Detecting down to 25mA-150mA on 0-20A line?

Jim Thompson wrote:

I may have made a mistake, but I'm not getting any small signal gain through that MOSFET, the only instability should arise from the bulk Ciss loading of the OA: View in a fixed-width font such as Courier.

. * . cutoff=150mA - Iadj x 168 . . .-----+-----------+---------. .------------. . | | | | | | . | | 9V ON | |_ | | . | --- | * |/| | | . | \\ / ~~ | [100k] | | . | --- | / | cutoff | | | . | | | | adj Ibatt --- | PS . | | | | | - LiPo ----------- . | [470] | [12k] | v | | _ _ | . |+9V | | | Iadj | | /| _| || .--- | | | | | | - | | || . - .--+----. | | v | | ' o - - | . | | | | /| | | ----------- . | --- | |/+|-[1K]-+-[50]+-----------+ | . | / \\--+---+ .--< | | | | . | --- | | | |\\-|----. | 1N5817 | | . | | /10k| | | \\| | | .-| | . | | | | | . | '--[100]----+-[1k]----+----------------+ . | | | | . | | |< | . | +-------| 2N2907 | . | |\\ | . | | | . | com | . | | . | .--------------------------' . | | . | | . | | +9V . | Vgs,th,min | |\\ | . | ---------- '--|-\\ . | 2 | >--+--[560]---+ . '--------[10k]------+-----|+/ | | OC to PIC . | |/| | ------- . | | | | a |/ | . | | | | k | | . +-[1M]------+ | | |> | . | 1n4148 ------- . +------| ---- = ------------ ---- = . dVgs 0.3Rbatt dVgs . . . =K[2Vds+2(Vgs-Vth)dVds/dVgs-2VdsdVds/dVgs] . . . = 2KVds + 2K((Vgs-Vth)-Vds)dVds/dVgs . . Ids . = 2KVds + --- dVds/dVgs . Vds . . . - 2KVds . dVds . ---- = --------------------- . dVgs Ids (Rbatt+0.3) . --- + ----------- . Vds 0.3Rbatt . . . . Ids Vref*0.3 -Vds(Rbatt+0.3) . substituting --- = ----------------------- . Vds Vds*0.3Rbatt

. . . - 2KVds . dVds . ---- = ---------- . dVgs Vref . -------- . VdsRbatt . . . dVds . ---- = - 2KVref*Rbatt at the cutoff point . dVgs . . . . taking K=5 to 6 A/V^2 and Vref=45mV and Rbatt~ milliohms . . . dVds . ----

Reply to
Fred Bloggs
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although

voltage

Whoops. Silly me. Nix that little complaint.

Reply to
Fritz Schlunder

For what it is worth, I happen to think your suggested solution is quite clever. Saving practically all of the power dissipation while eliminating the non-idealities of actual current measurements are real major advantages.

Reply to
Fritz Schlunder

Aagh... so many parts... just bridge the voltage regulator with a current source and detect when the voltage regulator drops out of regulation...

Thomas

Reply to
Zak

Hey, your ok Fred! :))

--
Real Programmers Do things like this.
http://webpages.charter.net/jamie_5
Reply to
Jamie

Did you zoom in on where the waveform reaches closure? I saw some fine "oscillation"/slew-hunting.

Good idea, Fred!

[snip]

...Jim Thompson

--
|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
|  Analog/Mixed-Signal ASIC\'s and Discrete Systems  |    manus    |
|  Phoenix, Arizona            Voice:(480)460-2350  |             |
|  E-mail Address at Website     Fax:(480)460-2142  |  Brass Rat  |
|       http://www.analog-innovations.com           |    1962     |
             
I love to cook with wine.      Sometimes I even put it in the food.
Reply to
Jim Thompson

You're welcome to post a Tech-chat schematic of your brilliant idea. Anyone can shoot their mouth off with a vague description of "just do this" or "just do that", and when we get down to it, most of the time they don't know their butt from a hole in the ground. So let's see a schematic of your brilliant simplification the rest of us incompetents overlooked.

Reply to
Fred Bloggs

I used the same LM324 model and I don't see it, the whole thing is way overdamped with a 250-350us charge-up of the '3704 gate no matter what time constant used for local feedback compensation. I did solve the problem of abusive battery plug-in with high surge current into the circuit, and that is to simply turn the MOSFET backwards, it makes a heck of a good clamp diode, and circuit dynamics are essentially unchanged. This thing will be finished with some interlock stuff to prevent discharge on power loss and reverse battery protection: View in a fixed-width font such as Courier.

. . . . .-----+-----------+ .------------. . | | | | - ----------- . | +---[100k]--|---------. | | _ _ | . |+9V | / | | | | /| _| || .--- | *cutoff | [6.8K] | | - | | || . - .--+--. adj | | | | ' o - - | . | | | | /| | | ----------- . | | [2.7k] |/+|-[1K]-+----[50]-+-------+ | . | | | .--< | | | | . | --- | | |\\-|----. | | | . | / \\----+ | | \\| | | | | . | --- | | | | | | IRF3704 | . | | | +-----||---+ | S| | . | | [2.7k] | | 15N | [0.3] -|| | . | | | | | | | '>||-+--. | . | | +----------[1K]--' | -|| | | | . | | | | | | D| | | | . | | [50] | | 44mV | | [1k] | | . | | | | | | | | | | . '--+-----+-----|--+-------------------+-------+----+--|----' . | | | | .--- | --- | .com | /// | . | | . | +9V | . | | | . | |/ | . | +-------| 2N2222 | . | | |> | . | | | | . '--[100]----+-[1k]----+----------------+ . | | | . | |< | to cutoff . +-------| 2N2907 | detect . |\\ | . | | | . com | \\ / . | . | . | . |

Reply to
Fred Bloggs

A solar cell can be modelled as a current source with a current proportional to the amount of light falling on it, in parallel with a silicon diode.

As such it tends to have an output voltage that is proportional to the logarithm of the solar intensity that drops with increasing temperature (~2mV per degree). The individual cell voltage will be a maximum of about 0.5V and because of the logarithmic effect of the diode does not drop greatly as the intensity reduces.

A typical solar array will consist of 36 cells in series giving an open circuit voltage of about 18V. When this is fed to a 12V lead acid battery through a diode the voltage will drop to about 13-14V depending on charge level - it will act as a constant current generator so the series diode will have little effect on charging current except at low intensity levels or high-temperatures.

At nighttime the solar panel will act act like 36 forward biased silicon diodes in series - with ~13V from the battery each diode will only have about 360mV across each and the current will be pretty low, usually the array temperature will be lowest at night so this increases the voltage required for the diodes to conduct helping reduce the current.

So it is usually acceptable to not have a reverse blocking diode but it will not significantly affect the charging current except when the array temperature is high (under midday sun).

I like to have a blocking diode as a safety measure to avoid dangerous currents from the battery if the array wiring that is exposed to the elements has a short ( you should put a fuse in series with the battery physically close to the battery as well - even a small battery can have short-circuit currents of hundreds of amps).

Some solar arrays do have diodes included in the array but they are usually in parallel rather than series - this is to avoid problems when arrays are placed in series and one array gets shaded while the others don't. Without the diode the shaded array would be reverse biased by the voltage from the other arrays and be damaged (a similar problem can occur with series cells in a battery when one is discharged before the others.)

It is usually advisable to have a charge controller - if the voltage goes above ~14V the electrolyte in the battery will be electrolyzed away pretty quickly - in the case of sealed lead acid batteries there is no way of replacing it and so the battery will be damaged.

There are some solar arrays available with fewer cells that claim to be self regulating - these usually just have fewer cells (33 rather than

36) so that the charging current drops off as the battery is charged. These are not usually completely effective.

These days small charge controllers only cost $20-30 and are easily available - my local Fry's has them.

kevin

Reply to
Fred Bloggs

Implementation depends on regulator, input voltage, etc.

If the regulator is switching the current could be detected by looking at duty cycle, frequency, or whatever. That would be teh simplest solution.

For a linear regulator: bypass the pass device with a current regulator set to the desired end point. Note: current source should have a disable input.

In the linear regulator, detect when the error amp output hits its limit. It may well be that this ouput goes to 0 v or negative. Detect with comparator, spare opamp, discrete transistor with base resistor, or even tie to logic with a resistor.

This output going 0 or negative has to shut off both the linear regulator and the current source. Shutting off the current source and reducing the output set voltage of the power supply is enough: the supply will stay out of regulation as the setpoint is lower, but charging has stopped.

Do not make the voltage too low: the supply has to come back into regulation when an empty battery is connected.

The reduced voltage will keep the supply out of regulation, giving a schmitt trigger effect.

Kill the current source by pulling a node to ground. Change the reference or feedback, depending which way you feel, to reduce teh voltage of the main regulator.

TO do it quick and dirty, one needs:

Three diodes, two resistors and a transistor for a (poor) current source.

Tie its base to ground to shut it down: 2nd transistor. Mayeb a diode in the ouput for protection.

3rd transistor + resistor is the required inverter that drives this from the op-amp output. Use the collector of this transistor as the ground of the feedback divider if you feel like doing things the dirty way (changing the supply output voltage).

You need some capacitance on the output (probably via a resistor) to keep everything from oscillating of no battery is connected.

I hope you enjoy your schematic drawing program, I'm looking forward to the picture.

Thomas

Reply to
Zak

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