Allison wrote:
whit3rd wrote:

...it may be more practical to use a current-sense amplifier like
http://datasheets.maximintegrated.com/en/ds/MAX4373-MAX4375.pdf

** You for real ?

Some oddball, one maker SMD chip...

... I just picked that part number because there's
about twenty thousand showing in stock...
LT6108, PAC1720, IR2175, FAN4010, INA300, are other manufacturers' functional
equivalents (but there's not much agreement on pinout and feature set).

so how do you guess which will be available in 10 years time?
comparitors and resitors seem to have long-term availability.

In best-manufacturing-practices style, you'd pick one or two from the list,
making multiple sections on the circuit board if necessary to accomodate.
It's usual for manufacturers or distributors to give warning before end-of-life
for the product, if they KNOW you're a customer, so buy a few. Or, a few thousand.

And the same goes for the reed-switch parts, except you'd have different turns
counts specified around the different available reed capsules.

If you'd designed in a '741 op amp in 1970, it would have been 'some oddball,
single manufacturer' chip, and you'd have a headache when the 14-pin package
was deprecated, but it still might be a good decision.

 

On Saturday, July 4, 2015 at 2:32:51 PM UTC-4, John Fields wrote:

Since there are no DC transformers, if you mean a transformer with a
12.6VRMS output, that's about 17.8 volts, peak, and subtracting out
about 1.5 volts for two diode drops through the bridge will leave
you with about 16 volts, DC across the cap.

Then, since your supply output is unregulated, instead of the 12
volts you've called out it'll be 16 volts, so we've got a whole new
ball game.

Well... damn. That plus the ripple mean I can scrap my power supply.

Realistically, I should just buy a 12v, 5A power supply off Amazon. Chop off the plug and done. It will be regulated and they've got one for $8, cheaper than I can get a 12v transformer 4A anyway.

My only hesitation is that I've used them for years, in the guise of laptop power bricks, and they seem to fry on the first surge (my house is prone to lightning) or unexpected load (my designs being very amatuer, I always expect I'll do something unexpected.) I've been using an unregulated trans, diode bridge and honking big cap since I was a kid, largely because even I don't seem able to kill them.

I want this to run reliably for years. What's my best bet for a 12v 5A regulated supply that thrives on abuse and won't catch fire if abused?

 

[John Fields]

On Sat, 4 Jul 2015 02:01:42 -0700 (PDT), scott.a.mayo@gmail.com
wrote:


>Here's what I came up with in Spice, starting from the previous design offered:

..
..
..

>Some notes - I don't know how to model the 400ma LED array as a load, so R11 stands in for that.

---
That'll work.
---

In reality it's inactive whenever the dc converter is; I just couldn't get a SPDT relay out of >spice.

I tweaked R7 to change the triggering current up around 600ma+. It looked like a voltage divider and fair game for adjusting. (In practice: 4.7k + 5k pot)

---
OK
---

>I don't know how to model a dc-dc converter, so R9 stands in for that. I've made sure the circuit simulates well for resistances between 2 ohms and 300 ohms, on the guess that the converter can't possibly be outside that range.

---
OK
---

>C1 and R10 are insurance - they cause a current spike when the relay first closes, large enough to guarantee the detector fires even if the converter doesn't draw as much as I think. They cause the detector to fire for at least 50ms, enough for the PIC downstream to see it. If the converter load load is large enough they'll be dropped.

---
OK
---

>The power supply is from a 12.6Vdc, 4A transformaer, taken through an 8A diode full wave bridge and a largish cap, unregulated.

---
Since there are no DC transformers, if you mean a transformer with a
12.6VRMS output, that's about 17.8 volts, peak, and subtracting out
about 1.5 volts for two diode drops through the bridge will leave
you with about 16 volts, DC across the cap.

Then, since your supply output is unregulated, instead of the 12
volts you've called out it'll be 16 volts, so we've got a whole new
ball game.
---

>Questions: if R6, R& and R8 just set the comparison voltage, would a different voltage regulator other than 5V have made sense at U2?

---
The 5V isn't holy, it just needs to be high enough to allow U1B's
trip point to be adjusted so that with the desired load current
through the shunt (R5) U1B's output will go high.

U1A is a difference amplifier with a gain of ten, so if your target
trip current is 600mA through the 0.1 ohm shunt, that'll drop 60
millivolts across the shunt. U1A multiplies that by ten, so for 600
mA through R5, 600 millivolts will show up on U1B+. Then, to make
U1B's output go high when U1B+ has >=600 millivolts on it, U1B- must
have a tiny bit less than 600 millivolts on it. That voltage is
generated by U2, a shunt reference, in conjunction with the voltage
divider R7R8, with the junction of R7 and R8 connected to U1B-. R6
is used to provide the operating current for R1, and doesn't
directly affect the voltage on U1B-.

Bottom line is that as long as the output of U2 is enough above 600
millivolts so that its output can be adjusted to 600 millivolts with
R7R8, it doesn't really matter what U2's output voltage is.

So, you could use pretty much any cheap garden-variety Zener there
as long as you fed it enough current to keep its output stable and
then use R7R8 (or a pot, for that matter) to adjust the voltage into
U1B- to 600 millivolts,
---

>When the dc converter is operational, it's probably going to draw at least 2A at 12v. Is power supply ripple going to be a problem?

---
90% yes.

To know for sure you have to decide on whether you want to/have to
regulate the raw supply output, and what the maximum supply current
needs to be.

 

[John Fields]

On Sat, 4 Jul 2015 12:30:45 -0700 (PDT), ikrastsphere@gmail.com
wrote:

On Saturday, July 4, 2015 at 2:32:51 PM UTC-4, John Fields wrote:

Since there are no DC transformers, if you mean a transformer with a
12.6VRMS output, that's about 17.8 volts, peak, and subtracting out
about 1.5 volts for two diode drops through the bridge will leave
you with about 16 volts, DC across the cap.

Then, since your supply output is unregulated, instead of the 12
volts you've called out it'll be 16 volts, so we've got a whole new
ball game.

Well... damn. That plus the ripple mean I can scrap my power supply.

---
If you have a brute force power supply on hand, maybe you can just
regulate its output?
---

Realistically, I should just buy a 12v, 5A power supply off Amazon. Chop off the plug and done. It will be regulated and they've got one for $8, cheaper than I can get a 12v transformer 4A anyway.

My only hesitation is that I've used them for years, in the guise of laptop power bricks, and they seem to fry on the first surge (my house is prone to lightning) or unexpected load (my designs being very amatuer, I always expect I'll do something unexpected.) I've been using an unregulated trans, diode bridge and honking big cap since I was a kid, largely because even I don't seem able to kill them.

I want this to run reliably for years. What's my best bet for a 12v 5A regulated supply that thrives on abuse and won't catch fire if abused?

---
Don't know of anything that _thrives_ on abuse , but this looks
pretty good:

http://www.amazon.com/gp/offer-listing/B00D7CWSCG/ref=dp_olp_new?ie=UTF8&condition=new

Read the reviews...

John Fields

 

Just a note of thanks; I just breadboarded up the suggested current detector (based on the LT1366) and it worked exactly as described on the first try.

 

[John Fields]

On Tue, 21 Jul 2015 22:38:51 -0700 (PDT), scott.a.mayo@gmail.com
wrote:

>Just a note of thanks; I just breadboarded up the suggested current detector (based on the LT1366) and it worked exactly as described on the first try.

---
YAY!

You're welcome.

John Fields