Likbez - How does the protection of the circuit elements.

<b>Likbez</b> - How does the protection of the circuit elements.

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            How does the defense.



   For those who do not know, but wants to know ...
The following chain of reasoning dliiiinnaya chewed percent to 
80 percent, and digested at 50. Enjoy your meal! Swallow ... 



                 Of the diode.


   Diode passes current if the anode is attached greater 
potential than the cathode, ie if the anode plus relative to 
the cathode, the which is negative (toward the anode). But

it's not all! diode is not ideal and leaves
a portion of the voltage - the loss, it is necessary to open 
the diode and the transmission toka.Poteri in silicon diodes 
(about 0.7V) is greater than that of germanium (about 0.3

B). Losses grow with increasing current through the
diode and germanium diodes losses grow
more linear (and more), so at some relatively high current loss
both types sravnivayutsya.T.e.pri low current loss on a 
germanium diode is less than on silicon. If the diode voltage 
is applied less than the stress of opening, it will not pass 
current, except that negligible. T.e.esli you leave a silicon 
diode polvolta all the losses, horseradish He opens 
you-miss-talk normalnoyvelichiny.Esli interesting - open 
textbook Physics of high school, there should

be characteristic of the diode: the dependence of current
through the diode voltage on it (or vice versa).



               Potentials.


   The highest potential - the "absolute
Plus - it's +12 V supply (except power supply). The lowest - an 
"absolute negative" - a common wire GND, common minus 5 and 12ti

volt power supply. Then what is the +5 V? - Plus
or a minus? 12B is a disadvantage with respect to -
t.k.menshy potential (5 <12) to common (GND) is a plus - 
t.k.bolshy potential (5 0). What is a stem 40 KR1818VG93 
controller? In the absence of current is 0V, ie, GND. In the 
presence of current is a positive potential is proportional to 
toku.Esli note that the current increases from 0, the initial 
state leg 40 - "an absolute negative." In fact, actually on it 
already may be some tension on the signals, or 5V, t.e.ot 
internal electronic communications chips. Similar property and 
other +12 consumers, such as drive and amplifier. They also 
have an initial 0-vym potential, which in turn

Supply is growing under the influence of the current block
supply.

   According to the accepted principle that the current flows
from a higher to a lower capacity (taken for convenience, in 
fact, determined by the type of carriers). Flows at

potential difference, with equal potential power there. The 
current direction of the plus to minus inherent in the 
designation of the diode and bipolar transistors. Triangular 
Anode diode forms an arrow showing the direction in which the 
diode current flows when properly attached thereto potentials. 



             Thoughts in a bunch ...


   To map the top diode is included in the direct
direction, its anode at the highest potential, and cathode at a 
lower - namely, a stem 40. 

   Under normal diet after operation,
voltage at the anode increases, while at the cathode
is the 0th - negative potential
stem 40. This potential difference provides the upper diode, 
and it skips to the current leg 40. The potential for this leg 
starts increase under the current through the diode,

but the potential of stem 40 - cathode
will always be less than the potential of +12 V power supply - 
an anode and, accordingly, will be a positive potential 
difference is needed for opening the upper dioda.Po essentially 
diode can not miss such a large current, which

voltage on its cathode, becomes too
close to the voltage at its anode and the difference they will 
be less need for very diode - it otkryvaniya.Diod no fool to

open and not keep the tension - and himself close. Therefore,
equilibrium is established: the voltage drop across the diode 
opens the diode so to talk through it gave such a potential

stem 40, the difference between the power supply voltage with
that would provide the necessary voltage drop across the diode 
- the losses. 

   In this second diode will be closed, because
at its cathode will be high (approximately 12V)
potential of stem 40 and the anode (lower
scheme) a lower potential power supply +5 V
This corresponds to the inverse diodes, and he does not pass 
current, except for insignificant low current utechki.Mozhno 
consider the situation differently. Leg 40 on the scheme is 
above +5 V, and in the presence of +12 V has a large

potential. According to the accepted principle
current flows from larger to smaller capacity, in our case, 
from top to bottom, and the lower triangle of the diode 
indicates the bottom up - incorrect inclusion of the lower 
diode is closed and the current will not pass (the upper diode 
indicates top down - that corresponds to a positive potential 
difference (from high to low, a greater potential negative lower

give a positive difference).

   Now imagine that +12 V
disappeared, or become isolated, in general, has become 0V.
Imagine that the scheme turned up
feet (diodes are reversed), and instead of +12 V should provide 
an inscription or 0V GND - it is now "an absolute negative." 
Some time (due to parasitic capacitance) at leg 40 will be a 
high potential that starts to fall. Because anode of the former 
top diode will be less capacity and a cathode,

due to parasitic capacitance - in more that
gives a negative potential difference. AND
vlyuchenie diode reverse: it is closed and not
will carry a current. Stress on the leg
40 will fall and at one point
is less than +5 V. Former lower diode (in
our view it is upside down the top) will be in direct 
inclusion, because on its anode is a great potential (+5 V)

than at the cathode - Leg 40. It will open and
will miss the current 5-volt power supply to the stem 40, so 
that its capacity will cease to fall, and then comes the 
balance of losses in the diode voltage, according description 
of the above equilibrium in the normal nutrition.


   We now assume that the +12 V power again appeared - back to 
the unturned scheme - if the top diode will be opened and

will raise the potential of stem 40 to
+12 V. And the lower diode is closed, since voltage on its 
cathode - foot 40 - will be rasti, and at the anode (+5 V 
supply) does not change As a result, the potential difference 
(anode +5 subtract the cathode +12 V) is negative, and a 
reverse voltage of diode closes. 


   Description of the above processes occur almost
instantly and stretched for ponimaniya.Vozmozhno that your PSU 
+12 V will grow more slowly than the plus 5V, while

process of inclusion will take place on
disaster scenario. Scheme as it will
turned from the very beginning, starting
"5 volt" LED, raising the potential of stem
40 to an emergency - about 5V - voltage, and
with the growth of +12 V scheme "roll over" back to normal, and 
there will be process, described above. But even with this

power it takes a split second, and the computer itself will be 
at the stage of discharge, and by the time the disk will be 
powered normal. This is possible if you have a comp

- Single 5-volt device with the internal shaper voltage of 5 - 
"12B specifically for the controller (and drive -

external device is powered separately.) In
this case, the protection circuit is required because
converter - a relatively complex device and can fail.


   But what if the +12 V drops not to
0V? The scheme does not "roll over" as described
above. In this case, it all depends on the ratio of potential 
supply voltage and the loss of the diodes. Since the potential 
of a free 40 feet can be considered as 0-vym or negative (it 
always flows from a certain supply current - positive source - 
any application of a positive building), then the source of 
stress which will be more, and will open its diode.

In case of equality of stresses will
one diode drop which will be less.

   If you gently lower the supply voltage
+12 V, then at about 5 V (depending on the type of
diodes and the scatter of their parameters) can find a voltage 
ratio, when both diode will be opened, and currents towards the 
legs 40 will be the sum of the currents through both diodes. As

Whatever it was in violation +12 - most importantly, that the 
controller will be left without food, t.k.budet eat either one 
or from both sources at once.



   Now imagine that disappeared 5V, and 12V
left. It is not dangerous. 12V - auxiliary power supply, it 
shifts the pathways operating in the circuit 5B, and unable to 
damage the chip.Eto control voltage, not dial-up. With regard 
to protection schemes, the situation with a normal diet did not 
change in the presence of 12 and the absence of 5B, t.k.nizhny

LED will remain closed even lower -
0-vym potentsalom, and the top will continue to otkryt.Esli 
after five volts will disappear more and 12B, received a 
classical off-state companies. 


   And what if after the disappearance of the +12 V will
fall and +5? This is probably the most unpleasant
case. Since the rest - 5B - the voltage across the lower diode 
protects mikruhu. When his work reducing the chip will

increasingly disturbed. But cascade controller powered by the 
same 5 different volt - are mitigated (we do not consider a 
break in the circuit protection circuit, and the overall 
decline in food com). It is known that the current through all 
the transistors in part proportional to the applied These 
napryazheniyu.A a MOS voltage applied to the gate, sets current 
fit and open, and at lower main power supply current is 
uncontrolled snizhatsya.No as accessory power will decrease, 
the violation of modes will increase, and uncontrolled current 
tozhe.V partly a result, the product of voltage at current 
yields power which, when intact protection scheme, there is no 
+12 V and lower +5 V, there will be little changed (most likely

slightly falling), and it will not lead to overheating of the 
controller. And at lower ground power less than 2-3V power 
consumption of the controller will plummet, and overheating 
will decrease, so as field effect transistors of any structure:

1) is normally operated at a sufficiently large (nominal in 
this case) voltage, 2) at lower voltage affects saturation 
(much greater than that of bipolar transistors) and resistive 
work area - transistors start behaving like resistors.



   Here, perhaps, and reviewed all the situations
protection, it remains to sum itog.Dlya What is the bottom LED 
on the circuit? We need to supply controller in the absence of 
the +12 V from another source - +5 V. But lock the leg

40 to +5 V can not, because normal diet
12 In this case, closes at 5 V and can burn the complex or unit 
pitaniya.Mozhno, of course, vlepit resistor, but it will eat a 
lot of current at normal pitanii.Tok will flow through the 
resistor from the source +12 V to +5. Energy loss on the 
resistor, the upper diode, taking into account losses have a 
more powerful, power supply is stronger - in general, some 
losses. The diode is ideal - it serves eletronnogo switch, not

squander the source of +12 V to +5 V and automatically feeds 
the controller with the loss of 12 V. And what is then needed 
top diode? Anything can happen, the worst -

is the closure of 12 V. If the upper diode
will happen and the closure, the bottom is not
save, rather, it depends on the capacity of the lower 
dioda.Moschny diode squandered +5 V on the circuit, and either 
burn down the fault, and tension, at least partially restored,

either dies or the power supply works
some of its protection from disconnection. Weak
diode starves himself (not withstand the fault current) and 
then either dies controller if diode to break the chain, or, in 
the case of closure in the diode, the situation is repeated 
with a powerful diode. But even if the circuit will not, the 
diode will nourish not only the controller, but

everything hangs on the +12 V line, and if he
weak, for him it may be equivalent to
closure (see above). The upper diode is needed,
that the current emergency power did not leave on
drive and power, for example (load
12V), and fed only the controller.
The lower diode is needed to supply the main current of +12 V 
is not dripping into a load of +5 V, perenapryagaya 
(increasing) its power from the +5 V to +12 V.




     Some information for absolute

     fools, and / or specific lazy:



   1. Postavish only the top diode (lower
zamknesh) - at best spalish diode
(If it is a low-power), and at worst (the powerful diode) 
spalish computer and power supply .... and drive, and 
AU-Schnick ... all that crap, what  ponavesil ... Here is such 
a OBLAMIS! 

   2. Postavish only the lower (upper zamknesh) - in the best 
case (emergency food) VG-scale will not burn, at worst - will 
strike a diode, and then see a worst case situation 1.



   So break off SEOs, it is easier
These two parts do not already.
However ... izvraschenstvu no limit? It seems
there might be such helluva lot of smart-genius who (quite 
logically) uzreyut two diodes pnp transistor structure ... 

   I specifically did not say what conclusions transistor what 
conclusions diodes match (not to provoke the slackers). And 
perverts will say that once a long time, when I started to 
learn the electronics, I've tried to make a transistor from the 
two diodes (Then there was a shortage of parts), and if you do

actually succeeded, I congratulate you!, but you are not of 
this world ... 

   If you vlepit transistor, at best,
If lift energy company, and
at worst Burn transistor, and ... sm.hudshy
case of situation 1.

   In fact, there are dual diode assembly with common cathode, 
but they are powerful, and physically awkward to use in our 
case. 



               Conclusion.


   When preparing this article, I looked
some publications (I have a few), and
found two mentions of protection. Just do not
there is always time to look at their own
drive (but there is always a chance to find them
something new:). Sometimes, it is quite realistic, people
publish the scheme run by themselves do not ponimayut.V my 
stories, I'm trying to justify everything. 

   I have a (purely practical) interest on the implementation 
of security with minimal loss of voltage. Of course, the scheme 
will not most reliable on the number of components.

I hope the relay will not think about ...


   According to the Timex / Sinclair Club, he used KD521 diodes 
and tested the scheme - it works as expected. Well, thanks to 
him for check, I would not dare ...



   Recently, a person who does not have competent 
representation of the company, even in general, glance at the 
square plastic case with the Pentagon, asked: what is it? I 
said that computer. So little? - He was surprised.

People, and yet many of us actually
did not notice!

   Spectrum is not dying, it progresses! AND
become more compact and economical in
Compared with modern microwave
bashnyami.Pro resource availability and the user can not tell 
... 



  The last days of June 2003. By KSA-7G.