My take on simple LED drivers... Starting your electronics hobby ? - Page 1
Forum section: electronics / beginners

author: RJSV
author_rank: Super Contributor
timestamp: March 18, 2021, 08:19:03 pm
content: Many folks start out pondering what a LED driver does, and this topic has been often covered. First, a driver transistor gives a boost, to a more feeble input.Like I said, hopefully readers can gain some additional insight.First, (see diagram), your input flows down through the transistor BASE, and encounters a sort-of BACK-VOLTAGE, or 'reaction voltage'. That is typically going to be near 0.6 volts, likely maybe 0.75 volts. So this is considered 'transistor ON' as that input current, and resulting 'back voltage'.Now, a question arises: What if the 0.6 volt input measured, base to ground, is not enough? Well, you work through the current GAIN situation, to check if there is strong enough input, to let an LED and load resistor conduct the full switched-on current, that it would naturally have if user had simply grounded theload resistor, without using a transistor switch.But that's a different case ? You might be thinking.It turns out, a fully conducting transistor output, viewed as 'Collector to Ground', is quite low, and is often ignored, in designing your GAIN and output requirements. Most circuits show maybe 0.010 volts there. Indeed: in finding a higher voltage, (collector terminal), such as 2.4 volts, you can start assuming that the transistor is either OFF, or very poorly conducting partially on, (linear region).So you use your transistor's typical or minimum GAIN, simply as a ratio: input to output (current).Problem remaining: You have to account for the small voltage loss, in the BASE to Emitter, (the typical 0..That's it, for example if your input voltage is 4 volts, then you will want to use (4.0 - .as the input resistorvoltage. Course, that then sets the input current, that then determines output current. Now if that output capability exceeds your connected load, then you can have confidence, that your circuit will pull AT LEAST what you want, and more if needed by some other load. This is giving a good definition of 'SATURATION',the transistor-LED circuit will be considered fully ON.Mostly, I'm not familiar with driving in the linear region, for brightness control: Most of what I have seen uses PWM or pulse-width to control LED brightness by average power delivered, by way of full-power pulsing ( the LED and it's current limiting resistor).By the way: a transistor in linear region is going to have issues with power / heat as it presents a voltage drop, exceeding that low value (0.010 volts) that a fullyturned on transistor has. (Remembering that power is voltage times current: Watts=V x I).-thanks, RJ

author: MikeK
author_rank: Super Contributor
timestamp: March 18, 2021, 11:06:09 pm
content: Quote from: RJHayward on March 18, 2021, 08:19:03 pmYour 2N2222 (which is an NPN, not PNP) has a Vce(sat) of about 0.4V (not 0.01V) at Ic=150mA and Vce(sat) of about 1.6V (also not 0.01V) at Ic=500mA. Per the datasheet.Calculating for saturation for this transistor is done with a gain of 10. Per the datasheet.

author: RJSV
author_rank: Super Contributor
timestamp: March 19, 2021, 01:01:23 am
content: Wow, aside from the typo, perhaps my assumptions are off, quite a bit: Thanks.But, I'm looking at typical low power, and maybe single LED circuits: Saturation at 5 ma, that sort of region, of operation. I'd like to see how low or how close my description was!And, plus: a half-amp (), I'm surprised the puny 2n2222 is rated that high, thinking more like 300mA max. Another favorite is NPN 2N2907, I think, or 2n2905: I will have to look that up.EDIT: It is 2n3904 which looks to be a better fit, having a gain peaking around 10 mA collector.I think I could get a 'pass' on most of this; but the typo is another error.THANKS: What did you think about the Vbe characteristic as that voltage detracts from the input voltage?And, for GAIN (at 10 mA output)? I've been using 40, minimum?

author: MikeK
author_rank: Super Contributor
timestamp: March 19, 2021, 01:58:47 pm
content: The gain of 10 is listed in the datasheet. It says something like Ib = 15mA, Ic = 150mA. That's a gain of 10. So setup your collector circuit with your proposed collector current (5mA). You need a collector resistor, of course. Divide the collector current by 10 (0.5mA). Then calc your base resistor: base voltage minus Vbe of about 0.65V and divide by the base current.

author: RJSV
author_rank: Super Contributor
timestamp: March 19, 2021, 07:08:41 pm
content: THANK YOU, MIKEK,Your feedback is just the thing, making discussion interesting!First, I am thinking, back to school days, when the analog descriptions of semiconductor behavior get into lots of heavy physics and math.Now, considering that a pure SINE wave (voltage at the input resistor) will cause the GAIN to vary according to current, I'm guessing: Is this part of a 'linearity' issue, and thus part of the reason for using negative feedback?(I'm going to have to crack open some old, dusty Electronics textbooks).Plus, another issue I want to check on, if there are different (lower) power handling ratings, on the PLASTIC CASE versions, as I noticed the Philips spec sheet shows the round metal can package.Thanks for bearing with me.Interesting: approaching a transistor circuit, starting at the output and working backwards to determine input resistor value. I think I've been starting at some nominal input (resistor, like 560 ohms) and then checking if output can be considered to be in saturation. This being designs for discrete switch.

author: Benta
author_rank: Super Contributor
timestamp: March 19, 2021, 07:25:28 pm
content: You're mixing up transistor gain in the linear region with test parameters in the saturated state. That is not the way to do it.

author: RJSV
author_rank: Super Contributor
timestamp: March 19, 2021, 07:32:55 pm
content: Not to be irritating, but I think you somehow accidently shifted, from 150 mA to 5 mA output...OK, my thoughts are around some little driver transistor you might see, when a processor chip won't handle more than 1 mA, with the transistor driver for a single large LED, (like a pwr on indicator).So, I calculated a base input resistor of 6.1 kOhms.R= V / I = {3.7 - .65} ÷ .5 mA = 6.1 K ohmsThat's actually quite small base current, but my feeling is that a 1 k resistor instead would make the transistor be 'even more' into saturation...Keeping in mind to not exceed max BASE current.Oh and also: I know that 5 mA collector current is going to exhibit a different GAIN, my guess is higher GAIN, at that specific current (5 mA).Thanks again: I am trying to learn the basics, for using as discrete switching, mainly micro-processor driven.MORE SOON

author: MikeK
author_rank: Super Contributor
timestamp: March 19, 2021, 08:33:36 pm
content: Note that hfe is unreliable. Which is why transistor amplifiers are designed with feedback, producing a gain that is almost entirely controlled by the feedback resistors. For transistors as switches you only need to get into saturation for the given collector current. You can use a 1K base resistor if you want, sure. But there's no benefit to the larger base current if the collector current is only 5mA. You're just wasting electrons. But they're yours to do with as you please.

author: RJSV
author_rank: Super Contributor
timestamp: March 19, 2021, 09:46:08 pm
content: My first concern would be to stay below maximum safe base current.Next concern is looking at creating that set of 'typical cases'. Then, the choice(s) go through a quick check, verifying they fit the situation. It's backwards from suggested formula for base resistor, but still verifies that {input volts-.7} ÷ Rbase = at least required minimum base current (course, after working thru output ÷ by GAIN.

author: RJSV
author_rank: Super Contributor
timestamp: March 22, 2021, 03:13:10 am
content: Sorry about delay, I am doing my due diligence, and (bewildered) google-study of the serious definitions around 'SATURATION'.Seems like a hard boundary, as rather than getting tiny tiny, or diminishing increases, in Collector current, as in asymptote, apparently the article claims that saturation is the end of it. But, confusingly, after taking the reader through a calculation for exactly where the transistor saturation happens, (330 ohms in the example, for the BASE resistor) the author starts talking about using a de-rating factor, to guarantee inside of saturation mode. (That producing a 200 ohm value).Honestly, I don't think a few percent into 'linear region's is gonna hurt, at that 15 mA or so through the LED and load / limiting resistor. Plus some of the enhancements, (slightly increased LED current) even can be noticed / discerned.I'm thinking only a tiny bit of extra mWatts of power would be in the transistor, in a case of a few percent outside the bounds, into linear operation of the BJT transistor. Gotta Study: more on this as I progress.

author: RJSV
author_rank: Super Contributor
timestamp: March 26, 2021, 10:39:09 am
content: I wanted to resort to a diagram having a special form of distortion, where LED circuit components assume a size according to voltage drop. So for a supply Vcc of 12 volts, the LED is shown, as connected to the + (12v), on the diagram left.For the case of 3-cell circuit power (Vcc= 3.7 v), on the diagram right side, an LED is connected in a similar manner, to the + supply (Vcc).This type of diagram can be seen in NEUROLOGY texts: where a face can be distorted according to what amount of 'brain matter' directly involved, such for the EYEs, sense of smell, etc.This diagram helps when considering how the switching transistor introduces some small voltage drop. In this common circuit, a transistor (not shown here) will switch a path to ground (minus supply), for LED on. The blue arrows show this small region, occupying approx. 0.05 to 0.20 Volts, in somewhat rough terms. But the proportions are illustrated, compared with each case of limiting resistor.The example, using either supply choice,(12 V or 3.7 V) is for a 'Power On' indicator that can also be blinked, under microprocessor control, for a simple fault indication, as an option in this example.For the point of connection, circuit board to LED, a red arrow shows, typically where the minus side of the LED connects, that point being pulled low to illuminate.