Make the parallel combination of R1//R2 have the same resistance as the inverting pin of U1 sees. This drops the bias current out of consideration.
Wire the (+) supply of the OP-AMP and the 12V zener with their own wire to the Rs current pin. Only R2 and the LM385 should go to the voltage leg of Rs. This keeps those DC currents out of the measurement circuit.
The 22K to the inverting input of the OP-AMP and the 0.01, C2 feedback on the op-amp look a bit too big to me as does R3. R3 and the gate capacitance of the IRF4905 make a time constant. You want the 22K and the C2 to make one that is bigger than that by perhaps a factor of 10 but not a factor of 100. I did not look up the MOSFET.
Thermally isolate the non-power parts away from the heat of the power section if your mechanical situation allows.
A constant current diode like the 1N5313 can be used in place of the 220 Ohm resistor. If not, I suggest making the 220 a 10W resistor.
If you add a Schottky across the 12V zener, the circuit will be fairly well protected against reversed supply leads, assuming the 0.1, the MOSFET and the output diode can all take the current while the 5A fuse blows.
You are the most worthless pseudo-intellectual p.o.s. on this NG- you don't even know what a load dump is.
Who says the transient response needs "improving"? What circuit are you referring to?- Certainly nothing posted by you, bullshit boy.
That's what you say, mouse, but as usual absolutely no details are forthcoming.
Isn't that what he's doing now? Why don't /you/ research them and tell him what they are- what kind of bs non-informational statement is "You will need to research them"...p.o.s.
Hmmm- another highly learned and considered response- what an incredibly complex mind you have...
Both of those parts are unsuitable for this app.
Oh- I'm sure WH was anxiously awaiting your bs approval on that parts list- what a p.o.s.
Oooh- here we go- more bs from the fake:
"suprisingly large"- well what kind of engineering terminology is that, fruity boy? I don't recall ever seeing "suprisingly" anything in any datasheet specification. What a total p.o.s....
Nope- you're wrong about that.
Really? How much transient power should this generic "zener" be able to absorb without damage? Or do you intend to sacrifice it at the first blast?
"willing"? There you go again with your "girlie man" descriptors. You think that damned circuit gives a damn about what he's "willing to see"? What a pathetic little worm joke and incapable piss ant you are! Unbelievable...
A power rectifier?
He should "satisfy himself"- what a waffling p.o.s. moron! And you're damn right you will have to do some research to find out what that "might" be...waffling p.o.s.
More pseudo- theoretical generalistic non-informational PISS. You haven't provided a stitch of specifics to back a single one of your bullshit "hints"- friggin wishy-washy pseudo-intellectual worm.
Yeah- right. GFOAD and stay off this NG if that is the kind of weak, pretentious, non-informational, non-engineering, flake post you offer.
I would not use a zener with a higher voltage than the legal limit on the gate of the MOSFET. If no load is connected, the op-amp will come to rest at the (-) rail.
Whew! That's supposed to be a simple way to get the moderately- precise current the O.P. needs? Ten resistors, a cap, a diode and an opamp to drive the FET? Nah, come-on man, let's get down!
The single-transistor regulator is a simple yet effective rough current source, requiring only two resistors with a transistor.
But this circuit needs about 0.65 to 0.7V across the current-sense resistor Rs, whereas we'd like that to be about 0.25V for 2.5A through a 0.1-ohm resistor, etc. So we need to subtract a little voltage from the Q1 Vbe drop. We can use an old Robert Widlar trick, and get this from a fraction of the pass-transistor's drive voltage, like this:
That's pretty simple, only three resistors. The current is given by
Io Rs = Vbe (1+R1/R2) - R1/R2 (Vgs + Io Rs)
R2/R1 = (Vgs - Vbe + Io Rs) / (Vbe - Io Rs)
Trying for a 2.5A CS, so the last term is 0.25V, assume Vbe = 0.65 and Vgs = 3V, so we can choose some values. R2 has 3 -0.65 + 0.25 = 2.6V across it and with 0.4V across R1 we have R2 = 6.5 R1, so if we pick R1 = 220 ohms, then R2 = 1430 ohms, which we can make from 1k plus a 1k trimpot.
The OP can assemble the circuit, and adjust R2 to get his desired Iout. There will be a little variation with load because Vgs changes a little with Id = Io and because the FET's junction is heating up, but it after trimming it may be all the OP needs.
Given that the 12V battery is probably pretty constant (near 12V if running alone, or near 13.8V if in an operating automobile) we could choose to ignore the issue of changing supply voltage. But if not, there's a simple one-resistor improvement to the circuit to cancel the transistor's change in Vbe due to varying Ic current through Rg.
Properly chosen, the higher voltage across Rc at high battery voltages cancels Q1's higher Vbe voltage at higher Ic currents. Readers who are familiar with re = kT/qIc, etc. (see AoE page 80), may recognize that choosing Rc = R2/R1 re, will do the trick.
Still pretty compact, only four resistors. Now that's getting down!
BTW, watch out. Anyone trying to simulate this circuit with spice should realize that most power MOSFET models will provide incorrect Vgs vs Id values at these low current densities.
Here's your chance to show how smart you really are, Fred. Beware, these are real questions and blowing them off with a bunch of insults will merely reveal (to the newcomers) that you are an electronics idiot.
Do you believe that your diode forward drop used as a reference will meet the OP's stated 5% accuracy requirement over the voltage and temperature range to be expected in an automotive application?
Do you believe that the current output of your circuit will remain within the OP's stated transient response requirement in the presence of load-dump transients?
--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.
Win, Have you had a chance to review the efficacy of the models I posted for you?
...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 | |
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| 1962 | I love to cook with wine. Sometimes I even put it in the food.
range 2-3 amps and compliance within 0.5V of the rail.
and really doesn't give a damn about any transient
I take it, then, that you believe the OP to be mistaken about his claim to need the current controlled within a +/-5% range, after being specifically asked about the transient response requirement.
I also take it that you believe your circuit will itself survive in the automotive environment, with nothing in the way of protection between it and the battery.
I have some advice to the OP: Take Fred's advice with a large grain of salt, and be sure not to choke on that "grain".
--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.
He doesn't require any precision whatsoever- and can hack together an acceptable current source with readily available Radio Shack and NTE stuff- a simplified circuit w/o protection would be like so: View in a fixed-width font such as Courier.
Yeah- it's all good enough for what the OP needs- an output current in the range 2-3 amps and compliance within 0.5V of the rail. The OP has no other requirements, is probably charging an accessory battery, and really doesn't give a damn about any transient response other than it doesn't blow his components.
range 2-3 amps and compliance within 0.5V of the rail.
and really doesn't give a damn about any transient
Yeah- that's right- he's trying to pretend he's something other than Harry-hobby-hacker, he has no real requirements.
I stated that the protection was omitted, the w/o means "without". It will just clutter the essential details of the current clamp.
Eh- you don't like it because you don't understand how it works. You can make it more of a textbook ideal by going to the LM385 or a zener for the voltage offset reference and adding one more resistor in the (-) input leg of the bridge to balance the impedances- making S(Io, Vbatt)=0- but like who cares. Go spend $30 on a bunch of unnecessary fluff instead of $3 on something of equal and indistinguishable performance modulo the application.
And don't forget offset- as I said- a few tens of milliamps here and there makes no difference for this circuit. You go to any ss OA datasheet from National and you will find that darned "ground referencing a voltage" example which has been there for the past 50 years- this is that ckt, with feedback loop closed by P-FET to balance the inputs through Io*Rs drop instead of locally and an offset inserted. You pick any group of 10KR out of the same package and they will all be within 100 ohms even at 5%.
In this classic circuit the Q1 stage gain is fairly high, e.g., with 9 to 11V across the Rg collector resistor, G = 40 V = 400. From that viewpoint, 5% regulation isn't an issue. But the circuit below isn't so well off. The added voltage-offsetting resistors create a feedback loop limiting the Q1 stage gain to R2/R1, or about 6.5 in this case.
This doesn't sound good.
Let's look at the circuit fleshed out with part values.
With a big FET like the IRF4905, we're into the constant-current "saturated" region if Vds is greater than about 500mV at Id = 2A (this big FET is happy conducting well over 100A, if given a little Vds to work with, so a low 2A is getting down into its high-gain subthreshold region). We can estimate the FET's transconductance to be about 10S at 2A and Vds, above 0.5 to 1V (datasheet fig 1), so the FET's source-follower gain, G = gm RL / (1 + gm RL), would be above 0.5, which isn't too bad...
Ooops, our poor loop gain drops to a little above 3. It's not the 0.1 ohms that's killing us so much as the R2/R1 feedback connection. But that's there to reduce the Rs voltage drop from 0.7 to 0.25V. After trimming R2 to establish the circuit's current, its stability will be determined by how well subsequent Vgs changes are corrected for by the Q1 error amplifier. If the circuit is operating with Vds above 0.5 to 1V (where the FET is acting like a current source on its own), the Vgs change with Vds won't be much, perhaps under 50mV, which would be 15mV on Rs, and regulate the current to under 6%. But oops, the FET's dissipation varys with Vds drop, so its temperature varys. Vgs has a negative tempco (the IRF4905's zero-tempco current = 22A), and looking at the datasheet figure 3, we see this can easily amount to more than 50mV with junction heating, so my circuit is toast. :>)
Playing with the value of Rs and its voltage drop, can't rescue the circuit. E.g., reducing Rs degrades the Q2 stage gain, but this is canceled by a corresponding increase in R2/R1 gain, so that's a wash.
We could add an LM385 and a resistor to get a stable voltage to which we can connect R2 and offset Q1, but that would destroy the circuit's simplicity.
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