 |
Search Sci.Electronics.Basics |
|
|
|
|
|
|
 |
|
|
Sci.Electronics.Basics -> Zener Diodes / Voltage Regulation
There are 15 messages in this thread.
You are currently looking at messages 1 to 15.
|
Author: FyberOpticDate: 01:01 22-07-07
|
|
I'm having some confusion regarding zener diodes which I hope somebody
much brighter on the subject can clarify somewhat, or at least point
me in the right direction!
The basics of diodes has always been "current can only flow one way",
and beginners are left to assume this means they'll always do just
that. But obviously that's not the case, since normal diodes will
stop doing such a thing if they're pushed too far, I understand. And
as a result, people take advantage of this effect more reliably with
the zener diode apparently, by making them "fail" at much lower rates,
it would seem.
So while this is possibly out of the scope of asking on a newsgroup,
what sorts of applications could they be used for reliably? I hear
they can be used as regulators, but only if the load is fairly
constant. So, for something like an electronic circuit, with chips
going on and off and constantly changing the load, would this be of
any use there? I've seen them used in situations such as when one is
drawing power from the PC parallel port. I'm assuming this isn't
related to actually getting that power, but keeping it from going too
high.
I'm curious because, despite my lack of knowledge regarding the
details of analog circuits, I have a pretty decent understanding of
digital ones, and find myself making and planning new things in that
realm all the time. I'd like to power some of these sorts of things
off of batteries sometime instead of a 7805 on an ac adapter, and am
interested in the best way to go about it. One project in particular
involves an 8052 microcontroller with an lcd which I'd like to make
portable, for example.
On a side note, how possible would it be to boost the voltage from a
couple of AA's to run 5v logic reliably? I can't really think of how
such a circuit would be made, though I know they surely exist. I
assume it uses capacitors and an oscillation of some kind, but I
dunno. When I think of changing voltages, I mostly just think of
transformers.
Anyhoo, any help or pointers would be greatly appreciated!
|
|
|
|
Author: John PopelishDate: 01:48 22-07-07
|
|
FyberOptic wrote:
(snip)
> The basics of diodes has always been "current can only flow one way",
> and beginners are left to assume this means they'll always do just
> that.
That rule applies below reverse break down voltage, and
zeners are operated above reverse breakdown.
> But obviously that's not the case, since normal diodes will
> stop doing such a thing if they're pushed too far, I understand. And
> as a result, people take advantage of this effect more reliably with
> the zener diode apparently, by making them "fail" at much lower rates,
> it would seem.
I'm not sure what you are trying to get across with that,
but the difference between normal, rectifier diodes and
zeners are that normal, rectifier diodes are designed to
have a minimum guaranteed reverse breakdown voltage and be
operated below that guaranteed minimum, while zeners are
designed to have specific reverse breakdown voltages and to
be operated with in a reverse breakdown situation.
Rectifier diodes have forward current ratings based on the
heat they can get rid of while dropping a volt or so in the
forward direction. Zener diodes are rated for the power
they can get rid of while leaking reverse current (and
dropping their rated zener breakdown voltage.
> So while this is possibly out of the scope of asking on a newsgroup,
> what sorts of applications could they be used for reliably?
Any application that keeps their die temperature below the
rated temperature. These include signal clamping, and
reference voltage generation and supply regulation.
> I hear
> they can be used as regulators, but only if the load is fairly
> constant.
They act as pretty fair regulators as long as the current
through them does not go too low (or the breakdown process
gets sloppy) or too high (and the zener overheats). The
zener shunt regulator must carry a minimum current when the
load current is at maximum (to avoid the first case) and
must carry that and the difference between the maximum and
minimum load current without overheating. In addition,
variations in the unregulated voltage that is current
limited by a series resistance into the parallel combination
of the zener and the load, must not push the zener operation
into either of those extremes through its expected variation.
> So, for something like an electronic circuit, with chips
> going on and off and constantly changing the load, would this be of
> any use there?
It depends on whether power is cheap or expensive and how
much the raw voltage is expected to vary, and how much
current variation the load will produce. there are some
pretty large zener diodes produced. But the most common
types are axial leaded devices with power ratings of 5 watts
or less. If the load power is much less than that, a zener
might be a practical regulator.
> I've seen them used in situations such as when one is
> drawing power from the PC parallel port. I'm assuming this isn't
> related to actually getting that power, but keeping it from going too
> high.
The power available through a parallel port is quite
limited, so a small zener is quite practical as a way to
limit the voltage from that source. Of course the limited
power also implies that the load is draws only a small current.
> I'm curious because, despite my lack of knowledge regarding the
> details of analog circuits, I have a pretty decent understanding of
> digital ones, and find myself making and planning new things in that
> realm all the time. I'd like to power some of these sorts of things
> off of batteries sometime instead of a 7805 on an ac adapter, and am
> interested in the best way to go about it.
Since zener regulators must consume some raw supply current,
just to keep them regulating, they are poor choices for most
battery applications. 3 terminal linear regulators, like
the 7805 have pretty low minimum current requirements, and
very much better (in that department) regulator types are
available. But best efficiency for higher current loads
involves switching the battery voltage off and on, as
completely as possible, and then filtering those pulses to
extract their average voltage. There are no intentional
losses in that process, so the power out equals the power
in, except for unavoidable losses. in many cases, the
output current exceeds the input current, because the
voltage has been converted down. The key words for this
sort of thing are [buck converter].
> One project in particular
> involves an 8052 microcontroller with an lcd which I'd like to make
> portable, for example.
>
>
> On a side note, how possible would it be to boost the voltage from a
> couple of AA's to run 5v logic reliably?
It is often done the same way that spark voltages are
generated by contact points and a coil. inductors have the
property that they generate voltage to fight sudden
variations in their current. You connect the battery across
an inductor and its current ramps smoothly up, as it
produces a voltage that almost holds back the battery
voltage. Then you switch that current off, and the inductor
produces a voltage that tries to keep its current going. If
the battery had been applying positive voltage to the
inductor, when the current is interrupted, the inductor will
produce a negative voltage to try to keep sucking that
current through the switch that is blocking the current. If
you provide an alternate path for that current driven by the
inductive so called flyback voltage reversal, you can
connect that negative voltage to a storage capacitor that
charges to an arbitrarily high negative voltage. If,
instead, you add a second winding to the inductor, making it
into a transformer, that flyback voltage can be reversed (by
swapping the winding ends) and you can use the diode to
connect the output to a capacitor that is charged to an
arbitrarily high positive voltage. Or add several windings
and produce several proportional voltages, either positive
or negative, all at the same time.
You also need a control circuit that varies the current ramp
up time , each cycle to produce the desired output voltage.
The key words for this sort of thing are [boost converter].
> I can't really think of how
> such a circuit would be made, though I know they surely exist. I
> assume it uses capacitors and an oscillation of some kind, but I
> dunno. When I think of changing voltages, I mostly just think of
> transformers.
There are ways to charge a pair of capacitors, in parallel,
with the battery voltage, and then reconnect them, so that
they are in series, doubling the battery voltage, and
connect that voltage to an output storage capacitor to act
as a power source while you go back and repeat the cycle.
The key words for this sort of thing are [charge pump].
> Anyhoo, any help or pointers would be greatly appreciated!
Here is a basic tutorial for the buck and boost converters:
http://www.national.com/appinfo/power/files/f5.pdf
|
|
|
|
Author: Jonathan KirwanDate: 02:18 22-07-07
|
|
On Sat, 21 Jul 2007 22:01:28 -0700, FyberOptic <fyberoptic@gmail.com>
wrote:
><snip>
>So while this is possibly out of the scope of asking on a newsgroup,
>what sorts of applications could they be used for reliably? I hear
>they can be used as regulators, but only if the load is fairly
>constant. So, for something like an electronic circuit, with chips
>going on and off and constantly changing the load, would this be of
>any use there? I've seen them used in situations such as when one is
>drawing power from the PC parallel port. I'm assuming this isn't
>related to actually getting that power, but keeping it from going too
>high.
A zener usually isn't used directly as a power supply regulator, per
se. It may be a good voltage reference, though, as input to a
comparator (which doesn't require much current.) If a zener is used
for supplying some current, a BJT voltage follower is often added (the
base doesn't require much current, once again.)
Take a basic circuit:
R1 / Z1
+Vsupply ----/\/\--+--|<|--------gnd
| /
|
'---- LOAD ---gnd
If you know your load requires something less than 20mA, for example,
you might set things up so that R1 passes that 20mA plus a little more
for the zener, itself. (The zener voltage will not remain the same as
current changes occur, but will stay somewhere in the same area.) So
let's say 25mA. This means that when your load is using 20mA, the
zener itself will get 5mA. But when the load is removed or otherwise
not taking much current, then Z1 will have to pick up the load of the
full 25mA. So say that Vsupply=21V and Z1 is a 6.2V variety. You
compute R1=(21-6.2)/25mA = 592 ohms. Make it 560, for a standard
value and accept the fact that the actual max current will be a little
more than 26mA. Now, the whole thing completely falls apart, though,
if your load requires almost all of that 26mA or more. Once that
happens, R1 will drop too much voltage and the zener won't 'zener' and
your voltage will go completely out of regulation.
Maybe not the very next step, but a small step past it, is to add a
BJT and another diode:
R1 / Z1 D1
+Vsupply ----/\/\--+--|<|---|>|--gnd
| /
|
|
Q1 --- NPN
/ v
+Vsupply -------c e---- LOAD ---gnd
The Q1 emitter will present about the zener voltage to the load. The
addition of D1 does at least two things -- one is that it 'jacks up'
the base voltage of Q1 because there is a base-emitter diode drop
there and D1 helps to set things up so that the emitter is about the
zener voltage. (If you didn't include D1, then the emitter would be
somewhat lower, by one diode drop, roughly); Another that maybe isn't
important for this discussion is temperature compensation.
If the load requires under 20mA, still, then the base of Q1 will
probably only need about 1/100th of that, or about 200uA. Now, if you
set up your zener current to be about 10mA for example, by calculating
R1 accordingly, then the variations over the entire 0mA-20mA load
possibilities will only present a variation of 0uA-200uA shift in the
zener current. That's only 1/50th of the 10mA, so we are talking
about 2% variation. This means the zener voltage will stay put pretty
well. The emitter voltage of Q1 will vary some if the emitter current
gets really tiny (rising by some 60mV per decade reduction in
current), but for anything in the probable range it will be
sufficiently fixed.
Note that for all of the above examples, to calculate R1 you need to
know what your Vsupply is. If that is an unregulated supply and
itself varies a lot, then you may need to calculate R1 on the basis of
the smallest likely voltage there and just realize that at other times
more current will be flowing into Z1 and that will impact the zener
voltage, somewhat. Or, you may just want it to work with any number
of those "wall warts" with different output voltages. So if you want
to set things up well for that more difficult situation, you need to
use a constant current source that doesn't vary over variations in the
unregulated supply. A constant current arrangement requires another
BJT. So it looks more like:
,-----e c--gnd
| v /
| --- Q1
| | PNP
| |
R2 | R1 | / Z1 D1
Vunreg---/\/\-+-/\/\--++--|<|---|>|---gnd
| /
|
| Q2
--- NPN
/ v
Vunreg--------------c e----- LOAD ----gnd
Q1 is now used to generate a constant current through R1, Z1 and D1.
The Vunreg supply can vary a lot, now, and it doesn't have much impact
on that zener current and therefore not much on the zener voltage,
which remains fairly stable regardless of what Vunreg's voltage is (so
long as it is several volts higher than the zener voltage.)
I've kept out things like capacitors and possibly some negative
feedback (which is usually a good idea) to control the output voltage
better, using the zener instead as a voltage reference for comparison
purposes and not as the input to an open-loop driver (no feedback.) If
you added something like that, it would appear more as:
C2
|| .5uF
,--------||-----,
| || |
| | Vunreg
| Vunreg | |
| | | |
| |\ | |
+--------|-\ | |/c Q1
| | >---+-----| NPN
| Vref---|+/ |>e
| |/ |
| | | Vout
| gnd '-------+----+----- LOAD ---gnd
| | |
| | |
| \ --- C1
| R1 / ---
| \ |
| / |
| | |
| | gnd
'-------------------------------+
|
|
\
/ R2
\
/
|
|
gnd
Here, you can see that an opamp is used to compare the Vref (output of
the constant current version of the zener regulator I just mentioned
above) with a portion of the output voltage (Vout.) The opamp then
drives Q1 as needed so that the difference is nil.
[That opamp might only be a differential pair of BJTs with some
additional BJTs for things like a constant current at the pair's
emitter or in the collector of one side as another constant current to
'stiffen' (make it much higher gain) it so that only one diff pair is
needed instead of, say, two pairs.]
Of course, you can just buy an IC that does all this and more for you.
Jon
|
|
|
|
Author: EeyoreDate: 11:16 22-07-07
|
|
FyberOptic wrote:
Could you perhaps try a post with a couple of straight questions ?
Graham
|
|
|
|
Author: Michael A. TerrellDate: 11:21 22-07-07
|
|
Eeyore wrote:
>
> FyberOptic wrote:
>
> Could you perhaps try a post with a couple of straight questions ?
>
> Graham
Dammit, you KNOW that this is news:sci.electronics.basics yet you
just have to slam the newbies.
--
Service to my country? Been there, Done that, and I've got my DD214 to
prove it.
Member of DAV #85.
Michael A. Terrell
Central Florida
|
|
|
|
Author: Michael BlackDate: 12:19 22-07-07
|
|
FyberOptic (fyberoptic@gmail.com) writes:
> So while this is possibly out of the scope of asking on a newsgroup,
> what sorts of applications could they be used for reliably? I hear
> they can be used as regulators, but only if the load is fairly
> constant. So, for something like an electronic circuit, with chips
> going on and off and constantly changing the load, would this be of
> any use there? I've seen them used in situations such as when one is
> drawing power from the PC parallel port. I'm assuming this isn't
> related to actually getting that power, but keeping it from going too
> high.
>
In the old days, voltage regulation wasn't common. There were a few
things that could make use of it, and you'd see voltage regulation
applied to that stage rather than to the whole equipment. Only exotic
lab equipment would have fully regulated power supplies.
So in the tube days, that voltage regulation would be done with VR tubes,
ie Voltage Regulation tubes. There'd be a current limiting resistor
from the power supply, and then the two terminal tubes would kick in
when the voltage went too high. Change the load, and you'd need to
change the current limiting resistor.
Transistors came along, and so did zener diodes. They were like
those VR tubes, except you could now get zeners that would
regulate at low voltage.
IN the early days of transistors, you would see those zener diodes
used basically like those VR tubes, regulating specific stages
rather than whole power supplies. They were lousy for general
supply use, since they tended to be low current devices and you
had to adjust the current limiting resistor as the load changed.
But they disappeared mostly after a decade. People made the
transisition to solid state devices, and realized things weren't
quite like the days of tubes. Tubes used such low current at
high voltage that the power supplies were pretty high impedance output.
But solid state needed high current at low voltages, and the old
high impedance output power supplies didn't really work so well. You'd
see people adding and adding large value electrolytics to the output
of the rectifier(s) in the solid state supplies, trying to get that
low impedance output. And it was never a complete success.
You'd start to see zener diodes used to supply a constant voltage,
rather than real power, so they'd be used with transistors that would
pass the actual current. And then you'd see supplies where there
was feedback, so no matter what the load on the supply, the voltage
was constant. Gradually these came into force, so fewer and fewer
solid state supplies had anything but a regulated supply. But they
weren't regulated because the devices had an absolute need
for an exact and constant voltage, but because the regulating state
allowed for that low impedance output that the devices wanted.
It was also a lot easier to make up a solid state regulator than
a tube type regulator, since large numbers of transistors would
fit in the space of a single vacuum tube.
Then a few years later, the discrete regulators basically disappeared.
IC regulators became common, especially 3 terminal regulators. It
became so easy to use them, it was hardly worth not using them. No
need to figure out the current limiting resistor, and no need to adjust
it every time the load changed. It was great.
You now rarely see zeners. They are relegated to where a low current
voltage source is needed. And where the drain of a 3 terminal regulator
is prohibitive.
Michael
|
|
|
|
Author: John PopelishDate: 13:27 22-07-07
|
|
PeteS wrote:
> Here's a little challenge for the OP. Given that
>
> Q=CV
> C(series) = 1/(1/C1 + 1/C2) (alternatively C1C2/(C1+C2))
> C(parallel) = C1 + C2
> Qt(series) = Q(C1) = Q(C2)
> Qt(Parallel) = Q(C1) + Q(C2)
>
> Then take two capacitors with a ratio of 10:1 (say 1uF and 0.1uF). Hook
> them up in parallel and charge them to some voltage (say 5V)
>
> Disconnect them and reconnect in series (a thought experiment, really)
> and calculate the total voltage across the pair. You might be surprised
> at the result.
Is the challenge to find the error in these equations?
|
|
|
|
Author: PeteSDate: 14:22 22-07-07
|
|
John Popelish wrote:
<snip>
>> I can't really think of how
>> such a circuit would be made, though I know they surely exist. I
>> assume it uses capacitors and an oscillation of some kind, but I
>> dunno. When I think of changing voltages, I mostly just think of
>> transformers.
>
> There are ways to charge a pair of capacitors, in parallel, with the
> battery voltage, and then reconnect them, so that they are in series,
> doubling the battery voltage, and connect that voltage to an output
> storage capacitor to act as a power source while you go back and repeat
> the cycle. The key words for this sort of thing are [charge pump].
Here's a little challenge for the OP. Given that
Q=CV
C(series) = 1/(1/C1 + 1/C2) (alternatively C1C2/(C1+C2))
C(parallel) = C1 + C2
Qt(series) = Q(C1) = Q(C2)
Qt(Parallel) = Q(C1) + Q(C2)
Then take two capacitors with a ratio of 10:1 (say 1uF and 0.1uF). Hook
them up in parallel and charge them to some voltage (say 5V)
Disconnect them and reconnect in series (a thought experiment, really)
and calculate the total voltage across the pair. You might be surprised
at the result.
Cheers
PeteS
>
>> Anyhoo, any help or pointers would be greatly appreciated!
>
> Here is a basic tutorial for the buck and boost converters:
> http://www.national.com/appinfo/power/files/f5.pdf
>
|
|
|
|
Author: John PopelishDate: 15:48 22-07-07
|
|
PeteS wrote:
> John Popelish wrote:
>> PeteS wrote:
>>
>>> Here's a little challenge for the OP. Given that
>>>
>>> Q=CV
>>> C(series) = 1/(1/C1 + 1/C2) (alternatively C1C2/(C1+C2))
>>> C(parallel) = C1 + C2
>>> Qt(series) = Q(C1) = Q(C2)
>>> Qt(Parallel) = Q(C1) + Q(C2)
>>>
>>> Then take two capacitors with a ratio of 10:1 (say 1uF and 0.1uF).
>>> Hook them up in parallel and charge them to some voltage (say 5V)
>>>
>>> Disconnect them and reconnect in series (a thought experiment,
>>> really) and calculate the total voltage across the pair. You might be
>>> surprised at the result.
>>
>> Is the challenge to find the error in these equations?
>
> Snort
>
> Do tell. Let's see
>
> Q = CV. I think we can agree on that. It was figured out a long time
> before we were born.
Needs some interpretation. Q is the charge transfered by a
change in voltage V.
> Series capacitance; reciprocal addition is the same (for two devices) as
> product over sum.
True but not terribly applicable to placing two charged
capacitors in series and measuring their open circuit voltage.
> Parallel capacitance is simply the addition of the capacitances.
True, and useful in determining how much total charge it
will take to produce a given change in voltage across the
parallel pair.
> The charge on any number of caps in series of any capacitance is identical
> The total charge on any caps in parallel is the sum of charges.
What does that have to do with the open circuit voltage of
an arbitrary string of precharged capacitors?
> I don't think I said anything different above.
The question is applicability of those somewhat simplified
equations to this problem.
I assert that if you charge a 1 uF cap and and a a 1 nF
cap, each to 12 volts and place them in series, the total
voltage across the pair can be either zero volts or 24
volts, depending on the relative orientations. How do your
generalities show this to be true or false?
|
|
|
|
Author: Charlie SiegristDate: 16:27 22-07-07
|
|
On Sat, 21 Jul 2007 22:01:28 -0700, in message
<1185080488.316208.174740@22g2000hsm.googlegroups.com>, FyberOptic
<fyberoptic@gmail.com> scribed:
>The basics of diodes has always been "current can only flow one way",
That's a bad assumption, and not the "basics" of diodes at all. Maybe a
Cliff Notes version.
>...the zener diode apparently [will] "fail" at much lower rates...
Start understanding better by using the proper term, which is not "fail,"
it is "breakdown."
>So while this is possibly out of the scope of asking on a newsgroup,
Not at all, your question is exactly what this newsgroup is for.
>what sorts of applications could they be used for reliably? I hear
>they can be used as regulators, but only if the load is fairly
>constant...
Actually, as long as the load and the supply are reliable within a fairly
large range, the zener can be used pretty safely, as long as the regulation
does not have to be tight. Zeners have quite a large +/- tolerance range,
on the order of 15%.
> I'd like to power some of these sorts of things
>off of batteries sometime instead of a 7805 [or] an ac adapter, and am
>interested in the best way to go about it.
The best way to go about it is to choose a battery value that most closely
matches the need of the load. Any voltage regulation using a zener will
waste power from the battery, and there are many, many batteries available
on the market to pretty much match any requirement.
>On a side note, how possible would it be to boost the voltage from a
>couple of AA's to run 5v logic reliably?
It's best done, as above, by choosing the battery voltage you need. Any
voltage conversion, up or down, again, will require some possibly
significant amount of power just for the conversion. Since getting 5V with
a battery of AA cells is not practical without conversion, due to the
voltage variation of differing types, you want to find a marketed 5V rated
battery.
To choose the right one, figure out the amperage requirement of your
project, then decide how long you want to run the project reliably. You
mentioned wanting to use AA batteries, which are in the range of 2000-2500
mAh for NiMH rechargeables. Say you are going to use 1A for 4 hours.
That's 4 amp-hours. Multiply by 125% and you arrive at the requirement for
a 5 amp-hour (5000mAh) battery, which is twice the typical capacity of the
NiMH AA. Now find a 5000mAh 5V (typically 4.8V, 1.2V per cell) battery
pack, which can be had for around $30 US.
> When I think of changing voltages, I mostly just think of
>transformers.
Which, of course, don't work well with DC. Hence the prevalence of AC in
power transmission.
>Anyhoo, any help or pointers would be greatly appreciated!
|
|
|
|
Author: Charlie SiegristDate: 17:20 22-07-07
|
|
On Sun, 22 Jul 2007 15:48:36 -0400, in message
<lb6dnRjElsiBKT7bnZ2dnUVZ_uygnZ2d@comcast.com>, John Popelish
<jpopelish@rica.net> scribed:
>What does that have to do with the open circuit voltage of
>an arbitrary string of precharged capacitors?
That's the question of the day. I'm thinking the answer depends highly
upon unstated conditions, and then after the stating of said conditions,
lots of sammy-the-snake squiggly lines and cuss words such as "dv/dt" and
"logarithmic graph paper." Either that, or magic.
|
|
|
|
Author: PeteSDate: 20:27 22-07-07
|
|
John Popelish wrote:
> PeteS wrote:
>
>> Here's a little challenge for the OP. Given that
>>
>> Q=CV
>> C(series) = 1/(1/C1 + 1/C2) (alternatively C1C2/(C1+C2))
>> C(parallel) = C1 + C2
>> Qt(series) = Q(C1) = Q(C2)
>> Qt(Parallel) = Q(C1) + Q(C2)
>>
>> Then take two capacitors with a ratio of 10:1 (say 1uF and 0.1uF).
>> Hook them up in parallel and charge them to some voltage (say 5V)
>>
>> Disconnect them and reconnect in series (a thought experiment, really)
>> and calculate the total voltage across the pair. You might be
>> surprised at the result.
>
> Is the challenge to find the error in these equations?
Snort
Do tell. Let's see
Q = CV. I think we can agree on that. It was figured out a long time
before we were born.
Series capacitance; reciprocal addition is the same (for two devices) as
product over sum.
Parallel capacitance is simply the addition of the capacitances.
The charge on any number of caps in series of any capacitance is identical
The total charge on any caps in parallel is the sum of charges.
I don't think I said anything different above.
Cheers
PeteS
|
|
|
|
Author: Michael A. TerrellDate: 11:57 24-07-07
|
|
TE Chea wrote:
>
> | Here is a basic tutorial for the buck and boost converters:
> | www.national.com/appinfo/power/files/f5.pdf
>
> This pdf says a * has an efficiency of 80% @12v input.
> Then if I replace my 1990 Honda engine's 4 resistors (
> each 5.6 ohm ) which cut voltage ( max 12.9 v ) to fuel
> injectors ( solenoid @ 2.1 ohm ) with a *, can I avoid
> these 4 resistors' wastage of current ( 15.71 w ) by using
> a * which presumably will use just 5.89w x ( 20% ÷ 80%
> ) = 1.47w ?
> Was any * available in 1990, to Honda or car owners ?
> This engine's ECU was made by NEC ( japan ), could
> NEC include a * in this ECU, so only max 3.51v is fed
> to these 4 injectors ( w-o any need for resistors ) ?
>
> [Image]
This is not a binaries newsgroup. You can post it to
news:alt.binaries.schematics.electronic by attaching it as a GIF, JPEG,
or other common image file. Then, post a message here giving the name
of the post. If you post a binary file here, most people will never see
it.
--
Service to my country? Been there, Done that, and I've got my DD214 to
prove it.
Member of DAV #85.
Michael A. Terrell
Central Florida
|
|
|
|
Author: defaultDate: 06:30 28-07-07
|
|
On Tue, 24 Jul 2007 20:00:12 +0800, "TE Chea" <4ws@gmail.com> wrote:
>| Here is a basic tutorial for the buck and boost converters:
>| www.national.com/appinfo/power/files/f5.pdf
>
>This pdf says a * has an efficiency of 80% @12v input.
>Then if I replace my 1990 Honda engine's 4 resistors (
>each 5.6 ohm ) which cut voltage ( max 12.9 v ) to fuel
>injectors ( solenoid @ 2.1 ohm ) with a *, can I avoid
>these 4 resistors' wastage of current ( 15.71 w ) by using
>a * which presumably will use just 5.89w x ( 20% ÷ 80%
>) = 1.47w ?
>Was any * available in 1990, to Honda or car owners ?
>This engine's ECU was made by NEC ( japan ), could
>NEC include a * in this ECU, so only max 3.51v is fed
>to these 4 injectors ( w-o any need for resistors ) ?
>
The four resistors would only "waste" 15 watts if they were pulling
current all the time. The fuel injectors only open very briefly to
allow a spray of fuel in.
The resistors may also serve another function - like modify the inrush
current or help absorb flyback current.
15 watts is a drop in the bucket when it comes to auto efficiency -
even if the injectors stayed open 100% of the time. Synthetic motor
oil, intelligent driving, tire pressure, etc., will all make a greater
impact.
--
----== Posted via Newsfeeds.Com - Unlimited-Unrestricted-Secure Usenet News==----
http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups
----= East and West-Coast Server Farms - Total Privacy via Encryption =----
|
|
|
|
Author: quietguyDate: 18:29 31-07-07
|
|
Just thought I'd mention that Jaycar (and others I presume) sell a device
that does this - ie a portable USB supply
David
FyberOptic wrote:
>
> On a side note, how possible would it be to boost the voltage from a
> couple of AA's to run 5v logic reliably? I can't really think of how
> such a circuit would be made, though I know they surely exist. I
> assume it uses capacitors and an oscillation of some kind, but I
> dunno. When I think of changing voltages, I mostly just think of
> transformers.
>
> Anyhoo, any help or pointers would be greatly appreciated!
|
|
|
|
1 |
|
|
|
|
Contact | Electronic Portal
|
|
|
