Chip with simple program for Toy

[Ben Stephens]

Kruminilius W. wrote:

I'm trying to find the type of dial that you can turn in one direction or
the other without it stopping (unlike a potentiometer). I have no idea what
this part is called. Any thoughts?

Thanks.

What about those wheels in a balled computer mouse??? they do what you

want to do, but they are fragile.

 

[Jonathan Kirwan]

On 26 Jun 2005 03:38:49 -0700, "PeteS" <ps@fleetwoodmobile.com> wrote:

On the original issue, be sure to put a clamp of some description ( a
decent TVS rated below the max coil voltage of the relays) across the
12V input line.

Vehicle supplies are not 'clean' in any sense of the word (60V
transients on 'batt+' for instance). In the words of a Linear Tech app
note 'the power supply from hell'. A typical automotive power feed can
be expected to drop to about 4V during cranking and jump to between 50
- 80V during load dump.

I design automotive equipment in my current incarnation, and the main
power feed is one of my biggest issues (not only for the great power
feed, but also because I have to meet pretty tight standards on
conducted and radiated emissions into the power system from my
equipment).

Cheers

PeteS

I had heard these things and, in fact, had found a technical paper
detailing the ranges of behavior. Nasty.

However, this does beg a question. I don't know if the relays in
question are ones that are already normally in service in a car and if
the OP is just rewiring the control or if they are just picked from
the usual electronic stock. But if they are normally in use, don't
they already have to deal with this "unclean" supply?

(Of course, any electronic circuit would also need to deal with these
issues, so that would, at least, suggest the use of a PNP with a
larger Vceo capability, yes? As well as concerns about how the +5V is
actually generated in the OP's circuit?)

Jon

 

[Morris Lee]

<junk1@davidbevan.co.uk> wrote in message
news:1119714098.857656.7340@g14g2000cwa.googlegroups.com...

1. Walk back to your car to charge the lipos (or the lead battery)

Charging from my car is an option I had thought about since it has got
a 12V output in the back. The thing that put me off was that the LiPos
are best charged outside of the car (since in certain circumstances
they can catch fire) and I didnt really want to leave 2 or three LiPos
(at Ł50 each) laying around on the floor where someone might walk off
with them.

3. How about some sort of foot pump like generator that you can pump the
small lead battery back up, or possibly a lipo directly to save a few
strokes, perhaps even while you are flying.

Not too sure if thats a serious suggestion?

4. Buy more lipos and pre-charge them?
At Ł50 each I have bought 3 batts and 2 chargers as this allows for

almost continuous flying.

Thanks

David Bevan
http://www.davidbevan.co.uk

If you're worried about a LiPo burning up your vehicle, make yourself an
extension cord so your charger and LiPo are away from it. If you're flying
from a grassy area, it would be a good idea to bring something like a bucket
of sand or some other nonflammable substance to put your pack on or in while
it's charging. You could also fix up a lockable top on the bucket and
chain it to your vehicle's frame if you're worried about theft! (just
kidding)

Morris

 

[Larry Brasfield]

"Jonathan Kirwan" <jkirwan@easystreet.com> wrote in message
news:nfksb1pm3armgkqa80m2n8go8ebjmch1v1@4ax.com...

On Sat, 25 Jun 2005 20:59:03 -0700, "Larry Brasfield"
donotspam_larry_brasfield@hotmail.com> wrote:
"Jonathan Kirwan" <jkirwan@easystreet.com> wrote in
message news:l68rb1tsks25l38tpvl8s4ilsagnuavg1m@4ax.com...
On 25 Jun 2005 10:34:26 -0700, cgwiita@ucla.edu wrote:
[snip]
Finally,
there may be a bit of voltage "ringing"/oscillation at the PNP's
collector as the solenoid's inductance "fights" with the non-linearity
of the diode near the point where the inductor's energy is nearly
spent.

I have looked at lots of diole-clamped solenoids without seeing
such a thing. The circuit will certainly not support oscillation. The
usual appearance is a more rapid drop toward zero Volts across
the diode as its current approaches zero and its impedance goes up.

[snip]

I first admit I'm just a hobbyist on this score. But I didn't need to
simulate it, as I've seen it before. Damped oscillations in the 50kHz
range. I chalked it up to that non-linearity issue. But, of course,
I could be wrong.

Ringing, if it occurs, is due to a passive LC resonator with low
enough losses that the resonance is underdamped. The diode's
contribution is to simply add its capacitance to the (now open)
collector's capacitance. Once the diode begins to turn off, its
resistive impedance is high compared to the solenoid coil's.
While the diode capacitance happens to be non-linear, that
does not contribute to the fact of ringing; it merely alters the
shape of the waveform.

So, I just simulated it. And there it is! The simulator seems to see
it, too.

I do not doubt that such can show up in a simulation, where
the Q of the ersatz solenoid can be very high. With a real
solenoid, there is no effort made to preserve high Q. The
magnetic structures are typically solid metal with high eddy
current losses at the frequencies set by solenoid inductance
and stray capacitance, tending to spoil resonance.

This is a case where simulation should be regarded with a
lot of skepticism. A real solenoid should be consulted.

As I stated, I've not observed such resonance in real
solenoids. Perhaps, if I had looked more closely at
the region where the diode has just turned off, some
short lived, small ringing could have been observed. I
doubt the circuit Q would be high enough to justify use
of the term "oscillation".

--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.

 

[Larry Brasfield]

"Jonathan Kirwan" <jkirwan@easystreet.com> wrote in message
news:39lsb15cti959e0h33rkatmqeg9517qakf@4ax.com...

On Sat, 25 Jun 2005 20:59:03 -0700, "Larry Brasfield"
donotspam_larry_brasfield@hotmail.com> wrote:

That is a common misconception. When the solenoid having L = Ls
and R = Rs is turned on by applying Vs, its current begins ramping at
Vs/Ls A/S and asymptotically approaches Vs/Rs with an exponential
decay time of Ls/Rs. When the switch is turned off, the current begins
ramping down at (Vs+Vd)/Ls A/S (where Vd is the forward voltage of
the clamp diode, treated as constant) and asymptotically approaches
-Vd/Rs with an exponential decay time of Ls/Rs. The current not only
ramps slightly faster, it is headed for a value on the other side of zero.
So the current decay is definitely faster than the onset.

If such a circuit is slower to release the solenoid than energize it, the
reason is either dissymmetry between pull-in and drop-out voltages,
or an increase in inductance when the moving element is in the
energized position. The clamp diode should only be blamed for
not dropping the voltage as fast as would be possible by applying
a larger reverse voltage to bring the current to zero.

If you doubt this, I urge you to simulate it and study the result.

I just did, and it shows something I didn't expect but also not what
you say above.

I also just did so, and it shows exactly what I say above.

The decay time appears to be quite similar to the rise
time. About twice as long, actually, for the decay as for the rise.

Then you are not measuring those times in a meaningful
way, or you have simulated something else.

With 18 mH and 12 Ohms, I see a 1% to 63% rise time
of 1.492 mS and a 99% to 37% fall time of 1.335 mS.
The initial current rise rate is 657 A/S and the initial current
fall rate is 698 A/S. Note that the ratio well reflects my
claim, with Vs = 12V and Vd = 720 mV averaged over the
relevant period. (698/657 = 1.062, (12+0.72)/12 = 1.06)

I chose 1% and 99% to ensure that transistor switching
was (mostly) completed, and 37% and 63% because of
their relation to the well known exponential decay time
and because they approximate pull-in and drop-out.

But not nearly as much longer as I suspected. I need to look more
into this.

Yes. If you have trouble sorting it out, you can post the
ASCII content of an LTSpice schematic.

--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.

 

[Jonathan Kirwan]

On Sun, 26 Jun 2005 10:51:24 -0700, "Larry Brasfield"
<donotspam_larry_brasfield@hotmail.com> wrote:

With 18 mH

I'm not using 18mH. I looked up on the web for automotive style
solenoids to get some idea and came up with something on the order of
50mH to 200mH. I will post my LTSpice schematic, if you'd like.

Jon

 

[Larry Brasfield]

"Jonathan Kirwan" <jkirwan@easystreet.com> wrote in
message news:martb1t4njf658h7r5c0sjstiqcgiala06@4ax.com...

On Sun, 26 Jun 2005 10:51:24 -0700, "Larry Brasfield"
donotspam_larry_brasfield@hotmail.com> wrote:

With 18 mH

I'm not using 18mH. I looked up on the web for automotive style
solenoids to get some idea and came up with something on the order of
50mH to 200mH.

The size of the inductance hardly matters. Changing it
will only scale the waveforms in time. Their shape is
not affected except at a transistor switching timescale.

I will post my LTSpice schematic, if you'd like.

Before doing that, I suggest you state your criteria for
rise time and fall time. Be sure you saturate the switch.

--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.

 

[SC]

(about 70Ah).
Good luck finding a car battery that small. They usually come in
ratings of 600Ah or more.
I think on this point your are way of the mark a 600Ah battery is massive
fork lift truck sized, the maximun sized battery fitted to my Vauxhall
Astra is 70Ah

FWIW - car batteries are supposed to be used to start engines and nothing
else.
The most important number therefore is the maximum current that can be
drawn..
Logically - thats the number placed in the big letters on the side of
batts

Biggest 'car' batts I have personally seen are 100AH capacity - these
delivered 700A
- at least thats what it said on the side Methinks if you were to drain
it in 8 mins it would be a bit warm!

SC

 

[Jonathan Kirwan]

On Sun, 26 Jun 2005 11:05:33 -0700, "Larry Brasfield"
<donotspam_larry_brasfield@hotmail.com> wrote:

"Jonathan Kirwan" <jkirwan@easystreet.com> wrote in
message news:martb1t4njf658h7r5c0sjstiqcgiala06@4ax.com...
On Sun, 26 Jun 2005 10:51:24 -0700, "Larry Brasfield"
donotspam_larry_brasfield@hotmail.com> wrote:

With 18 mH

I'm not using 18mH. I looked up on the web for automotive style
solenoids to get some idea and came up with something on the order of
50mH to 200mH.

The size of the inductance hardly matters. Changing it
will only scale the waveforms in time. Their shape is
not affected except at a transistor switching timescale.

I'm not disagreeing with this. Just pointing out what I used and why.

I will post my LTSpice schematic, if you'd like.

Before doing that, I suggest you state your criteria for
rise time and fall time. Be sure you saturate the switch.

For a control signal I chose a rise time of 20ns and a fall time,
similarly. This is the actual times I have measured from a PIC, some
time back. Yes, the switches are saturated. In fact, I've also used
your values for the resistors, as well. As the PIC is, indeed, CMOS
these days, and since the drive current is in the low hundreds of uA,
the output voltage will be dropped by no more than about 20mV at the
output from the supply rail. I've also used that figure.

Jon

 

[Larry Brasfield]

"Jonathan Kirwan" <jkirwan@easystreet.com> wrote in message
newsnutb1ht0mllevuna4qm68728se35qg7np@4ax.com...

On Sun, 26 Jun 2005 11:05:33 -0700, "Larry Brasfield"
donotspam_larry_brasfield@hotmail.com> wrote:

"Jonathan Kirwan" <jkirwan@easystreet.com> wrote in
message news:martb1t4njf658h7r5c0sjstiqcgiala06@4ax.com...
On Sun, 26 Jun 2005 10:51:24 -0700, "Larry Brasfield"
donotspam_larry_brasfield@hotmail.com> wrote:

With 18 mH

I'm not using 18mH. I looked up on the web for automotive style
solenoids to get some idea and came up with something on the order of
50mH to 200mH.

The size of the inductance hardly matters. Changing it
will only scale the waveforms in time. Their shape is
not affected except at a transistor switching timescale.

I'm not disagreeing with this. Just pointing out what I used and why.

I had guessed that you had some associated point to make.

I will post my LTSpice schematic, if you'd like.

Before doing that, I suggest you state your criteria for
rise time and fall time. Be sure you saturate the switch.

For a control signal I chose a rise time of 20ns and a fall time,
similarly. This is the actual times I have measured from a PIC, some
time back. Yes, the switches are saturated.

The criteria I mentioned relate to how you define the times
you reported when you posted: "The decay time appears to be
quite similar to the rise time. About twice as long, actually ..."
That result is contrary to what a simple analysis will predict
and contrary to what a simple simulation shows. So I wonder
how you have derived those times.

In fact, I've also used
your values for the resistors, as well. As the PIC is, indeed, CMOS
these days, and since the drive current is in the low hundreds of uA,
the output voltage will be dropped by no more than about 20mV at the
output from the supply rail. I've also used that figure.

For that low an input drive, I would revise that circuit, probably
with a NMOSFET or Darlington in the first stage.

--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.

 

[Fred Abse]

On Fri, 24 Jun 2005 10:58:33 -0700, Dave wrote:

One would think Hollanders were pacifists, guess not - Suriname or
Indonesia.

That makes them Indianoceanists, I think

--
"Electricity is of two kinds, positive and negative. The difference
is, I presume, that one comes a little more expensive, but is more
durable; the other is a cheaper thing, but the moths get into it."
(Stephen Leacock)

 

[John Fields]

On 26 Jun 2005 13:30:35 -0700, "JennaMyria" <likestowalk@hotmail.com>
wrote:

Thank you all so far! I am however still uncertain.
I do not intend to get the kids into electronis, but I really intend
the stuff to be used in crafts, i.e. glue them to cardboard or make
jewelery out of the stuff!!!
And I am afraid that I don't really know what "phenolic Bakelite" is,
although I will go and google it immediately. I really just want to
know if it is safe to have kids touching those little resistors or
having them wear a piece on a string around their neck.

---
It's safe.

--
John Fields
Professional Circuit Designer

 

[Jonathan Kirwan]

On Sun, 26 Jun 2005 12:21:23 -0700, "Larry Brasfield"
<donotspam_larry_brasfield@hotmail.com> wrote:

"Jonathan Kirwan" <jkirwan@easystreet.com> wrote in message
newsnutb1ht0mllevuna4qm68728se35qg7np@4ax.com...
On Sun, 26 Jun 2005 11:05:33 -0700, "Larry Brasfield"
donotspam_larry_brasfield@hotmail.com> wrote:

"Jonathan Kirwan" <jkirwan@easystreet.com> wrote in
message news:martb1t4njf658h7r5c0sjstiqcgiala06@4ax.com...
On Sun, 26 Jun 2005 10:51:24 -0700, "Larry Brasfield"
donotspam_larry_brasfield@hotmail.com> wrote:

With 18 mH

I'm not using 18mH. I looked up on the web for automotive style
solenoids to get some idea and came up with something on the order of
50mH to 200mH.

The size of the inductance hardly matters. Changing it
will only scale the waveforms in time. Their shape is
not affected except at a transistor switching timescale.

I'm not disagreeing with this. Just pointing out what I used and why.

I had guessed that you had some associated point to make.

I will post my LTSpice schematic, if you'd like.

Before doing that, I suggest you state your criteria for
rise time and fall time. Be sure you saturate the switch.

For a control signal I chose a rise time of 20ns and a fall time,
similarly. This is the actual times I have measured from a PIC, some
time back. Yes, the switches are saturated.

The criteria I mentioned relate to how you define the times
you reported when you posted: "The decay time appears to be
quite similar to the rise time. About twice as long, actually ..."
That result is contrary to what a simple analysis will predict
and contrary to what a simple simulation shows. So I wonder
how you have derived those times.

Simply used the cursors in LTSpice. Quite easy to do and it reads off
directly.

In fact, I've also used
your values for the resistors, as well. As the PIC is, indeed, CMOS
these days, and since the drive current is in the low hundreds of uA,
the output voltage will be dropped by no more than about 20mV at the
output from the supply rail. I've also used that figure.

For that low an input drive, I would revise that circuit, probably
with a NMOSFET or Darlington in the first stage.

I meant only that this is all the current that is actually used by the
NPN. A PIC can drive much more, as its high side impedance is along
the lines of 90-120 ohms and its low side impedance something a little
lower, perhaps 65-70 ohms. However, the circuit really doesn't need
that much, since the NPN's beta divides down the base drive to the PNP
by quite a bit. With a decent PNP, at a little more than an amp for
the collector current, the base really only needs some 50th of that (I
tried planning a variety of base currents based on /20, /30, /50, and
even /100, and the Zetex has a Vce of less than 0.2V across the range
of them. So the NPN collector really only needs to supply a little
more than that (just enough extra for the resistor to +12.) Divided
by its own beta, which because it is not operating saturated is rather
high, leaves very little need for base drive to the NPN.

Jon

 

[Larry Brasfield]

"Jonathan Kirwan" <jkirwan@easystreet.com> wrote in message
news:sjdub15i7g81h0910aqsclbqmucc7cmtao@4ax.com...

On Sun, 26 Jun 2005 12:21:23 -0700, "Larry Brasfield"
donotspam_larry_brasfield@hotmail.com> wrote:
....
I will post my LTSpice schematic, if you'd like.

The time has come for that, IMHO.

Before doing that, I suggest you state your criteria for
rise time and fall time.
....
For a control signal I chose a rise time of 20ns and a fall time,
similarly. This is the actual times I have measured from a PIC, some
time back.
....
The criteria I mentioned relate to how you define the times
you reported when you posted: "The decay time appears to be
quite similar to the rise time. About twice as long, actually ..."
That result is contrary to what a simple analysis will predict
and contrary to what a simple simulation shows. So I wonder
how you have derived those times.

Simply used the cursors in LTSpice. Quite easy to do and it reads off
directly.

"Criterion" means some rule for deciding something.
What I'm asking is: How do you decide where to place
those cursors before you read off their time difference?
In another post, I mentioned the 99% to 37% fall time
and the 1% to 63% rise time. The criteria for how one
would measure those are evident. (I suppose I should
mention that they relate to current changes, not voltage.)
Your criteria remain a complete mystery. What precisely
happens between the start of your "rise time" and the end
of it? Repeat the question for your "decay time".

....

and since the drive current is in the low hundreds of uA,
the output voltage will be dropped by no more than about 20mV at the
output from the supply rail. I've also used that figure.

For that low an input drive, I would revise that circuit, probably
with a NMOSFET or Darlington in the first stage.

I meant only that this is all the current that is actually used by the
NPN. A PIC can drive much more, as its high side impedance is along
the lines of 90-120 ohms and its low side impedance something a little
lower, perhaps 65-70 ohms. However, the circuit really doesn't need
that much,

Ok, I misunderstood you.

--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.

 

[Jonathan Kirwan]

On Sun, 26 Jun 2005 16:38:37 -0700, "Larry Brasfield"
<donotspam_larry_brasfield@hotmail.com> wrote:

"Jonathan Kirwan" <jkirwan@easystreet.com> wrote in message
news:sjdub15i7g81h0910aqsclbqmucc7cmtao@4ax.com...
On Sun, 26 Jun 2005 12:21:23 -0700, "Larry Brasfield"
donotspam_larry_brasfield@hotmail.com> wrote:
...
I will post my LTSpice schematic, if you'd like.

The time has come for that, IMHO.

I have a set of them, by now. But I can include the one that is based
on your schematic (uses the Zetex model and a 10 ohm, 200mH relay) as
a starting point. Would that be okay? (See bottom of post.)

Before doing that, I suggest you state your criteria for
rise time and fall time.
...
For a control signal I chose a rise time of 20ns and a fall time,
similarly. This is the actual times I have measured from a PIC, some
time back.
...
The criteria I mentioned relate to how you define the times
you reported when you posted: "The decay time appears to be
quite similar to the rise time. About twice as long, actually ..."
That result is contrary to what a simple analysis will predict
and contrary to what a simple simulation shows. So I wonder
how you have derived those times.

Simply used the cursors in LTSpice. Quite easy to do and it reads off
directly.

"Criterion" means some rule for deciding something.
What I'm asking is: How do you decide where to place
those cursors before you read off their time difference?
In another post, I mentioned the 99% to 37% fall time
and the 1% to 63% rise time. The criteria for how one
would measure those are evident. (I suppose I should
mention that they relate to current changes, not voltage.)
Your criteria remain a complete mystery. What precisely
happens between the start of your "rise time" and the end
of it? Repeat the question for your "decay time".

I used the same measuring rules for both directions. I will leave it
to you to look at when you load up the LTSpice schematic. I think you
will be able to see that the fall time is at least ... longer ...
regardless of your standard chosen.

and since the drive current is in the low hundreds of uA,
the output voltage will be dropped by no more than about 20mV at the
output from the supply rail. I've also used that figure.

For that low an input drive, I would revise that circuit, probably
with a NMOSFET or Darlington in the first stage.

I meant only that this is all the current that is actually used by the
NPN. A PIC can drive much more, as its high side impedance is along
the lines of 90-120 ohms and its low side impedance something a little
lower, perhaps 65-70 ohms. However, the circuit really doesn't need
that much,

Ok, I misunderstood you.

No problem. It seemed that way, which is why I posted a more detailed
description of my thinking.

Jon

P.S. Note that the last two TEXT lines describe the two models I
included, one for the diode, the other for the PNP transistor. I'm
not sure how my email posting program will wrap these lines versus how
LTSpice wants them. But you should be able to edit them in,
correctly, with only a small bother at most.



Version 4
SHEET 1 880 680
WIRE -512 -720 -512 -736
WIRE -512 -608 -512 -640
WIRE -416 -736 -512 -736
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SYMATTR InstName Q1
SYMATTR Value ZTX790A
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SYMATTR InstName R1
SYMATTR Value 1000
SYMBOL npn -224 -304 R0
SYMATTR InstName Q2
SYMATTR Value 2N3904
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SYMATTR Value 130
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SYMATTR Value {Vbb}
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SYMATTR Value PULSE(0 5 500u 20n 20n 100m 200m)
SYMBOL res -320 -752 R90
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WINDOW 3 32 56 VTop 0
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SYMATTR Value 120
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SYMATTR Value 200m
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SYMATTR Value 1N4002
TEXT -882 -96 Left 0 !.tran 1
TEXT -880 -640 Left 0 !.param ILoad=1A
TEXT -880 -704 Left 0 !.param Vbb=12V
TEXT -880 -672 Left 0 !.param Vcc=5V
TEXT 128 -160 Left 0 ;BUZZER
TEXT -896 -736 Left 0 ;PRIMARY SPECIFICATIONS
TEXT -896 -576 Left 0 ;DESIGN ESTIMATIONS
TEXT -880 -512 Left 0 !.param Q2Vce=0.2V Q2Vbe=0.85V Q2beta=30
TEXT -880 -544 Left 0 !.param Q1Vbe=0.75V Q1beta=200
TEXT -896 -416 Left 0 ;DERIVATIVE CALCULATIONS
TEXT -880 -224 Left 0 !.param RLoad=(Vbb-Q2Vce)/ILoad
TEXT -880 -384 Left 0 !.param Q2Ic=ILoad Q2Ib=ILoad/Q2beta
TEXT -880 -480 Left 0 !.param IR2=0.02*Q2Ib
TEXT -880 -352 Left 0 !.param Q1Ic=Q2Ib+IR2 Q1Ib=Q1Ic/Q1beta
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TEXT -880 -160 Left 0 !.param R1=(Vcc-Q1Vbe)/Q1Ie
TEXT -880 112 Left 0 !.MODEL ZTX790A PNP(IS=1.09684E-12 NF=1.0102
BF=650 IKF=1.7 VAF=23.5\n+ISE=9.88593E-14 NE=1.47256 NR=1.00391 BR=270
IKR=0.2 VAR=30\n+ISC=5.4933E-14 NC=1.07427 RB=0.055 RE=0.049 RC=0.078
CJC=96E-12\n+MJC=0.495 VJC=0.67 CJE=275E-12 TF=0.75E-9 TR=10.8E-9)
TEXT -880 16 Left 0 !.MODEL 1N4002 D(IS=14.11E-9 N=1.984 RS=33.89E-3
IKF=94.81 XTI=3\n+ EG=1.110 CJO=51.17E-12 M=.2762 VJ=.3905 FC=.5
ISR=100.0E-12\n+ NR=2 BV=100.1 IBV=10 TT=4.761E-6)

 

[Mike Norton]

I bought a charger for 69 dollars from Sears. It charges anything that is
12V lead acid under 200 AH, including small gel cells. You can connect the
battery up backwards with no harm to charger or battery. It can also
provide a 100A burst to help start an engine (car engine, that is).

I have been charging 7AH gel cells with no problem after every use. It
normally takes less than a minute. I suspect most of them die from
overcharge or being left in a discharged state. I have also been charging
30 AH deep-cycle batteries for field use.

I know that it is a lot of bucks for a charger, but I wanted an automotive
charger that worked better than the one I had. It is quite an amazing
gadget.

-- Mike Norton

"Funfly3" <dontemailme@ntlworld.com> wrote in message
news:0Adve.3790$rz1.692@newsfe5-gui.ntli.net...

"John Miller" <me2@privacy.net> wrote in message
news:d9jhq2$5ge$2@n4vu2.n4vu.com...
Funfly3 wrote:
buy a cheap car charger from a car shop and buy a car battery from Macro
for
Ł17/Ł19 and its got 3 a year guarantee that's what I use

Beware of cheap (unregulated) chargers used routinely. They will boil
your electrolyte, and generate excess hydrogen while they're at it.

--
John Miller
email domain: n4vu.com; username: jsm(@)
What's the price of a regulated charger that will charge a car battery
routinely??? a 10AH charger is around Ł45 so a 50AH is going to be a lot
more and the battery has a 3 year warranty all for Ł17 cook it then take
it back

 

[Larry Brasfield]

"Jonathan Kirwan" <jkirwan@easystreet.com> wrote in message
news:sigub1lj498b6jl0n4366mht3570qgvqiv@4ax.com...

On Sun, 26 Jun 2005 16:38:37 -0700, "Larry Brasfield"
donotspam_larry_brasfield@hotmail.com> wrote:

"Jonathan Kirwan" <jkirwan@easystreet.com> wrote in message
news:sjdub15i7g81h0910aqsclbqmucc7cmtao@4ax.com...
On Sun, 26 Jun 2005 12:21:23 -0700, "Larry Brasfield"
donotspam_larry_brasfield@hotmail.com> wrote:
...
I will post my LTSpice schematic, if you'd like.

The time has come for that, IMHO.

I have a set of them, by now. But I can include the one that is based
on your schematic (uses the Zetex model and a 10 ohm, 200mH relay) as
a starting point. Would that be okay? (See bottom of post.)

Sure, if it works.

Before doing that, I suggest you state your criteria for
rise time and fall time.
...
For a control signal I chose a rise time of 20ns and a fall time,
similarly. This is the actual times I have measured from a PIC, some
time back.
...
The criteria I mentioned relate to how you define the times
you reported when you posted: "The decay time appears to be
quite similar to the rise time. About twice as long, actually ..."
That result is contrary to what a simple analysis will predict
and contrary to what a simple simulation shows. So I wonder
how you have derived those times.

Simply used the cursors in LTSpice. Quite easy to do and it reads off
directly.

"Criterion" means some rule for deciding something.
What I'm asking is: How do you decide where to place
those cursors before you read off their time difference?
In another post, I mentioned the 99% to 37% fall time
and the 1% to 63% rise time. The criteria for how one
would measure those are evident. (I suppose I should
mention that they relate to current changes, not voltage.)
Your criteria remain a complete mystery. What precisely
happens between the start of your "rise time" and the end
of it? Repeat the question for your "decay time".

I used the same measuring rules for both directions.

You appear determined to be mysterious about them.

I will leave it
to you to look at when you load up the LTSpice schematic. I think you
will be able to see that the fall time is at least ... longer ...
regardless of your standard chosen.

(LTSpice schematic and related notes cut.)

I ran your simulation with these results:
current rise time 0% to 90% = 44.25 mS
current fall time 100% to 10% = 36.33 mS
current rise time 1% to 63% = 19.21 mS
current fall time 99% to 37% = 17.35 mS
(The cycle time is not adequate to let the on current
settle to the steady state, being only about 10 L/R.
This has reduced the expected 1% to 63% rise time
from the expected 19.8 mS to the value seen above.)
To me, it appears that the fall time is uniformly shorter.
I will leave it to you to figure out why your method for
measuring rise and fall time leads to inexplicable results.

--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.

 

[ehsjr]

Mike Norton wrote:

I bought a charger for 69 dollars from Sears. It charges anything that is
12V lead acid under 200 AH, including small gel cells. You can connect the
battery up backwards with no harm to charger or battery. It can also
provide a 100A burst to help start an engine (car engine, that is).

I have been charging 7AH gel cells with no problem after every use. It
normally takes less than a minute.

Something is very wrong.
A battery that charges in less than a minute either
did not need to be charged, was charged *WAY* too fast,
or wasn't really charged, due to the charger shutting
off when it shouldn't.

Check the instructions for your charger - is there something
different you need to do when charging gel cells vs car
batteries?

Ed

<snip>

-- Mike Norton

 

[Jonathan Kirwan]

On Sun, 26 Jun 2005 19:25:50 -0700, "Larry Brasfield"
<donotspam_larry_brasfield@hotmail.com> wrote:

snip of other material, until later
(LTSpice schematic and related notes cut.)

I ran your simulation with these results:
current rise time 0% to 90% = 44.25 mS
current fall time 100% to 10% = 36.33 mS
current rise time 1% to 63% = 19.21 mS
current fall time 99% to 37% = 17.35 mS
(The cycle time is not adequate to let the on current
settle to the steady state, being only about 10 L/R.
This has reduced the expected 1% to 63% rise time
from the expected 19.8 mS to the value seen above.)
To me, it appears that the fall time is uniformly shorter.
I will leave it to you to figure out why your method for
measuring rise and fall time leads to inexplicable results.

To the issue mentioned above, looking at the curve for I(L1), at about
0.2s on my .tran display, I see it just start to rise. (You are right
about not providing enough time, but I don't think this is material to
the discussion. It does rise to about 1.17A and must ramp down from
that point and is able to completely go back to zero, so the situation
should still be worth discussing.) I used your figure of 63% from the
earlier post and this is about .737A. I get about 19.72ms for the
time, as close as I can make it out zooming out the display. At 0.3s,
the solenoid looses its supply and the current ramps down. 37% is
..433A. I see this take place 18.1ms later. Looking back over my
notes, I see where I'd confused two pairs of numbers taken with
different values for the solenoid. My mistake! Thanks.

I'm curious to see if you found the oscillation in the simulation and
might address yourself to that.

Jon

 

[Larry Brasfield]

"Jonathan Kirwan" <jkirwan@easystreet.com> wrote in message
news:021vb11ct2oe2a3hd7u37rs5v4vmji0r7o@4ax.com...

On Sun, 26 Jun 2005 19:25:50 -0700, "Larry Brasfield"
donotspam_larry_brasfield@hotmail.com> wrote:
snip of other material, until later
(I, Brasfield, did not write the above line.)

....

I ran your simulation with these results:
current rise time 0% to 90% = 44.25 mS
current fall time 100% to 10% = 36.33 mS
current rise time 1% to 63% = 19.21 mS
current fall time 99% to 37% = 17.35 mS
(The cycle time is not adequate to let the on current
settle to the steady state, being only about 10 L/R.
This has reduced the expected 1% to 63% rise time
from the expected 19.8 mS to the value seen above.)
To me, it appears that the fall time is uniformly shorter.
I will leave it to you to figure out why your method for
measuring rise and fall time leads to inexplicable results.

To the issue mentioned above, looking at the curve for I(L1), at about
0.2s on my .tran display, I see it just start to rise. (You are right
about not providing enough time, but I don't think this is material to
the discussion.

It is only material in that it explains the deviation in rise time
from what would be predicted using the component values.

[snip]

I'm curious to see if you found the oscillation in the simulation and
might address yourself to that.

My first post in this thread on June 26 addressed that.
With the addition of a few elements to simulate the effect
of eddy current losses in the solenoid magnetic structure,
your simulation shows behavior at diode turn-off that is
much closer to what can be observed with real solenoids.

In this simulation, (included below), there is a small amount
of ringing, barely visible without zooming in on it. Without
considerably more work, I cannot be sure that those added
elements accurately model the layered single turns of steel
that are present in most real solenoids, acting as parasitic
secondaries that absorb energy at higher frequencies. To
get that close would require looking at some actual parts.

Version 4
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TEXT -882 -96 Left 0 !.tran 0 210m 0 1u
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TEXT 128 -160 Left 0 ;BUZZER
TEXT -896 -736 Left 0 ;PRIMARY SPECIFICATIONS
TEXT -896 -576 Left 0 ;DESIGN ESTIMATIONS
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TEXT -880 -544 Left 0 !.param Q1Vbe=0.75V Q1beta=200
TEXT -896 -416 Left 0 ;DERIVATIVE CALCULATIONS
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TEXT -896 -256 Left 0 ;RESISTOR VALUES
TEXT -880 -192 Left 0 !.param R2=Q2Vbe/IR2
TEXT -880 -160 Left 0 !.param R1=(Vcc-Q1Vbe)/Q1Ie
TEXT -880 112 Left 0 !.MODEL ZTX790A PNP(IS=1.09684E-12 NF=1.0102 BF=650 IKF=1.7 VAF=23.5 ISE=9.88593E-14 NE=1.47256 NR=1.00391
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TR=10.8E-9)
TEXT -880 16 Left 0 !.MODEL 1N4002 D(IS=14.11E-9 N=1.984 RS=33.89E-3 IKF=94.81 XTI=3 EG=1.110 CJO=51.17E-12 M=.2762 VJ=.3905
FC=.5 ISR=100.0E-12 NR=2 BV=100.1 IBV=10 TT=4.761E-6)
TEXT 264 -368 Left 0 !K12 L1 L2 0.5
TEXT 248 -184 Left 0 !K23 L2 L3 0.5
TEXT 144 -424 Left 0 !.param Nts=100, Ls=200m

--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.