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continuity across discretes

S

sudheervemana

Jan 1, 1970
0
Hi all,

Can anyone please clarify the following doubts.

1: When will Discretes(like capasitors,diodes and resistors) gets
opened or short in the circuit.

2:Continuity across the pins of discretes is seen only when they enter
in to what state like (open or short).

3:when i tried to measure the value of capasitors like 10uf on the
board ,it was showing in some nano farads,even when i changed the
capacitor it was showing very low value ,what could be the reason.

4.When discretes like (capasitor ,diode and resistor) gets
destroyed,to what state they will enter in to.

Thanks in advance,

With Regards,

sudheervemana.
 
R

Robert Baer

Jan 1, 1970
0
sudheervemana said:
Hi all,

Can anyone please clarify the following doubts.

1: When will Discretes(like capasitors,diodes and resistors) gets
opened or short in the circuit.

2:Continuity across the pins of discretes is seen only when they enter
in to what state like (open or short).

3:when i tried to measure the value of capasitors like 10uf on the
board ,it was showing in some nano farads,even when i changed the
capacitor it was showing very low value ,what could be the reason.

4.When discretes like (capasitor ,diode and resistor) gets
destroyed,to what state they will enter in to.

Thanks in advance,

With Regards,

sudheervemana.

The questions almost sound like they are from a class test.
Any electronic component can fail open, fail to a total short, or
(partly) fail so that the resistance is more or less than normal; and
some failures cannot be detected by resistance readings.
In-circuit readings of a given component will typically be different
than an out-of-circuit reading, due to shunting by other components.
Likewise, using a meter to measure circuit voltages and currents can
result in values that are not representative of actual (unloaded)
circuit values.
Voltmeters add a shunt resistance between the nodes, and current
meters add resistance in a line.
Sometimes, knowledge of the circuit *and* test equipment can be used
to calculate correct values from loaded values.
Other times, known value high-value resistors may need to be added in
series with a voltmeter to reduce loading to a tolerable level; then the
reading must be used along with those added resistors and meter
resistance, to calculate the circuit loaded value.
If the meter does not have the sensitivity to allow reasonable
accuracy in calculated circuit values, then the "slide-back" method may
become the method of last resort.
A good electronic technician always knows that there is an equivalent
of the Heisenberg's Principle involved with measurements, and takes into
account the characteristics of the measuring equipment.
 
N

N. Thornton

Jan 1, 1970
0
1: When will Discretes(like capasitors,diodes and resistors) gets
opened or short in the circuit.

When you tread on them, put too much V i or P thru them, overheat
them, or keep bending the leads till they break.

I cant help thinking thats not what you wanted, but I cant pull
anything more useful out of your question.

2:Continuity across the pins of discretes is seen only when they enter
in to what state like (open or short).

failure. Or normal funtion in a low value R or L.
3:when i tried to measure the value of capasitors like 10uf on the
board ,it was showing in some nano farads,even when i changed the
capacitor it was showing very low value ,what could be the reason.

misreading the meter, misreading the capacitor, failed cap, or
circuitry across the cap that stops the meter working.

4.When discretes like (capasitor ,diode and resistor) gets
destroyed,to what state they will enter in to.

Any. A small carbon reistor can make a nice carbon arc light. Its in
negative resistance then.


If you get your questions clearer perhaps we could be more helpful.


Regards, NT
 
P

Paul Burridge

Jan 1, 1970
0
Any. A small carbon reistor can make a nice carbon arc light. Its in
negative resistance then.

How can you have negative resistance? That implies a value of <0 ohms.
Even superconductors can't do that.
 
T

Tim Auton

Jan 1, 1970
0
Paul Burridge said:
How can you have negative resistance? That implies a value of <0 ohms.
Even superconductors can't do that.

*In* negative resistance. Look up tunnel diodes, they have negative
incremental resistance in a certain range too.


Tim
 
J

John Woodgate

Jan 1, 1970
0
I read in sci.electronics.design that Paul Burridge
How can you have negative resistance? That implies a value of <0 ohms.
Even superconductors can't do that.

This is *incremental* negative resistance, and a lot of electronic
devices exhibit it, from tetrode valves/tubes to tunnel diodes. There
are two sorts, with different shapes of V/I curve - N type and S-type

Use Courier font


| | /
| | /
C| | /
u| | \ Neg
r| / | \ res
r| / | \
e| Neg / | \
n| res / | /
t| /\ / | /
| / \ / | /
| / \ / | /
| / \/ | /
| / |/
________________________________ ____________________________
Voltage Voltage
 
N

N. Thornton

Jan 1, 1970
0
Paul Burridge said:
How can you have negative resistance? That implies a value of <0 ohms.
Even superconductors can't do that.

A possibly misleading term, but thats what its called.

Regards, NT
 
P

Paul Burridge

Jan 1, 1970
0
This is *incremental* negative resistance, and a lot of electronic
devices exhibit it, from tetrode valves/tubes to tunnel diodes. There
are two sorts, with different shapes of V/I curve - N type and S-type

Use Courier font


| | /
| | /
C| | /
u| | \ Neg
r| / | \ res
r| / | \
e| Neg / | \
n| res / | /
t| /\ / | /
| / \ / | /
| / \ / | /
| / \/ | /
| / |/
________________________________ ____________________________

Thanks, John. I'll investigate further in the text books...
 
R

Robert Monsen

Jan 1, 1970
0
Paul Burridge said:
Thanks, John. I'll investigate further in the text books...
Winston Churchill

http://www.patchn.com/tunneldiode.htm

note how current goes down as voltage increases. Also, for fun, look up
"negative-impedance converter" in AoE... Its a way of building a negative
resistance with an opamp... :)

Regards,
Bob Monsen
 
P

Paul Burridge

Jan 1, 1970
0
A possibly misleading term, but thats what its called.

Indeed. So all it basically amounts to is that there are regions of
operation where the usual Ohm's law rules don't apply but it doesn't
mean negative resistance in absolute terms even though that's the
implication of the dumb term. Goddit. :)
 
K

Keith R. Williams

Jan 1, 1970
0
Indeed. So all it basically amounts to is that there are regions of
operation where the usual Ohm's law rules don't apply but it doesn't
mean negative resistance in absolute terms even though that's the
implication of the dumb term. Goddit. :)

Ohm's law still applies to an incremental negative impedance. An
incremental change in voltage will cause a negative change in
current (slope of the curve is negative), for example. The
negative resistance (slope of the curve) is useful in analyzing
oscillators, flip-flops, single-shots, etc.

Of course negative real impedances exist too (second and fourth
quadrants in the curves above). Power has to be applied from the
outside though.
 
P

Paul Burridge

Jan 1, 1970
0
Ohm's law still applies to an incremental negative impedance. An
incremental change in voltage will cause a negative change in
current (slope of the curve is negative), for example. The
negative resistance (slope of the curve) is useful in analyzing
oscillators, flip-flops, single-shots, etc.

Of course negative real impedances exist too (second and fourth
quadrants in the curves above). Power has to be applied from the
outside though.

Oh bollocks. Well I thought I'd understood it, but I'm back to square
one again. :-( Maybe next year...
 
R

Robert Monsen

Jan 1, 1970
0
Paul Burridge said:
Oh bollocks. Well I thought I'd understood it, but I'm back to square
one again. :-( Maybe next year...

According to ohm law, resistance is V/I. However, thats assuming a linear
relationship between the quantities. This is never true in real life, which
is why ohm's law isn't a 'real' law, its an approximation.

If you generalize the definition of R at a point on V(I) to be dV/dI, then
you can see that R is really just the slope of the graph of measured V to I.
For a resistor, the slope is pretty much constant (and positive) for a
particular interval, so you say that R is constant, and equals
(approimately) (V(I) - 0) /(I - 0). Which is to say, R = V/I, which means
our new definition is a generalization of Ohm's law for resistors.

That graph can take on various shapes. The one described by the 'negative
resistance' portion of the tunnel diode graph has a negative slope, and thus
dV/dI (and our expanded definition of R) is negative there. Thats what they
mean by negative resistance.

Now, suppose there existed a device such that 1) it obeyed ohms law, and 2)
R < 0. Then, when V = 0, I = 0 (by 1), and when I increased, V across the
device decreased (by 2). This would be represented by a line going through
the origin with a negative slope on the V vs I graph. A device that
implements this is called a negative-impedance converter. Here is the
schematic:

___
+----|___|-----+
| R1 |
| |
| |\ |
in ----+----|-\ |
out | >------+
+----|+/ |
| |/ |
| |
| ___ |
+----|___|-----+
| R1
|
.-.
| |
| |R2
'-'
|
|
===
GND
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

At Vin = 0V, its obvious that Iin is also 0 (since the current across R2
must be then 0), so the graph of V(I) starts at (0,0).

Since R1 is equal on the top and bottom, and since there is negative
feedback, its clear that the current from the output of the opamp to either
junction is equal, and is equal to the current through R2, and is the
negative of the current flowing into the inverting junction.

Thus, if I is the input current, then Vin = V+ = -I*R2.

If we now apply ohms law, then the 'resistance' of the circuit to ground is
equal to -R2!

Regards,
Bob Monsen
 
P

Paul Burridge

Jan 1, 1970
0
According to ohm law, resistance is V/I. However, thats assuming a linear
relationship between the quantities. This is never true in real life, which
is why ohm's law isn't a 'real' law, its an approximation.

If you generalize the definition of R at a point on V(I) to be dV/dI, then
you can see that R is really just the slope of the graph of measured V to I.
For a resistor, the slope is pretty much constant (and positive) for a
particular interval, so you say that R is constant, and equals
(approimately) (V(I) - 0) /(I - 0). Which is to say, R = V/I, which means
our new definition is a generalization of Ohm's law for resistors.

That graph can take on various shapes. The one described by the 'negative
resistance' portion of the tunnel diode graph has a negative slope, and thus
dV/dI (and our expanded definition of R) is negative there. Thats what they
mean by negative resistance.

Now, suppose there existed a device such that 1) it obeyed ohms law, and 2)
R < 0. Then, when V = 0, I = 0 (by 1), and when I increased, V across the
device decreased (by 2). This would be represented by a line going through
the origin with a negative slope on the V vs I graph. A device that
implements this is called a negative-impedance converter. Here is the
schematic:

___
+----|___|-----+
| R1 |
| |
| |\ |
in ----+----|-\ |
out | >------+
+----|+/ |
| |/ |
| |
| ___ |
+----|___|-----+
| R1
|
.-.
| |
| |R2
'-'
|
|
===
GND
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

At Vin = 0V, its obvious that Iin is also 0 (since the current across R2
must be then 0), so the graph of V(I) starts at (0,0).

Since R1 is equal on the top and bottom, and since there is negative
feedback, its clear that the current from the output of the opamp to either
junction is equal, and is equal to the current through R2, and is the
negative of the current flowing into the inverting junction.

Thus, if I is the input current, then Vin = V+ = -I*R2.

If we now apply ohms law, then the 'resistance' of the circuit to ground is
equal to -R2!

Oh, FFS, I just got back from the pub. Gimme a break, will ya?
Seriously, I thank you sincerely for your input which I shall examine
in great detail 2morrow, when I may be sober (God forbid).
 
N

N. Thornton

Jan 1, 1970
0
Paul Burridge said:
Oh, FFS, I just got back from the pub. Gimme a break, will ya?
hehe

yes but it isnt, and thats the sticking point with the term 'negative
resistance'.

If I run a neg R device directly coupled, it will show resistance. It
will always show positive resistance according to R=V/I.

OTOH if I separate the ac and dc paths through it, the dc path shows
positive resistance and the ac path shows negative resistance. For an
ac signal it can thus behave like a neg R. It kinda does - but
according to the basic R formula R=V/I, it still has positive
resistance.


Regards, NT
 
K

Keith R. Williams

Jan 1, 1970
0
Oh bollocks. Well I thought I'd understood it, but I'm back to square
one again. :-( Maybe next year...

Oh Paul! Let me know when you're ready to discuss. I just got
back from the pub myself (football and all) ;-)
 
R

Robert Monsen

Jan 1, 1970
0
N. Thornton said:
yes but it isnt, and thats the sticking point with the term 'negative
resistance'.

If I run a neg R device directly coupled, it will show resistance. It
will always show positive resistance according to R=V/I.

I understand what you are saying. If you _define_ resistance to be V/I, then
given any V and I, the value of R is fixed. For a tunnel diode, this value
varies with V and I, and is always positive if V and I are positive. From
this point of view, the idea that resistance can be negative when V and I
are positive makes no sense. Thats where Paul's confusion lies.
OTOH if I separate the ac and dc paths through it, the dc path shows
positive resistance and the ac path shows negative resistance. For an
ac signal it can thus behave like a neg R. It kinda does - but
according to the basic R formula R=V/I, it still has positive
resistance.

(I don't understand your dc vs. ac example.)

If you increase the voltage, the current through a tunnel diode decreases
once its over its hump.

http://www.americanmicrosemi.com/tutorials/tunneldiode.htm

The region of the graph of I vs. V between this hump in current, and the
voltage where current starts increasing again, is _defined_ to be the area
of negative resistance.

Since V/I (or I/V) is never negative when V and I are both positive, how can
this be an 'area of negative resistance', unless resistance, as defined by
the creators of this graph, isn't really defined by V/I, but by the slope of
the graph (which is what is 'negative') in that region? That would mean they
think of resistance as the slope of the function V(I). Since the graph is
really I(V), I'll take it to be conductance, so dI/dV is negative in this
area.

If you then go on to _define_ conductance as dI/dV, the generalization is
consistent for devices like resistors where I/V is 'constant'. And, it fits
with the idea of negative resistance, as described by this tunnel diode
graph. If you integrate dI/dV from 0 to V, it gives you I. If you multiply
1/R by V, you also get I. Given a value for dI/dV, you can predict where I
will go as V changes in a region 'near' V by pretending that the graph near
that V is linear. Thus, for resistors, the definition is consistent, and for
tunnel diodes, it explains the notion of 'negative resistance'.

On another point, I just realized what a tunnel diode is good for. If you
slowly increase I, at some point, it'll jump from the 'hump' to the other
part of the graph, giving a really quick increase in V. Then, when you
slowly decrease the current, the voltage will again jump from one point on
the graph to the other, giving a quick decrease in V. I wish they were
readily available, I'd like to play with one.
Regards, NT

Regards,
Bob Monsen
 
J

John Woodgate

Jan 1, 1970
0
I read in sci.electronics.design that Robert Monsen
I wish they were
readily available, I'd like to play with one.

A small neon tube has the same sort of characteristic and is much more
readily obtained. Just use a suitably higher voltage.
 
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