Fuse and LED design question

J

Junkmail789

Guest
I have a fuse that cannot be checked by visual, it must be ohm'd out. If I
put an LED and resistor in parallel with it and run this second leg to
ground, the LED should light when the fuse blows and NOT supply power to the
other side of the fuse.

Am I correct?

Thanks,

Matt
 
It will supply the LED current to the rest of the circuit.

You will have to know the circuit voltage and I assume this is a DC circuit
fused.

"Junkmail789" <junkmail789@comcast.net> wrote in message
news:1uydnT5d-8NIfB6iRVn-tg@giganews.com...
I have a fuse that cannot be checked by visual, it must be ohm'd out. If I
put an LED and resistor in parallel with it and run this second leg to
ground, the LED should light when the fuse blows and NOT supply power to
the
other side of the fuse.

Am I correct?

Thanks,

Matt
Click to expand...
--
*
| __O Thomas C. Sefranek WA1RHP@ARRL.net
|_-\<,_ Amateur Radio Operator: WA1RHP
(*)/ (*) Bicycle mobile on 145.41, 448.625 MHz

http://hamradio.cmcorp.com/inventory/Inventory.html
http://www.harvardrepeater.org
 
I do not understand what you mean by "second leg to ground".
The Led and resistor should just be across the fuse.
This was a common ploy back in the High voltage days of tubes and a neon was
used instead of a LED.
You will have to know the voltage and the effect of the LED current on the
circuit after it has blown the fuse.
Tell us a few facts and we might offer a bit of advice.
--
John G

Wot's Your Real Problem?

"Junkmail789" <junkmail789@comcast.net> wrote in message
news:1uydnT5d-8NIfB6iRVn-tg@giganews.com...
I have a fuse that cannot be checked by visual, it must be ohm'd out. If I
put an LED and resistor in parallel with it and run this second leg to
ground, the LED should light when the fuse blows and NOT supply power to
the
other side of the fuse.

Am I correct?

Thanks,

Matt
Click to expand...
 
On Wed, 8 Oct 2003 06:07:03 -0400, "Junkmail789"
<junkmail789@comcast.net> wrote:

I have a fuse that cannot be checked by visual, it must be ohm'd out. If I
put an LED and resistor in parallel with it and run this second leg to
ground, the LED should light when the fuse blows and NOT supply power to the
other side of the fuse.

Am I correct?
Click to expand...
---
Probably not.


Why not do this:

DC>---[FUSE]--+------+
| |
[R] |
| [LOAD]
[LED] |
| |
GND>----------+------+

The LED will go out when the fuse blows.

--
John Fields
 
"Junkmail789" <junkmail789@comcast.net> wrote in message
news:1uydnT5d-8NIfB6iRVn-tg@giganews.com...
I have a fuse that cannot be checked by visual, it must be ohm'd out. If I
put an LED and resistor in parallel with it and run this second leg to
ground, the LED should light when the fuse blows and NOT supply power to
the
other side of the fuse.

Am I correct?

Thanks,

Matt
Click to expand...
You are looking at a circuit like this:

|
|
|
+----+------+
| |
.-. |
( X )LED |
'-' |
| |
| |
.-. ,-.
| | |||
| |R ||| Fuse
'-' '-'
| |
+---+-------+
|
|
===
GND
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

If so, then yes, the LED will not light unless the fuse blows. However,
there WILL be current through the LED/resistor pair after the fuse blows.
You should limit it by making the resistor as large as it can be while still
glowing sufficiently under failure conditions.

If you use an NPN transistor where the base connects to the + side of the
fuse, you can separate the two circuits more cleanly as so:

VCC
+
|
+---------------+
| |
.-. |
| |
| | Your Load Goes Here
'-'
| |
| |
\| ___ |
|----|___|- --+
<| |
| ,-.
| |||
.-. |||
( X ) '-'
'-' |
| |
+---------------+
|
===
GND
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

That way, when the fuse blows, the transistor will turn on, but your load
won't be powered through the LED/resistor pair. The base resistor should be
chosen low enough so that when the fuse blows, the voltage drop across the
base generates enough current to saturate the transistor, but not low enough
to cause the PN junction to pass too much current and fry either the LED or
transistor.

Regards,
Bob Monsen
 
On Wed, 08 Oct 2003 16:22:44 GMT, "Robert Monsen"
<postmaster@BulkingPro.com> wrote:


VCC
+
|
+---------------+
| |
.-. |
| |
| | Your Load Goes Here
'-'
| |
| |
\| ___ |
|----|___|- --+
| |
| ,-.
| |||
.-. |||
( X ) '-'
'-' |
| |
+---------------+
|
===
GND
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

That way, when the fuse blows, the transistor will turn on, but your load
won't be powered through the LED/resistor pair. The base resistor should be
chosen low enough so that when the fuse blows, the voltage drop across the
base generates enough current to saturate the transistor, but not low enough
to cause the PN junction to pass too much current and fry either the LED or
transistor.
Click to expand...
---
Usually, when the fuse blows, it's desirable to remove the source of
power from the load instead of just removing the load from ground.

That can be done like this, and have the LED light when the fuse blows
as well:

+DC>--+--[FUSE]--+------+
| | |
E | |
PNP B--------+ |
C | |
| | [LOAD]
[LED] | |
| [R] |
[R] | |
| | |
GND>--+----------+------+

--
John Fields
 
A much better idea is to take it from the fused side (load side, not supply side)
thru the resistor to return (ground).

This is called a pilot light.
The indication will be the inverse of what Matt wanted,
but the results will be more predictable.
 
---
Usually, when the fuse blows, it's desirable to remove the source of
power from the load instead of just removing the load from ground.

That can be done like this, and have the LED light when the fuse blows
as well:

+DC>--+--[FUSE]--+------+
| | |
E | |
PNP B--------+ |
C | |
| | [LOAD]
[LED] | |
| [R] |
[R] | |
| | |
GND>--+----------+------+

--
John Fields
Click to expand...
Maybe it's wise to put a diode between B (A) and E (K) of the transistor, to
protect it if the suddenly drop of current causes high(er) voltage spike at
the load.

Al
 
On Thu, 9 Oct 2003 00:26:58 +0200, "The Al Bundy"
<the_al_bundy@*^no-spam!^*hotmail.com> wrote:

---
Usually, when the fuse blows, it's desirable to remove the source of
power from the load instead of just removing the load from ground.

That can be done like this, and have the LED light when the fuse blows
as well:

+DC>--+--[FUSE]--+------+
| | |
E | |
PNP B--------+ |
C | |
| | [LOAD]
[LED] | |
| [R] |
[R] | |
| | |
GND>--+----------+------+

--
John Fields

Maybe it's wise to put a diode between B (A) and E (K) of the transistor, to
protect it if the suddenly drop of current causes high(er) voltage spike at
the load.
Click to expand...
---
Good idea, but since the only way the load could generate a spike is if
it looked inductive when the fuse blew, this might work better:

+DC>--+--[FUSE]--+------+---------+
| | | |
E | | |
PNP B--------+ | |
C | | |K
| | [LOAD] [DIODE]
[LED] | | |
| [R] | |
[R] | | |
| | | |
GND>--+----------+------+---------+

--
John Fields
 
Let me clarify a few points, now that I see the other ideas presented.

1. I don't want any load on the circuit at all. That's why I was thinking of
doing it in parallel, but I was unsure of the size of the resistor in order
to get the least amount of current, but then if you get the resistor high
enough so that the LED doesn't light does this leg actually conduct?

Didn't want to go across the fuse itself because then the circuit would
still get power through the R and LED.

Didn't want to go in series because of #1.

So let's try this question--

Is there any way to check the status of a fuse without putting a load into
the circuit?

To give you guys an idea of my problem. We use gas analyzers that contain
thermal fuses. I don't want to interfere with any of the internal circuitry
but would like to be able to have a LED or alarm like a piezo to indicate
that the fuse has blown open. As it stands we have to remove the cover,
unplug the fuse and ohm it out.

Thanks in advance you guys are a great help.


"John Fields" <jfields@austininstruments.com> wrote in message
news:jt49ovkbhmgemvrkurstseu4sd4peq8fes@4ax.com...
On Thu, 9 Oct 2003 00:26:58 +0200, "The Al Bundy"
the_al_bundy@*^no-spam!^*hotmail.com> wrote:

---
Usually, when the fuse blows, it's desirable to remove the source of
power from the load instead of just removing the load from ground.

That can be done like this, and have the LED light when the fuse blows
as well:

+DC>--+--[FUSE]--+------+
| | |
E | |
PNP B--------+ |
C | |
| | [LOAD]
[LED] | |
| [R] |
[R] | |
| | |
GND>--+----------+------+

--
John Fields

Maybe it's wise to put a diode between B (A) and E (K) of the transistor,
to
protect it if the suddenly drop of current causes high(er) voltage spike
at
the load.

---
Good idea, but since the only way the load could generate a spike is if
it looked inductive when the fuse blew, this might work better:

+DC>--+--[FUSE]--+------+---------+
| | | |
E | | |
PNP B--------+ | |
C | | |K
| | [LOAD] [DIODE]
[LED] | | |
| [R] | |
[R] | | |
| | | |
GND>--+----------+------+---------+

--
John Fields
Click to expand...
 
On Wed, 8 Oct 2003 20:04:44 -0400, "Junkmail789"
<junkmail789@comcast.net> wrote:

Let me clarify a few points, now that I see the other ideas presented.

1. I don't want any load on the circuit at all. That's why I was thinking of
doing it in parallel, but I was unsure of the size of the resistor in order
to get the least amount of current, but then if you get the resistor high
enough so that the LED doesn't light does this leg actually conduct?

Didn't want to go across the fuse itself because then the circuit would
still get power through the R and LED.

Didn't want to go in series because of #1.

So let's try this question--

Is there any way to check the status of a fuse without putting a load into
the circuit?

To give you guys an idea of my problem. We use gas analyzers that contain
thermal fuses. I don't want to interfere with any of the internal circuitry
but would like to be able to have a LED or alarm like a piezo to indicate
that the fuse has blown open. As it stands we have to remove the cover,
unplug the fuse and ohm it out.

Thanks in advance you guys are a great help.
Click to expand...
---
I'm confused with what you mean by "parallel" and "series" and about
what you mean when you say that that you don't want to interfere with
the circuitry. If you want to modify the instrument to include this
blown fuse detector you're going to have to mess with _something_, no?

If you can add circuitry around the fuse you should be able to use the
circuit shown below which will put about an extra 120ľA load on the
supply when the fuse is intact and about a 2mA load on the supply when
the fuse blows if you use a high-efficiency LED like an HLMP-4700.

Plus, when the fuse blows and the LED comes on there will be absolutely
_no_ current flowing into the load.

Some other info would be handy, like how big is the fuse (amps) and is
it really on the DC side of the supply?



+DC>--+--[FUSE]--+------+
| | |
S | |
PCH G--------+ |
D | |
| | [LOAD]
[LED] | |
| [100K] |
[R] | |
| | |
GND>--+----------+------+

--
John Fields
 
"Junkmail789" <junkmail789@comcast.net> wrote in message
news:1uydnT5d-8NIfB6iRVn-tg@giganews.com...
I have a fuse that cannot be checked by visual, it must be ohm'd out. If I
put an LED and resistor in parallel with it and run this second leg to
ground, the LED should light when the fuse blows and NOT supply power to
the
other side of the fuse.

Am I correct?

Thanks,

Matt
Click to expand...

Will this type of circuit do this work? Does it have too much overhead
in parts? I am a newbie and would like to know. This looks like it
should light up the LED on event failure (blown fuse). Could another
device like a FET or Transistor do the same thing (allow no current
to flow until the base looses power) as the relay?




o----------------------------o---------o
| |
| LED/LAMP |
| ___ .---. |
GND>--o---|___|--| X |o-----o |
R1 '---' | |
| | | Relay
o /o _|_
/ -|_\_|
/ |
o |
| |
o------------------o |
| |
| ____ |
+DC>------o--|_--_|-o---------------o---------o

Fuse

created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de


ChadMan
 
"ChadMan" <chadhammer@nospamcharter.net> wrote in message
news:vobb3l6vbsu572@corp.supernews.com...
Will this type of circuit do this work? Does it have too much overhead
in parts? I am a newbie and would like to know. This looks like it
should light up the LED on event failure (blown fuse). Could another
device like a FET or Transistor do the same thing (allow no current
to flow until the base looses power) as the relay?




o----------------------------o---------o
| |
| LED/LAMP |
| ___ .---. |
GND>--o---|___|--| X |o-----o |
R1 '---' | |
| | | Relay
o /o _|_
/ -|_\_|
/ |
o |
| |
o------------------o |
| |
| ____ |
+DC>------o--|_--_|-o---------------o---------o

Fuse

created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de


ChadMan
Click to expand...
Well it would work, only the relay should be activated at all time. Only
when the fuse blows it goes off and switches on the LED. In normal operation
(fuse ok) the relay draws some current and generates heat. I guess you do
not want that.
So at any time this circuit will draw a current.


The idea with a FET that John Fields drawed is a good one (last posting of
him). Only some diodes must be placed to prevent that the FET goes up in
smoke when the fuse blows, beteen G-S and between the Gate/resistor
connection and the load.
When the load has a huge capacitance (buffer) it is possible that the load
voltage drops slower when the fuse just went. This means a high voltage
across the G-S of the FET appears and can damage it. The G-S diode protects
it from high voltage spikes caused by interrupting a high current.

In this schematic the resistor between the gate/load and GND defines the
current that will be drawn through the fuse. This is much less then a coil
of a relay.
When a Mosfet is turned off the resistance between Drain and Source is very
high (>10meg), so no current goes through. (Be aware of the body diode of
the mosfet, between Drain-Source)
So in total only the current through the resistor OR through the LED is
drawn.

Hope this helps.

Al
 
"John Fields" <jfields@austininstruments.com> wrote:
news:jt49ovkbhmgemvrkurstseu4sd4peq8fes@4ax.com...
---
Good idea, but since the only way the load could generate a spike is if
it looked inductive when the fuse blew, this might work better:

+DC>--+--[FUSE]--+------+---------+
| | | |
E | | |
PNP B--------+ | |
C | | |K
| | [LOAD] [DIODE]
[LED] | | |
| [R] | |
[R] | | |
| | | |
GND>--+----------+------+---------+

--
John Fields
Click to expand...
In this way it protects that no reverse voltage can appear across the load.
However when there is some inductance in the load (wires, etc) and the
current is suddenly interrupted, the inductance generates a voltage(spike)
on top of the voltage that is still across the load. On the right side of
the fuse a higher voltage (spike) appears then the input DC voltage, and so
a too high reverse B-E voltage at the transistor. The diode across the load
wont stop this.

Al
 
On Thu, 9 Oct 2003 23:46:54 +0200, "The Al Bundy"
<the_al_bundy@*^no-spam!^*hotmail.com> wrote:

"John Fields" <jfields@austininstruments.com> wrote:
news:jt49ovkbhmgemvrkurstseu4sd4peq8fes@4ax.com...

---
Good idea, but since the only way the load could generate a spike is if
it looked inductive when the fuse blew, this might work better:

+DC>--+--[FUSE]--+------+---------+
| | | |
E | | |
PNP B--------+ | |
C | | |K
| | [LOAD] [DIODE]
[LED] | | |
| [R] | |
[R] | | |
| | | |
GND>--+----------+------+---------+

--
John Fields

In this way it protects that no reverse voltage can appear across the load.
However when there is some inductance in the load (wires, etc) and the
current is suddenly interrupted, the inductance generates a voltage(spike)
on top of the voltage that is still across the load. On the right side of
the fuse a higher voltage (spike) appears then the input DC voltage, and so
a too high reverse B-E voltage at the transistor. The diode across the load
wont stop this.
Click to expand...
---
Yes, it will, since the high voltage spike will be negative and will
largely dissipate in the diode. During the time the diode is conducting
there will be about -1V applied to the base because of the drop across
the diode, but all this will do is turn the PNP on harder while any
inductance in the load is discharging.

--
John Fields
 
On Thu, 9 Oct 2003 23:39:40 +0200, "The Al Bundy"
<the_al_bundy@*^no-spam!^*hotmail.com> wrote:


The idea with a FET that John Fields drawed is a good one (last posting of
him). Only some diodes must be placed to prevent that the FET goes up in
smoke when the fuse blows, beteen G-S and between the Gate/resistor
connection and the load.
When the load has a huge capacitance (buffer) it is possible that the load
voltage drops slower when the fuse just went. This means a high voltage
across the G-S of the FET appears and can damage it. The G-S diode protects
it from high voltage spikes caused by interrupting a high current.
Click to expand...
---
No. If the load looks partly like a huge capacitance in parallel with
the rest of the load there is no way a voltage more positive than +DC
can appear on the gate when the fuse blows.

+DC>--+--[FUSE]--+------+--------+
| | | |
S | | |
PCH G--------+ | |
D | | |+
| | [LOAD] [BFC]
[LED] | | |
| [100K] | |
[R] | | |
| | | |
GND>--+----------+------+--------+

Notice that while the fuse is intact the BFC will be charged up to +DC.
The instant the fuse blows the BFC will still be charged up to +DC, and
cannot become more positive than +DC. When the fuse blows, only the BFC
will be supplying operating current for the load and for the 100K ohm
gate resistor. As current is drawn from the BFC the voltage across it
will decay until the point is reached where its voltage will no longer
be sufficient to keep the P-Channel FET turned OFF, then as the gate
voltage falls closer and closer to GND the FET's channel resistance will
decrease until eventually, when the gate-to-source capacitance is fully
charged, the drain-to-source resistance will fall to as low a value as
it can.

In this schematic the resistor between the gate/load and GND defines the
current that will be drawn through the fuse. This is much less then a coil
of a relay.
When a Mosfet is turned off the resistance between Drain and Source is very
high (>10meg), so no current goes through. (Be aware of the body diode of
the mosfet, between Drain-Source)
So in total only the current through the resistor OR through the LED is
drawn.

Hope this helps.

Al
Click to expand...
--
John Fields
 
"Junkmail789" <junkmail789@comcast.net> wrote in message news:<FtydnSNlRJK3OxmiRVn-uA@giganews.com>...

To give you guys an idea of my problem. We use gas analyzers that contain
thermal fuses. I don't want to interfere with any of the internal circuitry
but would like to be able to have a LED or alarm like a piezo to indicate
that the fuse has blown open. As it stands we have to remove the cover,
unplug the fuse and ohm it out.
Click to expand...
LEd/resistor across the fuse is almost certainly good.

If thats no go for some odd reason, try an opamp monitoring the V at
each end of the fuse.

Regards, NT
 
"John Fields" <jfields@austininstruments.com> wrote
news:qmobov4sjfelesr9f7o8vbpsga667chqap@4ax.com...
---
No. If the load looks partly like a huge capacitance in parallel with
the rest of the load there is no way a voltage more positive than +DC
can appear on the gate when the fuse blows.

+DC>--+--[FUSE]--+------+--------+
| | | |
S | | |
PCH G--------+ | |
D | | |+
| | [LOAD] [BFC]
[LED] | | |
| [100K] | |
[R] | | |
| | | |
GND>--+----------+------+--------+

Notice that while the fuse is intact the BFC will be charged up to +DC.
The instant the fuse blows the BFC will still be charged up to +DC, and
cannot become more positive than +DC. When the fuse blows, only the BFC
will be supplying operating current for the load and for the 100K ohm
gate resistor. As current is drawn from the BFC the voltage across it
will decay until the point is reached where its voltage will no longer
be sufficient to keep the P-Channel FET turned OFF, then as the gate
voltage falls closer and closer to GND the FET's channel resistance will
decrease until eventually, when the gate-to-source capacitance is fully
charged, the drain-to-source resistance will fall to as low a value as
it can.

--
John Fields
Click to expand...
But with this circuit is it possible (long wires, high current) that
_spikes_ appear when the fuse blows:

+DC>--+--[FUSE]--+-////-+--------+
| | | |
S | | |
PCH G--------+ | |
D | | |+
| | [LOAD] [BFC]
[LED] | | |
| [100K] | |
[R] | | |
| | | |
GND>--+----------+-////-+--------+

//// = wire inductance, long wire

These will be higher the the applied DC voltage (which respect to ground).


Al
 
On Fri, 10 Oct 2003 22:45:16 +0200, "The Al Bundy"
<the_al_bundy@*^no-spam!^*hotmail.com> wrote:


But with this circuit is it possible (long wires, high current) that
_spikes_ appear when the fuse blows:

+DC>--+--[FUSE]--+-////-+--------+
| | | |
S | | |
PCH G--------+ | |
D | | |+
| | [LOAD] [BFC]
[LED] | | |
| [100K] | |
[R] | | |
| | | |
GND>--+----------+-////-+--------+

//// = wire inductance, long wire

These will be higher the the applied DC voltage (which respect to ground).
Click to expand...
---
Yes, of course it's possible, but the spikes won't be _higher_ than +DC,
they'll be more _negative_ , and the way to cure that problem is to do
this:

+DC>--+--[FUSE]--+-----+---////-+--------+
| | | | |
S | | | |
PCH G--------+ | | |
D | |K | |+
| | [DIODE] [LOAD] [BFC]
[LED] | | | |
| [100K] | | |
[R] | | | |
| | | | |
GND>--+----------+-----+---////-+--------+



I just tried it using a couple of transformers for inductors, like this:

HV
____ |
+DC------O O---+--+ +----
S1 )||(
)||(
)||(
+--+ +----
|
|
[8R]
|
|
+--+ +----
)||(
)||(
)||(
GND>---------------+ +----

I adjusted +DC until I got 1A flowing through the circuit with S1
closed, then when I opened S1, HV went to about -300V!

Then I connected a diode from HV to GND with the anode to GND and closed
the switch. This time when I opened the switch HV went to about -1V.

--
John Fields
 
On Fri, 10 Oct 2003 16:54:29 -0500, John Fields
<jfields@austininstruments.com> wrote:


I just tried it using a couple of transformers for inductors, like this:
Click to expand...

Actually, more like this:

HV
____ |
+DC------O O---+--------+ +----
S1 )||(
)||(
)||(
+--------+ +----
| |
| |+
[8R] [10000ľF]
| |
| |
+--------+ +----
)||(
)||(
)||(
GND>---------------------+ +----

--
John Fields
 
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