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.
OK, lets look at your problem.
First, it sounds like you want to measure the current that would normally go across that fuse. Therefore, you don't want to add additional losses from your own circuit.
But then, you want to power something from losses, with 6V. To get 6 volts, you have two choices - you put in a lot of loss to get that voltage drop across the contacts - not acceptable - OR you have to provide a separate power connection. You can just provide a ground lead that will let you have a separate power supply in your device, or make it battery powered so it is completely independent of your measured supply.
-- Charlie -- Edmondson Engineering Unique Solutions to Unusual Problems
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
-- --Larry Brasfield email: donotspam_larry_brasfield@hotmail.com Above views may belong only to me.
Wouldn't it just be easier to ask your wife where she's been ?
Gibbo
It's not clear what you want.
Would you use this device to replace a fuse in an active circuit or would you put it in an unused circuit?
When the circuit is unused, there is no power available across the fusee terminals, so the task is impossible.
If the circuit is in use, then you might be able to get some power, but this would have an adverse effect on whatever was connected to the circuit. For example, if you tried this on your headlight (do they call them headlamps in GB?) fuse it would definitely dim the headlights.
In addition, if you used something like this in place of a fuse on an active circuit, then you would of course be leaving that circuit unprotected. This could create a risk of fire in the event of a short circuit.
I've never been inside of a car that was on fire, but it strikes me as undesirable. ;-)
--Mac
How much current normally goes through this fuse? Is it ever zero?
Battery voltages vary considerably, arguably from 10 to 14.5V, so most applications for these voltages should be able to live with a say 0.7 to 0.8 volt loss, without knowing this has happened.
There's another way. Add a diode drop (either 0.7V silicon or 0.4V Schottky) and power the instrument from the drop with a low-voltage dc-dc converter to make 5V, like we recently discussed here on s.e.d. (The current sense resistor's drop can be kept under 0.1V, etc.)
.. fuse substitute .. batt --- (O)--o-//-o--|>|--o--(O) ---- load .. | Rs | | .. | ___|___ | .. | | | | .. o--| sense | | .. | |_______| | .. __|______________|__ .. | | .. | low-V power | .. | converter | .. |____________________|
--
Thanks,
- Win
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.
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
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