Post by thread:[EE]:NPN LED current drive

I'v added [EE]: to the topic... First, I think you should put the "R" between the LED and the collector. And connect the emitter directly to ground. Then. put an "R" between the PIC port and the transistor base to limit the base current to ( [LED current] / [gain of the transistor] ). There is no reason to drive more current inte the base then is needed to get the wanted current through the LED. That's wasted current. It's the gain of the transistor that let's you drive a LED, or anything else that needs a higher current then the PIC port can stand. And hould the LED be missing, the base resistor will limit the current into the base anyway. So putting a resistor between the PIC port and the base, more or less answers all your questions. Is there any special reason why you thought that connecting the PIC port directly to the base was a good thing ? Jan-Erik Svderholm S:t Anna Data tel : +46 121 42161 mob : +46 70 5241690 -- http://www.piclist.com hint: To leave the PICList spam_OUTpiclist-unsubscribe-requestTakeThisOuTmitvma.mit.edu

>> I'v added [EE]: to the topic... Woops, I was thinking about this as I filled in the address, but I still forgot:) >> Is there any special reason why you thought that connecting the PIC port >> directly to the base was a good thing ? Yeah, the idea, (as suggested to me recently) was that configuration you can drive the LED from a higher, unregulated supply and still get a fairly constant current through the LED. In the standard config that you suggest, the base current is constant but the collector current will vary with the unregulated supply voltage (I think because the gain of the transistor will vary based on the Vce). My product has two supply options. 1) unregulated wall wart (when the lower pwr LCD option is installed), 2) external regulated 5V (when the higher pwr VFD option is installed). With the external supply option, there is plenty of juice available because the PS is sized for the high startup current of the VFD. But with the unregulated supply, I am thinking that it is better to take the current surges off the regulated supply. Since the LED (its an IR transmitter) is on for only short pulses, I am thinking that a better alternative might be to put a large cap next to the LED supply so the current pulses are supplied by it. If I do this, should the cap be physically close to the LED, or can I just make the PS filter cap larger? (which is several inches from the LED). I am thinking that the cap should be low ESR and the size would be, worst case, the drive current times the total on times for a signal times the a acceptable voltage drop (or something like that). Instead of just putting the cap and the LED on the supply rail, would it be better to isolate them with a resistor like this... Vcc -------R-----.--------. | | C LED | | | R | | | C | B----R--- uP output | E | | GND -------------'--------. The idea being that this would isolate the supply rail from the current surge so that the LED drive voltage might dip during a on pulse, but the supply rail would not. --BobG {Original Message removed}

> First, I think you should put the "R" between the LED and the collector. > And connect the emitter directly to ground. > > Then. put an "R" between the PIC port and the transistor base to limit the > base current to ( [LED current] / [gain of the transistor] ). There is no reason > to drive more current inte the base then is needed to get the wanted current > through the LED. That's wasted current. Often systems have a regulated supply the PIC is running from derived from a higher voltage, often unregulated, supply. In such cases it can be useful to connect the PIC output to the base of an NPN transistor, emitter to ground via a resistor, and LED between the unregulated supply and collector. When the PIC output goes high, the collector acts pretty much like a controlled current sink, with current = (PIC output - Vbe) / R. R is then adjusted to get the desired LED current. The advantage of this scheme is that the LED current is drawn from the unregulated supply, but is still nicely regulated itself. Often LEDs are the major current sink of a supply, so keeping them from the regulated supply can make quite a difference in the power the regulator must dissipate. The drawback is that a supply must be available that is at least the PIC power voltage plus the LED on voltage. ***************************************************************** Embed Inc, embedded system specialists in Littleton Massachusetts (978) 742-9014, http://www.embedinc.com -- http://www.piclist.com hint: To leave the PICList .....piclist-unsubscribe-requestKILLspam@spam@mitvma.mit.edu

Often systems have a regulated supply the PIC is running from derived from a higher voltage, often unregulated, supply. In such cases it can be useful to connect the PIC output to the base of an NPN transistor, emitter to ground via a resistor, and LED between the unregulated supply and collector. When the PIC output goes high, the collector acts pretty much like a controlled current sink, with current = (PIC output - Vbe) / R. R is then adjusted to get the desired LED current. This WILL cause the transistor to be operating in it's linear region, since otherwise the voltages at the emitter would be Vunreg-Vled-Vcesat, which could easilly be greater than Vout from the PIC, right? This shouldn't be a problem for typical LED currents, but it does mean that for higher current loads (light bulbs?) you'll probably end up needing to go back to base and collector resistors instead of the emitter resistor (or risk blowing up your transistor due to excessive power dissipation.) BillW -- http://www.piclist.com hint: To leave the PICList piclist-unsubscribe-requestKILLspammitvma.mit.edu

> This WILL cause the transistor to be operating in it's linear region, Yes. > since > otherwise the voltages at the emitter would be Vunreg-Vled-Vcesat, which > could easilly be greater than Vout from the PIC, right? The voltage at the emitter is driven by the PIC voltage. It will be the PIC voltage minus the B-E junction drop. The transistor is acting as an emitter follower driving the resistor voltage from the PIC output voltage. Since the emitter load is a resistor, the emitter current will be the emitter voltage divided by the emitter resistor value. Assuming reasonable gain for a small signal transistor, this also becomes the collector current for practical purposes. > This shouldn't be > a problem for typical LED currents, but it does mean that for higher > current loads (light bulbs?) you'll probably end up needing to go back to > base and collector resistors instead of the emitter resistor (or risk > blowing up your transistor due to excessive power dissipation.) Let's say the unregulated supply was 20V (pretty high for linearly regulating to 5V), the LED requires 2V and 20mA, and the B-E drop is 600mV. The C-E voltage accross the transistor will be: Vce = 20 - 2 - 4.4 = 13.6V At 20mA the power dissipation in the transistor is 270mW. Yup, that's pushing a TO-92 package but still doable. However that represents another 300mW that the linear regulator doesn't have to dissipate. Especially with multiple LEDs, this can make a big difference in the regulator requirements. The wasted power to run the LEDs is spread over multiple small transistors and resistors instead of being concentrated in one package. Also note that 20mA LED current is rarely needed anymore. Nowadays for simple indicators I use decent LEDs and run them at 2 - 10 mA even though they are rated at 20mA. ***************************************************************** Embed Inc, embedded system specialists in Littleton Massachusetts (978) 742-9014, http://www.embedinc.com -- http://www.piclist.com hint: PICList Posts must start with ONE topic: [PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads

William Chops Westfield <.....billwKILLspam.....CISCO.COM> wrote: > > In such cases it can be useful to connect the PIC output to the base of > > an NPN transistor, emitter to ground via a resistor, and LED between the > > unregulated supply and collector. > > This WILL cause the transistor to be operating in it's linear region, Not necessarily. Yes for a bare LED, but maybe not for a bulb. > otherwise the voltages at the emitter would be Vunreg-Vled-Vcesat, which > could easilly be greater than Vout from the PIC, right? This shouldn't be > a problem for typical LED currents, but it does mean that for higher > current loads (light bulbs?) you'll probably end up needing to go back to > base and collector resistors instead of the emitter resistor (or risk > blowing up your transistor due to excessive power dissipation.) A light bulb will drop much more voltage than a LED, and in fact may drop more than the available Vunreg - Vpic. Consider a bulb rated at 50 mA at 12V, fed from Olin's 20V supply, with a 5V PIC controlling the transistor. If the emitter resistor is set for 50 mA, then the transistor will be in current limiting mode, with 12V across the bulb and 3.7V across the transistor, which will be dissipating 185 mW. However, if the unregulated voltage is less than 17V, then the transistor WILL be saturated, and dissipating very little power. Furthermore, this type of drive circuit provides a "soft start" for the bulb by limiting the cold-inrush current. This should extend its life. With an LED, you can always add a series resistor at the collector of the transistor, which won't change the LED current (assuming it isn't too big). It'll merely shift some of the power dissipation away from the transistor. This retains all of the other benefits of using a current source and the unregulated supply. -- Dave Tweed -- http://www.piclist.com hint: PICList Posts must start with ONE topic: [PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads

At 09:18 AM 10/19/02 -0700, Dave Tweed wrote: >A light bulb will drop much more voltage than a LED, and in fact may drop >more than the available Vunreg - Vpic. > >Consider a bulb rated at 50 mA at 12V, fed from Olin's 20V supply, with a >5V PIC controlling the transistor. If the emitter resistor is set for 50 mA, >then the transistor will be in current limiting mode, with 12V across the >bulb and 3.7V across the transistor, which will be dissipating 185 mW. > >However, if the unregulated voltage is less than 17V, then the transistor >WILL be saturated, and dissipating very little power. This is where the current sink falls to pieces. The transistor may be dissipating very little power but the poor PIC and its power supply is gonna be unhappy. You see, as soon as the collector current drops because the transistor is saturating, the base current INCREASES. The best way to visualize this is to simply disconnect the collector of the transistor. The E-B junction looks like a forward biased diode and the base current is (Vdd-Vbe) / R . In other words, this type of current sink maintains constant EMITTER current. If the collector current is too low, the remaining current comes from the base. This can be alleviated by adding a resistor in series with the base. As long as the transistor is operating in the linear region, the base current is negligible and there is only a small voltage drop on that base resistor. If the transistor saturates or its load becomes disconnected, the base resistor limits the current that the PIC has to supply. One final note: as soon as you add that series base resistor, it becomes real easy to add one more resistor so that the base is driven from a voltage divider that is driven by the PIC. This has a couple of benefits: less voltage drop across the emitter resistor which can mean a smaller (lower power) resistor, and more headroom (less chance of saturating the transistor). This is a real advantage when dealing with an un-regulated supply that has LOTS of ripple. dwayne -- Dwayne Reid <EraseMEdwaynerspam_OUTTakeThisOuTplanet.eon.net> Trinity Electronics Systems Ltd Edmonton, AB, CANADA (780) 489-3199 voice (780) 487-6397 fax Celebrating 18 years of Engineering Innovation (1984 - 2002) .-. .-. .-. .-. .-. .-. .-. .-. .-. .- `-' `-' `-' `-' `-' `-' `-' `-' `-' Do NOT send unsolicited commercial email to this email address. This message neither grants consent to receive unsolicited commercial email nor is intended to solicit commercial email. -- http://www.piclist.com hint: PICList Posts must start with ONE topic: [PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads

Dave Tweed wrote: {Quote hidden}> William Chops Westfield <billwspam_OUTCISCO.COM> wrote: >>> In such cases it can be useful to connect the PIC output to the >>> base of an NPN transistor, emitter to ground via a resistor, and >>> LED between the unregulated supply and collector. >> >> This WILL cause the transistor to be operating in it's linear region, > > Not necessarily. Yes for a bare LED, but maybe not for a bulb. > >> otherwise the voltages at the emitter would be Vunreg-Vled-Vcesat, >> which could easilly be greater than Vout from the PIC, right? This >> shouldn't be a problem for typical LED currents, but it does mean >> that for higher current loads (light bulbs?) you'll probably end up >> needing to go back to base and collector resistors instead of the >> emitter resistor (or risk blowing up your transistor due to >> excessive power dissipation.) > > A light bulb will drop much more voltage than a LED, and in fact may > drop more than the available Vunreg - Vpic. > > Consider a bulb rated at 50 mA at 12V, fed from Olin's 20V supply, > with a 5V PIC controlling the transistor. If the emitter resistor is > set for 50 mA, then the transistor will be in current limiting mode, > with 12V across the bulb and 3.7V across the transistor, which will > be dissipating 185 mW. > > However, if the unregulated voltage is less than 17V, then the > transistor WILL be saturated, and dissipating very little power. > > Furthermore, this type of drive circuit provides a "soft start" for > the bulb by limiting the cold-inrush current. This should extend its > life. > > With an LED, you can always add a series resistor at the collector of > the transistor, which won't change the LED current (assuming it isn't > too big). It'll merely shift some of the power dissipation away from > the transistor. This retains all of the other benefits of using a > current source and the unregulated supply. > > -- Dave Tweed Just to add some old/new words; As a LED switcher, a small transistor should not to be seen as a current limiter or extra power heatsink, for this job a much lower cost resistor is recommended, except of course if the supplied voltage is not known, so the transistor should be used also as a current limit device. As even unregulated power supply voltages can not change so much, few volts as expected, the resulting current change will be in the same order, or almost non perceptible by the user. I would say that even when the supplied voltage changes from 14 to 17V you still can use a simple current limiting resistor in series with the LED and forgeting about constant current controlling. In this case, a small NPN transistor setup as a simple switcher, emitter to ground, series resistor at base, led and resistor at collector will be most than enough. The good point of such simple and very common setup, is that you can substitute the transistor by ANY logic gate (74HC00 family) without change nothing, except if the supplied voltage is higher than the logic VCC, in this case, a 7406/7 gate is in order. The limiting current resistor calculation is easy as (VPower - 2 / 0.01), 14V-2V/0.01 = 1kohms 1/4 or 1/8W is enough. W46NER -- http://www.piclist.com hint: PICList Posts must start with ONE topic: [PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads

Doug Butler wrote: >> Just to add some old/new words; >> >> As a LED switcher, a small transistor should not to be seen as a >> current limiter or extra power heatsink, for this job a much lower >> cost resistor is >> recommended, except of course if the supplied voltage is not known, >> so the transistor should be used also as a current limit device. > > Actually as long as you need a transistor, the incremental cost of > making it a bigger transistor to handle the power as a current > regulator is small. It may be cheaper to use a bigger transistor > than to add a resistor as a new component. Also the bean counters > tell us that reducing parts count improves reliability (though I have > my doubts that is a usefull rule). > > Doug Butler > Sherpa Engineering As far as I understand, it will be no extra resistor, since the emitter resistor would go to the collector, except of course if you are just connecting the NPN base directly to the uC port pin, what I would not recommend at all - a simple short into the transistor and you will end up with VDC-VLED volts via collector -base short to the uC port pin, bang and smoke. For this solution I use to choose a small FET unit, no need any gate-uC resistor (it is already isolated), and the rON is lower than a 2N2222. Of course that if any 7406/7 gate is available at the board, then no transistor is needed (if the VDC is higher than 5V - or if 5V than any 74HC04 gate does the job). W46NER -- http://www.piclist.com hint: PICList Posts must start with ONE topic: [PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads

> As a LED switcher, a small transistor should not to be seen as a current > limiter or extra power heatsink, for this job a much lower cost resistor is > recommended, except of course if the supplied voltage is not known, so the > transistor should be used also as a current limit device. The place where a transistor current source IS a good idea, and the one which I based my original comment on, is where the unregulated supply varies for some reason. A very good example is with a battery power supply. The processor is powered by a voltage regulator but the LED is run at constant current directly from the battery.As the battery voltage falls the LED remains at constant brightness. For example, a 9v "manganese alkaline" battery will change from slightly over 9v to about 5.5v (0.9v endpoint per cell) across its lifetime. This would lead to a factor of almost 2 change in LED current and 4 times change in power dissipation in the LED circuit across voltage with a 1.5v LED. Without constant current drive the LED can be run at only about 53% of its rated current at battery endpoint. 9V batt I = (9-1.5)/R = 7.5/R 5.5V batt I = (5.5-1.5)/R = 4/R I change = 4/7.5 = 53.3% A specific example of the above is in typical IR remote controls where the LEDs typically draw very large currents (maybe several hundred mA) for very short periods. Here the constant current supply allows the LED to be run near its destruction point as the battery fades from new to dead. Without the current control capability the only choice is to spec operating current that will be safe with a new battery and suffer performance degradation as the battery voltage drops. Examination of most IR remote controls will show that the current limiting resistor is indeed in the drive transistor's emitter (whereas, if constant current was not a design aim, it would be in the collector,) Russell McMahon > > As even unregulated power supply voltages can not change so much, few volts > as expected, the resulting current change will be in the same order, or > almost non perceptible by the user. I would say that even when the > supplied voltage changes from 14 to 17V you still can use a simple current > limiting resistor in series with the LED and forgeting about constant > current controlling. > > In this case, a small NPN transistor setup as a simple switcher, emitter to > ground, series resistor at base, led and resistor at collector will be most {Quote hidden}> than enough. > > The good point of such simple and very common setup, is that you can > substitute the transistor by ANY logic gate (74HC00 family) without change > nothing, except if the supplied voltage is higher than the logic VCC, in > this case, a 7406/7 gate is in order. > > The limiting current resistor calculation is easy as (VPower - 2 / 0.01), > 14V-2V/0.01 = 1kohms 1/4 or 1/8W is enough. > > W46NER > > -- > http://www.piclist.com hint: PICList Posts must start with ONE topic: > [PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads > > > -- http://www.piclist.com hint: The list server can filter out subtopics (like ads or off topics) for you. See http://www.piclist.com/#topics

Russell McMahon wrote: [snip] > A specific example of the above is in typical IR remote controls > where the LEDs typically draw very large currents (maybe several > hundred mA) for very short periods. Here the constant current supply > allows the LED to be run near its destruction point as the battery > fades from new to dead. Without the current control capability the > only choice is to spec operating current that will be safe with a new > battery and suffer performance degradation as the battery voltage > drops. Examination of most IR remote controls will show that the > current limiting resistor is indeed in the drive transistor's emitter > (whereas, if constant current was not a design aim, it would be in > the collector,) That's 100% correct, where you need to force a constant current. There is of course, some drawbacks with the emitter resistor technique, when dealing with battery fading voltage. Suppose you are using a 9V battery in a circuit fed directly by this battery. This circuit will drives a transistor with a constant current emitter resistor and a LED at collector. Lets see what happens when the battery is 9V or 5.5V; Battery 5.5 or 9V o | | .-------o--------. | | | | .-----. D _V_ _--> | | | | DRV | C | |---R1-----B[NPN] | | A B E '-----' | | R2 | | _|_ _|_ NPN Beta = 100 ID = 20mA 1 IE = 20mA x (1+ -----) Beta IE = 20mA x 1.01 = 20.2mA 20mA IB = ------ = 200uA Beta R2 = 47 Ohms IR2 = 20.2mA VR2 = 0.9494 V VBE = 0.6V VB = VR2B = 1.594 V IB = 200uA VR2A = 5.4 or 8.9V hmmmm.... When VCC = 5.5V; VR2A = 5.4V VR2 = VR2A - VR2B = 5.4 - 1.594 = 3.806V R2 = 3.806 / 200uA = 19030 ohms. If R2 = 19030 ohms when VCC = 8.9V; VR2A = 8.9V 8.9V - Vbe IB = ------------------ 19030 + 47 x Beta IB = 7.5V / 23730 = 316uA IE = 316uA x 100 = 31.6mA VR2 = 47 x 31.6mA = 1.4852 V hmmmm It shows that constant current transistor configuration only works ok if the base current is also constant, of course, it relies on that. The above driver circuit is dependent of battery voltage (and it fades), doesn't matter so much if using emitter resistor to fix current or not, the R2 voltage will change along with battery voltage. The only way to keep the constant current would be to generate a fixed driver voltage at the driver output, what seems not to be the case. Some drivers use an internal double diode to ground to clip the output driver voltage to around 1.2V, even with a 3V battery application. It keeps VR2 fixed along with the LED current. In a 3V application, as remote controls, VR2 should be as small as possible, because when batteries fade out to less than 2.2V, less the transistor VCE, a high VR2 can kill the IR signal not because the batteries are low, but simple because there is not enough voltage around the IR LED. In this cases, the VBE and the known BETA will be used as the constant current controller. With a knowing Beta of 100, driving 2V to a 7kohms base resistor, will provide a 200uA base current, what will open 20mA collector current for the LED. As a small NPN can goes down to a 0.3VCE without saturating, it means the battery can fades down as much as VLED+0.3V, with the same IR emission level. It means what? that if you have a constant base current, it will result in a constant collector current, with a little more worries about making sure all the transistor batch shows a relative small beta variations. Using a very small REmitter (R2) will bend the constant current control to the Beta and VBE factors, since now they can almost dictate IC, being VR2 a small portion of it. It means that using an emitter resistor to fix current, it should develop a significative emitter-ground voltage, at least higher than 0.6V, so variations in VBE will represent small variations at the constant collector current. The same effect can be seem on operation amplifiers as constant current generators. In the feedback network, if the compared voltage is very small, the relation between the Compared Voltage and OPOffset (changes with temperature) will be small, causing the constant current very instable and sensible to temperature. __ Control > 3V __| |__ o | .------------. | 0.10909V | | R 1k | / \ | | | / \ _V_ Laser | 100k V / \ | 21.4mA o----R----o------(+)___(-) | Const | | | | Current _V_ 1.2 R 10k '--R10k---o | VREF | 0.10909V | _|_ _|_ R 5.1ohms _|_ In the above (ideal) circuit, a probable offset of 5mV at the op-amp will generate a change in the constant current, in the order of 5mV/109mV = 4.6% or almost 1mA. If by lack of VCC you need to reduce the compared voltage to less than 50mV, then the ratio gets worse. W46NER -- http://www.piclist.com hint: The list server can filter out subtopics (like ads or off topics) for you. See http://www.piclist.com/#topics

At 07:24 PM 10/18/02 +0200, Jan-Erik Soderholm wrote: >I'v added [EE]: to the topic... > >First, I think you should put the "R" between the LED and the collector. >And connect the emitter directly to ground. The configuration as drawn forms a constant current sink. The LED current is held relatively constant and is calculated by (Vdd-Vbe)/R. The transistor operates in the LINEAR region in this mode and the base current is approximately Iled / hfe. The only downside to this configuration is that the base current rises to about Iled if the LED becomes disconnected or if the unregulated supply drops too low. >Is there any special reason why you thought that connecting the PIC port >directly to the base was a good thing ? It is a great method for reducing the regulated power supply requirements. While it does not suit all methods of driving LEDs, it works well when you have only a few LEDs to drive. dwayne -- Dwayne Reid <@spam@dwaynerKILLspamplanet.eon.net> Trinity Electronics Systems Ltd Edmonton, AB, CANADA (780) 489-3199 voice (780) 487-6397 fax Celebrating 18 years of Engineering Innovation (1984 - 2002) .-. .-. .-. .-. .-. .-. .-. .-. .-. .- `-' `-' `-' `-' `-' `-' `-' `-' `-' Do NOT send unsolicited commercial email to this email address. This message neither grants consent to receive unsolicited commercial email nor is intended to solicit commercial email. -- http://www.piclist.com hint: The PICList is archived three different ways. See http://www.piclist.com/#archives for details.