Tricky problem....

[Richard Nienhuis]

This is not homework so don't worry. I want to be able to power a
little transmitter and current meter. The problem is I want to be
able to just plug it into an automotive fuse socket in a car and have
it power its self from that socket. Now if I was just measuring I
would use a hall sensor or a current sense resistor. But how can I
generate enough voltage to run 6V and .5ma worth of stuff? I have to
keep the heat disipation way down as well.

 

[ChrisGibboGibson]

(Richard Nienhuis) wrote:

This is not homework so don't worry. I want to be able to power a
little transmitter and current meter. The problem is I want to be
able to just plug it into an automotive fuse socket in a car and have
it power its self from that socket. Now if I was just measuring I
would use a hall sensor or a current sense resistor. But how can I
generate enough voltage to run 6V and .5ma worth of stuff? I have to
keep the heat disipation way down as well.

Wouldn't it just be easier to ask your wife where she's been ?

Gibbo

 

[Larry Brasfield]

"Richard Nienhuis" <richardnienhuis@email.grcc.edu> wrote in message news:df29d053.0412030952.44e4927a@posting.google.com...

This is not homework so don't worry. I want to be able to power a
little transmitter and current meter. The problem is I want to be
able to just plug it into an automotive fuse socket in a car and have
it power its self from that socket. Now if I was just measuring I
would use a hall sensor or a current sense resistor. But how can I
generate enough voltage to run 6V and .5ma worth of stuff? I have to
keep the heat disipation way down as well.

Can the loads on that fuse circuit tolerate a voltage down to, say,
10V? (I would think so, but that's a guess.) Does that load, when
non-zero, exceed 2 mA? (Again, I would guess so.) Is it OK for
the transmitter to indicate 0 current by being off? (One might hope
since you are into the realm of tricky.)

If those suppositions are true, you should be able to insert a little
power converter in the fused circuit which would run your circuit.
If it dropped 2V, it would need something like 2 mA to supply
your stated load. There are IC's intended to convert single-cell
battery voltage to more easily used levels. For example, see
http://www.maxim-ic.com/quick_view2.cfm/qv_pk/1210
This device is fairly efficient, so heat would not be a problem
for the light load you mentioned.

--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.

 

[Fred Bloggs]

Richard Nienhuis wrote:

This is not homework so don't worry. I want to be able to power a
little transmitter and current meter. The problem is I want to be
able to just plug it into an automotive fuse socket in a car and have
it power its self from that socket. Now if I was just measuring I
would use a hall sensor or a current sense resistor. But how can I
generate enough voltage to run 6V and .5ma worth of stuff? I have to
keep the heat disipation way down as well.

It's nothing that you can do easily- as already suggested - the simplest
is to add a 3rd terminal to your circuit for BATT or GND as the case may
be and regulate it down to 6V across your circuit.

 

[Ken Smith]

In article <df29d053.0412030952.44e4927a@posting.google.com>,
Richard Nienhuis <richardnienhuis@email.grcc.edu> wrote:

This is not homework so don't worry. I want to be able to power a
little transmitter and current meter. The problem is I want to be
able to just plug it into an automotive fuse socket in a car and have
it power its self from that socket. Now if I was just measuring I
would use a hall sensor or a current sense resistor. But how can I
generate enough voltage to run 6V and .5ma worth of stuff? I have to
keep the heat disipation way down as well.

The basic idea is a circuit like this:

L1 D1
-----+------((((((--------+----->!-----+---
! ! !
---C1 o Q1 --- C2
--- / - ---
! ! : !
-----+--------------------+------------+--
:
:
-----------
! Control !
-----------

The control electronics monitors the voltage on C2 whenever it gets to be
about enough, it closes the switch Q1 (MOSFET?). When the voltage on C2
starts to get too low, Q1 is opened. The circuit works as a booster
regulator. The inputs side voltage rises if the load on C2 rises.

It is hard to use a PWM converter chip for this job because the control of
Q1 is the exact inverse of the normal case. The more time it spends off
the higher the output voltage.

If you do use a PWM chip getting the servo loop to be stable is a bit of a
trick because of the funny extra zero that moves around. In the short
term, increasing the duty cycle raises the output voltage as the energy in
C1 is transfered to C2. This means you start off with a 180 degree phase
lag at high frequencies before you deal with any of the Ls and Cs and
integrators. Making it stable for a wide range of currents can be done
but it requires that the servo controller's phase be controlled over a
wide frequency range. You also have to make C2 quite large.

--
--
kensmith@rahul.net forging knowledge