remote device blowing fuse/power supply

[thomas]

I have a remote device at the end of 100' of 22 AWG wire. The 5V remote
device has a 2200uf capacitor and draws 30mA and up to 1A intermittently.
My local 9V power supply is rated at 2.4A and I fused it with a 1.5A fast
blow.

When I connected the remote device to the power wire it blew both the fuse
and power supply. The long wire and large remote capacitor were just too
much for it.(?) Was my mistake in making the connection with the power
supply hot? If I make the connection with the power supply off won't it
cause the same problem upon powering up? What can I do next time?

 

[Robert Stankowic]

thomas wrote:

I have a remote device at the end of 100' of 22 AWG wire. The 5V
remote device has a 2200uf capacitor and draws 30mA and up to 1A
intermittently. My local 9V power supply is rated at 2.4A and I
fused it with a 1.5A fast blow.

When I connected the remote device to the power wire it blew both
the fuse
and power supply. The long wire and large remote capacitor were
just too
much for it.(?) Was my mistake in making the connection with the
power
supply hot? If I make the connection with the power supply off
won't it
cause the same problem upon powering up? What can I do next time?

The long wir should not be the problem, it just increases the load
resistance, thus decreasing the load current. The capacitor of course
will cause a high current peak, which could cause trouble.
Fuses don't really protect electronics - the electronic usually is
dead looong before the wire in the fuse melts.
Check your wire for a shortcut and provide some soft - start at your
load.

 

[JeffM]

I have a remote device at the end of 100' of 22 AWG wire. The 5V remote
device has a 2200uf capacitor and draws 30mA and up to 1A intermittently.
My local 9V power supply is rated at 2.4A and I fused it with a 1.5A fast
blow.
When I connected the remote device to the power wire it blew both the fuse
and power supply.
thomas

You seem to be saying that you're running a 5V device on a 9V supply.
Is there something you're not telling us?

 

[thomas]

Yes, I get a couple volt drop from the wire length, then I run it through a
regulator to get 5V.


"JeffM" <jeffm_@email.com> wrote in message
news:f8b945bc.0407231120.404e830b@posting.google.com...

I have a remote device at the end of 100' of 22 AWG wire. The 5V remote
device has a 2200uf capacitor and draws 30mA and up to 1A intermittently.
My local 9V power supply is rated at 2.4A and I fused it with a 1.5A fast
blow.
When I connected the remote device to the power wire it blew both the
fuse
and power supply.
thomas

You seem to be saying that you're running a 5V device on a 9V supply.
Is there something you're not telling us?

 

[Robert C Monsen]

"thomas" <none@yahoo.com> wrote in message
news:nlYLc.5056$oB6.1332@fe37.usenetserver.com...

I have a remote device at the end of 100' of 22 AWG wire. The 5V
remote
device has a 2200uf capacitor and draws 30mA and up to 1A
intermittently.
My local 9V power supply is rated at 2.4A and I fused it with a 1.5A
fast
blow.

When I connected the remote device to the power wire it blew both
the fuse
and power supply. The long wire and large remote capacitor were
just too
much for it.(?) Was my mistake in making the connection with the
power
supply hot? If I make the connection with the power supply off
won't it
cause the same problem upon powering up? What can I do next time?

I'm not sure what the problem is, but if there isn't any soft start,
the current to charge that 2200uF cap will be about 3.3A.

Here is one simple-minded soft-start circuit:

ie 10 ohms/10W
IN V ___
,----|___|---,
| | |
| | |
| | P-MOSFET |
| | |
'----o-o----+^+-----o-------,
| ||| |
| === |
--- | |
C --- | |
| | |
'------o |
| OUT V
.-.
| | R
| |
'-'
|
GND

If the Vgs(th) of the P-MOSFET is Vx, and
the supply voltage is Vcc, then the P-MOSFET
will turn on after approximately

T = R*C*ln(1 - Vx/Vcc)

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

The idea is that the RC circuit is pulled up to Vin initially, and
then relaxes down to GND. On the way, when the voltage passes
12+Vgs(th) of the P-MOSFET, it'll turn on the mosfet, bypassing the 10
ohm resistor.

Regards,
Bob Monsen

 

[thomas]

I'm not sure what the problem is, but if there isn't any soft start,
the current to charge that 2200uF cap will be about 3.3A.

Here is one simple-minded soft-start circuit:

ie 10 ohms/10W
IN V ___
,----|___|---,
| | |
| | |
| | P-MOSFET |
| | |
'----o-o----+^+-----o-------,
| ||| |
| === |
--- | |
C --- | |
| | |
'------o |
| OUT V
.-.
| | R
| |
'-'
|
GND

If the Vgs(th) of the P-MOSFET is Vx, and
the supply voltage is Vcc, then the P-MOSFET
will turn on after approximately

T = R*C*ln(1 - Vx/Vcc)

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

The idea is that the RC circuit is pulled up to Vin initially, and
then relaxes down to GND. On the way, when the voltage passes
12+Vgs(th) of the P-MOSFET, it'll turn on the mosfet, bypassing the 10
ohm resistor.

Regards,
Bob Monsen

This simple (minded) soft-start is probably just what I need. The web is
full of soft-start MOSFET references but practically nothing on the simple
basics.

I found the IRF9Z34 MOSFET which looks like it will work. It is rated at
55V, .1ohm RDS, 17A ID Cont. and has a Vgs(th) of 2-4V.
With a Vgs(th) of 2-4V does this mean I'll be slowing down just the initial
~3V charging of the 2200uf cap?

If I want to slow by .5 seconds the initial charging of the 2200uf cap then
RC=~1.4, giving something like 10uf and 140R, or 2.2uf and 630R. Am I going
about this right?

Do you think .5s delay is enough?

 

[Robert C Monsen]

"thomas" <none@yahoo.com> wrote in message
news:A2UMc.6130$oB6.1842@fe37.usenetserver.com...

I'm not sure what the problem is, but if there isn't any soft
start,
the current to charge that 2200uF cap will be about 3.3A.

Here is one simple-minded soft-start circuit:

ie 10 ohms/10W
IN V ___
,----|___|---,
| | |
| | |
| | P-MOSFET |
| | |
'----o-o----+^+-----o-------,
| ||| |
| === |
--- | |
C --- | |
| | |
'------o |
| OUT V
.-.
| | R
| |
'-'
|
GND

If the Vgs(th) of the P-MOSFET is Vx, and
the supply voltage is Vcc, then the P-MOSFET
will turn on after approximately

T = R*C*ln(1 - Vx/Vcc)

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

The idea is that the RC circuit is pulled up to Vin initially, and
then relaxes down to GND. On the way, when the voltage passes
12+Vgs(th) of the P-MOSFET, it'll turn on the mosfet, bypassing
the 10
ohm resistor.

Regards,
Bob Monsen


This simple (minded) soft-start is probably just what I need. The
web is
full of soft-start MOSFET references but practically nothing on the
simple
basics.

I found the IRF9Z34 MOSFET which looks like it will work. It is
rated at
55V, .1ohm RDS, 17A ID Cont. and has a Vgs(th) of 2-4V.
With a Vgs(th) of 2-4V does this mean I'll be slowing down just the
initial
~3V charging of the 2200uf cap?

No, the fact that the resistor is switched in is only dependent on
what the resistor/cap combination connected to the gate are doing, not
what the right side connected tot he load is doing.

If I want to slow by .5 seconds the initial charging of the 2200uf
cap then
RC=~1.4, giving something like 10uf and 140R, or 2.2uf and 630R. Am
I going
about this right?

You dropped a factor of 1000. For 1/2 second, you need

1/2 = -R*C*ln(1-4/9)

So if you pick a 10uF cap, your resistor should be something like
82,000 (to pick a standard value)

Do you think .5s delay is enough?

Yes, in fact, I bet 22 ms is enough. You just need to charge up that
2200uF cap on the other side of the wire. The time constant for a 10
ohm resistor and a 2200uF cap is 22ms. OTOH, .5s won't hurt. The cap
should be completely charged when the circuit switches the 10 ohm
resistor out.

Note that the power dissipated in that 10 ohm resistor will be I^2 *
10, so if the initial voltage across the cap is 0, you have an
instantaneous power of about 8 watts dissipated by the resistor. That
number will drop quickly, as the cap charges, but it means that you
probably shouldn't use a 1/4W resistor; a 1W resistor will probably be
ok, a 2W would be better.

Regards,
Bob Monsen

 

[JTM]

With a Vgs(th) of 2-4V does this mean I'll be slowing down just the
initial
~3V charging of the 2200uf cap?
No, the fact that the resistor is switched in is only dependent on
what the resistor/cap combination connected to the gate are doing, not
what the right side connected to the load is doing.
I see

I going about this right?
You dropped a factor of 1000.
oops...uF is 1/1,000,000F

For 1/2 second, you need 1/2 = -R*C*ln(1-4/9)
So if you pick a 10uF cap, your resistor should be something like
82,000 (to pick a standard value)
Using the formula: .5 = R*C*ln(1--3/7) with Vx=-3V (midpoint of the Vgs(th)

range) and Vcc=7V since there is a couple volt drop across the wire.
This time I calculate R*C=1.4, giving 10uF and 140K.

Do you think .5s delay is enough?
Yes, in fact, I bet 22 ms is enough. You just need to charge up that
2200uF cap on the other side of the wire. The time constant for a 10
ohm resistor and a 2200uF cap is 22ms. OTOH, .5s won't hurt. The cap
should be completely charged when the circuit switches the 10 ohm
resistor out.
Yes, I understand. Maybe I'll go with 1/4s.

Note that the power dissipated in that 10 ohm resistor will be I^2 *
10, so if the initial voltage across the cap is 0, you have an
instantaneous power of about 8 watts dissipated by the resistor. That
number will drop quickly, as the cap charges, but it means that you
probably shouldn't use a 1/4W resistor; a 1W resistor will probably be
ok, a 2W would be better.
Ok. I have some 2W 6.8R around. With 7V and 2200uF the time constant is

15ms, @1A.
Thanks, Bob