Den mandag den 10. februar 2014 00.08.22 UTC+1 skrev Jim Thompson:
arkin
typical jfet.
de spread.
es up linearly
list for experimenting some day.
0 mA with Vgs at 0V?effects,
6
so you got the patent that is what matters I guess
but tell me, you don't think that if the voltage is higher than needed the obvious thing to do is to lower it?
some audio amplifers been doing it since forever
-Lasse
That looks to be a couple of transistors more complicated than needed.
Without monolithic matched transistors, current mirrors aren't worth much. When headroom was very limited (LiIon battery supply and 325nm IGaN LEDs), I did have good results with an NFET and bipolar current source. That was a very special case, though. If a 5V supply were available, I'd never go to such extremes. Too expensive.
...
But, if you don't care about accuracy, you can get 10% programming of the current easily. Even unmatched transistors, glued together, work fine. I'd not try to use them without emitter resistors, though, because you need as much current into the input transistor of the mirror as into the output transistor. A 1:30 ratio makes it possible to get 15 mA into the LED without overloading a lowly CMOS +5V logic output).
Obviously. The trick is to know _when_ to lower it (or raise it). There is a very subtle, but very accurate way to know that a CMOS mirror is about to fall out of the saturation region... and it can be done on a cycle-by-cycle basis. That is what my patent application is all about. 'Nuff said. No more until it issues. ...Jim Thompson
--
| James E.Thompson | mens |
| Analog Innovations | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| San Tan Valley, AZ 85142 Skype: Contacts Only | |
| Voice:(480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |
I love to cook with wine. Sometimes I even put it in the food.
That sounds sensible. If the variation in LED voltage drop is moderate and power consumption isn't critical, one LED supply voltage would work. +4, something like that, or just +5.
-- John Larkin Highland Technology Inc www.highlandtechnology.com jlarkin at highlandtechnology dot com Precision electronic instrumentation
I just realized I wasn't clear, "Bye" doesn't mean I'm going bye-bye. it means PAL is going bye-bye >:-} ...Jim Thompson
--
| James E.Thompson | mens |
| Analog Innovations | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| San Tan Valley, AZ 85142 Skype: Contacts Only | |
| Voice:(480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |
I love to cook with wine. Sometimes I even put it in the food.
Dunno, it's probably talked about on Wiki/Charge Conservation? Work it out yourself: a capacitor is always charged by either a resistor (lossy) or something fancier (reactance and fancy switching). Set up as many test cases as you like, observe the exponential RC curve and evaluate the resistor loss.
Charge pumps, BTW, aren't exceptionally lossy because the "conserve charge, halve energy" case is only true of discharging a capacitor into an uncharged one. When the difference is small, the loss is small (indeed, half the difference!). The overall effect is, the switch resistance simply looks like series resistance in the supply, which is handy. Of course, multiplied by the number of stages, if it's a step-up/down converter.
Well what the hell... it was on my screen, shouldn't that be enough? :^)
Ahyep... do this instead, drop the anode voltage to something convenient (4 or 5V?) and use CC sink drivers (one of those ICs, or generic logic with suitable current limiting resistors for each color). You'll take a particularly large hit on the red LEDs, but the blues won't be terrifying (better than 50% efficiency?).
If you don't have an anode per color, then... you could instead have a common 'ground' rail per color. This will require a separate switch for each, which should still be current limited (say, BJT switch + current limiting resistor). You'd use, say, +2.5V for the common red rail, +1.5V for green, and +1V or so for blue. These would be boost type switchers, since you're taking power in (the leftover from the LEDs, however many are on) and converting it back into the +5V rail. A synchronous converter would be best for compensation (continuous current mode) and efficiency (no diode drop to waste). You only need three converters (not really anything in common between them, unless you want to cook your own and use the same clock oscillator or whatever) for each module (which might be... a handful of matrices?).
I don't see how you'd get 24 of anything but the simplest circuitry alongside a matrix (how much of that was the matrix itself..?). I wouldn't want too anyway, it's such a pain by sheer numbers alone.
Yeah, that's the way to go; you might still be able to use whichever approach with the lifted-ground-rail idea. Mind the supply voltage requirement, dropout voltage (in the CCS case), and logic level translation to the outside world: at the very least, you'll need a series resistor between surrounding HC logic and the input pins of the lifted-ground devices; a resistor divider to +V, so that V_OL = Vgnd(lifted), would be even better.
Tim
-- Seven Transistor Labs Electrical Engineering Consultation Website:
On a sunny day (Sun, 09 Feb 2014 12:59:58 -0800) it happened John Larkin wrote in :
Nice, lower than I expected.
Got it, yes, almost looks like emitter follower, except the feed comes from the output, that is good.
Maybe he will post more info, Anyways, thanks for the help.
On a sunny day (Sun, 9 Feb 2014 12:46:05 -0800 (PST)) it happened Lasse Langwadt Christensen wrote in :
That may work, but better run the switchers from the same clock, I had weird coclor effcets when making this:
On a sunny day (Sun, 09 Feb 2014 18:23:23 -0500) it happened snipped-for-privacy@attt.bizz wrote in :
True, and this is not very efficient either, 2x current, Will have to see what exactly the OP is doing. But this cicuit indeed has a good range for voltage. Jim could make a chip with several times this in it, and IIRC there was a way to inncrease gain in a current mirror, but I have zero IC design experience.
On a sunny day (Sun, 09 Feb 2014 16:04:56 -0700) it happened Jim Thompson wrote in :
Actually you do not even need that, a battery with a high Zi will do, that is what is in those small keychain lights: a 3V cell and a LED.
Yea, but a bit over-complicated in my view.
Could not find the video input, erased it again.
Well, I dunno who you are ranting to, but you seem to have problems being confronted, When was the last time anybody did a design review of what you made? I mean other than you duh. :-)
Was it a _good_ buy?
Plus you can run your LED off 500V if you like. ;)
Cheers
Phil Hobbs
-- Dr Philip C D Hobbs Principal Consultant ElectroOptical Innovations LLC Optics, Electro-optics, Photonics, Analog Electronics 160 North State Road #203 Briarcliff Manor NY 10510 hobbs at electrooptical dot net http://electrooptical.net
If I asked in s.e.b, then it would be over complicated. I was asking because I was thinking about a sophisticated/optimal solution, not a easy-but-inefficient solution. A resistor or high Zi is not an answer to my question.
On a sunny day (Mon, 10 Feb 2014 08:39:41 -0800) it happened Daniel Pitts wrote in :
I have drawn two circuits that ARE an answer to a constant current source and are MUCH simpler than Jimmy boy's, and non-switching, read the thread.
On a chip, it's an entirely different game. Matching is then no problem and "Gain" is easy. Just scale the transistors (emitter). A
*lot* of chip techniques become ridiculous when you go to board designs.
I have read the thread. Have you read my posts? I didn't want a constant current source. I wanted a constant *switching* current source, to keep the wasted power to a minimum. I thought this circuit might be the basis for that.
Why do you want it switching? Increased EMI? ;-) If you're looking to improve efficiency, you need a storage element (inductor).
simple CC sources will always be inefficient due to Vbe drops.
Simple switcher with comparator built into to a Buck is far better for efficiency, continuous current, variable and efficient with inductor storage.
The power consumed by a LED is VF x ILED, and the power delivered by the 5V supply source, VSRC at current ISRC, is VSRC x ISRC. Because the exact tim ing of the current is a less than straightforward calculation, matters are simplified by looking at energies, the integrals of the power equations. In tegrating over any time frame that is much greater than the switching frequ ency of operation, say the lifetime of the LEDs, these become energy consum ed by LED is ELED= VF x QLED, and energy delivered by source is ESC=VSR C x QSRC, making for a lifetime efficiency of (VF x QLED)/(VSRC x QSRC). Th is ratio is bounded from above by (VF x QLED)/(VSRC x QLED)=VF/VSRC becau se the best you can do with your class of circuit is to circulate ever elec tron from the source through the LED. And this is not wildly oversimplified because your bias and other operating currents /are/ fairly small compared to the LED currents. So the upper limit on efficiency for the 3.3V LED is
3.3/5=66% and for the 2.2V LED it i 2.2/5=44%. These are not very good numbers. It's fair to say that all you've done is use a bunch of parts unne cessarily to end up with a circuit that is less efficient than a simple res istor current limiter, because the simple resistor circuit doesn't have the bias overhead of the capacitor circuit stealing current from VSRC that nev er gets delivered to the LEDs.On a sunny day (Mon, 10 Feb 2014 10:35:36 -0800) it happened Daniel Pitts wrote in :
Well, add a 555 to my current mirror, or the source of the JFET or use any any logic signal that changes from your system. Simple. :-)
ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.