cyamamot@nunki.usc.edu (Cliff Yamamoto) (07/13/89)
Greetings,
I know this is a simple circuit, but I can't seem to get the bias
right or something. The circuit is as follows :
C----bulb---> + 12 volts
TTL---1 Kohm----B NPN transistor
E---+
|
GND
The problem is when the TTL is high, the bulb barely lights up. This circuit
is for automotive use so the bulb is just an instrument panel bulb (not a high
powered spot light). Isn't 2N2222 or 2N3904 good enough for this? How does
one go about selecting a transistor for this? And lastly, how does one figure
out the bias needed to fully turn it on?
Thanks for all your EE101 advice! :)
Cliff Yamamotojohn@stiatl.UUCP (John DeArmond) (07/14/89)
In article <4363@merlin.usc.edu> cyamamot@nunki.usc.edu (Cliff Yamamoto) writes: >Greetings, > I know this is a simple circuit, but I can't seem to get the bias >right or something. The circuit is as follows : > > C----bulb---> + 12 volts >TTL---1 Kohm----B NPN transistor > E---+ > | > GND > >The problem is when the TTL is high, the bulb barely lights up. This circuit >is for automotive use so the bulb is just an instrument panel bulb (not a high >powered spot light). Isn't 2N2222 or 2N3904 good enough for this? How does >one go about selecting a transistor for this? And lastly, how does one figure >out the bias needed to fully turn it on? > Your ciruit would work if the load was not a bulb. The problem is the characteristics of a tungsten bulb. The resistance of tungsten varys widely between ambient and operating temperature. The net effect is that the bulb draws much more current cold than after it is lit. While your dashboard bulb may only be a 6 watt bulb (indicating about 500 ma), it will probably momentarily draw 10 amps or so when you apply power. Normally this is not a problem. Most circuits have enough reserve, either from adequate capacity, stray capacitance or inductance to provide this boost. The problem arises when you drive a bulb with a transistor. It helps to view a transistor as a variable constant current regulator. The current passed from collector to emitter is related to the current flowing from the base to emitter. This relationship is roughly the beta or gain of the transistor. This relationship implies that the collector-emitter current flow is independent of the applied voltage. This, in practice is roughly true. So a 2n2222 biased via the base resistor to handle the aforementioned 500 ma could not supply enough inrush current to cause the initial heating of the filiment. Even driven hard, a small signal transistor probably cannot handle the initial or inrush current due to collector saturation. In your circuit, a 1k resistor with 4.3 volts (5 volts - .7 volt base-emitter drop) passes 4.3 MA. This translates into a collector current of 430 ma, assuming a beta of 100. In other words, not nearly enough. A more insidious problem exists in trying to drive this device from TTL. Assume a beta or current gain of 100. That means that 1 ma of base current causes 100 ma of collector current to flow. For a collector current of 10 amps (inrush, remember), a base current of 100 ma would have to be supplied to the base. TTL can supply (hmm, my memory is foggy here) at most, about 10 ma or so. In any event, a TTL output is current-limited. Thus, even if your transistor had the collector current rating to handle the inrush, your TTL gate could not drive it. There are several solutions here. A darlington pair transistor effectively multiplies the gain of 2 transistors. A gain of 1000 or more is not uncommon. With this kind of gain, you can safely drive from TTL. There are incadescent lamp driver transistors available which combine a high current die in darlington configuration in a small package. This device has the necessary gain and can handle the momentary inrush current. Sorry, can't remember any part numbers off the top of my head. I do software here at work and hardware at home :-) A second alternative is the Sprague lamp driver ICs. These devices combine 6 high current drivers with the necessary TTL interface in one 16 pin IC. I believe the number is UNL1007 but I'm not sure. This is a very slick device. Contact sprague for a catalog. The last alternative is to use a power FET. This is a transconductance device meaning that its drain current is proportional to its gate VOLTAGE. No current is drawn from the driver. TTL can drive these directly. Again, don't remember a part number off the top of my head but I do know you can get suitable devices from Radio Shack. For sizing the device, a good rule of thumb is to assume an inrush to operating current ratio of about 10:1. The ratio may be lower for rugged filaments such as headlights. If you want to actually measure the cold resistance, you must use a low current ohmmeter. No, your Simpson 260 won't cut it. The problem with VOMs is that the current from the meter will heat the filament. The temperature vs resistance curve for tungsten is fairly steep so no heating is tolerated. A DVM on the low current ranges (ranges designed not to turn a diode junction on) will do pretty well. I've played around with this a bit. I've measured the effect of self-heating by carefully breaking a bulb and measuring the resistance of the filament with a DVM while the filament is immersed in oil as a cooling media. The difference is not great but is measurable. Hope this helps.. John -- John De Armond, WD4OQC | Manual? ... What manual ?!? Sales Technologies, Inc. Atlanta, GA | This is Unix, My son, You ...!gatech!stiatl!john **I am the NRA** | just GOTTA Know!!!
wiz@xroads.UUCP (Mike Carter) (07/14/89)
In article <4363@merlin.usc.edu>, cyamamot@nunki.usc.edu (Cliff Yamamoto) writes: > Greetings, > I know this is a simple circuit, but I can't seem to get the bias > right or something. The circuit is as follows : > > C----bulb---> + 12 volts > TTL---1 Kohm----B NPN transistor > E---+ > | > GND > > The problem is when the TTL is high, the bulb barely lights up. This circuit > is for automotive use so the bulb is just an instrument panel bulb (not a high > powered spot light). Isn't 2N2222 or 2N3904 good enough for this? How does > one go about selecting a transistor for this? And lastly, how does one figure > out the bias needed to fully turn it on? > > Thanks for all your EE101 advice! :) > Cliff Yamamoto Cliff; Try this circuit: +3-18 VDC | | > < > 330ohm - 1Kohm < / ------------------(*)------- / C lamp | TTL-----------I | I B | \ | \ E | \ | | | | | ----------------------------- GND The "> " and "<" are a resistor 2N2222A Is sufficient. Depending on the current needed to turn the lamp on will dictate the Ohm value. When the transistor is negatively biased at the base, the lamp will be in the ON state in this scheme. If (like you asked ) a HIGH is used to activate the lamp then move the connection of the lamp at the "C" collector to the "E" emitter and place the lamp in SERIES from emitter to ground. SInce I haven't actually built this (Just a quik draw here) , Don't get upset if you fry a few 2N2222's ...Since the transistor might buckle under the low Ohm to ground resistance , safety says place a equal value resistor in series with emitter to ground in the PICTURED drawing...experiment with differing values. All you need is the transistor to act like a switch..saturated ON and completely cut off. Just limit the amount of electrons flowing from Ground to suplly through the tranny and you'll do fine. IN my other suggestion, only ONE resistor is required and I have used them to activate relays with about 30ohm coils to 5V without using resistors. Your circuit fails here because you are limiting the base current with a 1 K in series. Sometimes the TTL voltage will be pulled down when the tranny "turns on" (depending on the TTL DRIVE circuit...ULN2003's are excellent when this occurs) take a 2.2K resistor from the TTL input at the base to the supply voltage. (On active highs) ..this "pulls up" the input. If you wish to learn Transistor biasing, you really need to get it from a work book that has Questions and Answers for a variety of problems and not take it in pieces from the net. Enjoy. P.S. What kind of automotive doo-hickey are you planning? My favorite past-time is making gadgets for my truck...I *love* gadgets with L.E.D's (slobber slobber. drool drool). :-) -- ============================================================================= = Mike Carter N7GYX, Phoenix AZ| Q: Why did the Chicken cross the road ? = = hplabs!hp-sdd!crash!xroads!wiz| A: To ESCape the Main Menu . = =============================================================================
brianr@tekig5.PEN.TEK.COM (Brian Rhodefer) (07/15/89)
When comiserating with my co-workers after committing some engineering boner, I used to lament that "I'm not competent to control a ##@$#%$ing light bulb". I *thought* the expression was hyperbole. In article <5769@stiatl.UUCP> john@stiatl.UUCP (John DeArmond) writes: >may only be a 6 watt bulb (indicating about 500 ma), it will probably >momentarily draw 10 amps or so when you apply power. A quick scan through our component database turns up four 12V, bayonet-base incandescent lamps. The highest operating current in the bunch is 200mA, and the others are in the 50 to 60 mA range. I don't have a manufacturer's catalog in front of me, but if we're talking about T1 3/4 - sized "pilot lamps", 5W seems way out of line for power dissipation. >Normally this is not a problem. Most circuits have enough reserve, either >from adequate capacity, stray capacitance or inductance to provide this boost. Stray capacitance, which is seldom larger than a few tens of picofarads, is NOT going to help light up an incandescent lamp. And NO amount of inductance will help in this application. >Even driven hard, a small signal >transistor probably cannot handle the initial or inrush current due to >collector saturation. In your circuit, a 1k resistor with 4.3 volts >(5 volts - .7 volt base-emitter drop) passes 4.3 MA. This translates Fine, except that the problem is the transistor coming OUT of saturation, and into its constant-current mode. Also, TTL gates can't pull all the way to +5V while sourcing useful current. My old TI databook has VoutHigh dropping to 2.5V at an output current of 10mA for vanilla TTL. Fiat Lux! Brian Rhodefer
roy@phri.UUCP (Roy Smith) (07/15/89)
In <4363@merlin.usc.edu> cyamamot@nunki.usc.edu (Cliff Yamamoto) writes: > I know this is a simple circuit, but I can't seem to get the bias > right or something. The circuit is as follows : > > C----bulb---> + 12 volts > TTL---1 Kohm----B NPN transistor > E---+ > | > GND > > The problem is when the TTL is high, the bulb barely lights up. The problem is that TTL can sink current, but not source it. With TTL you really have to think current sinks instead of voltage sources. For example, the typical circuit to light an LED: TTL---1-kOhm---LED---+5V When the TTL output is high, no current flows and the LED is dark. When the TTL output is low, you draw O(3mA) and the LED lights. Off the top of my head, I would suggest something like the following for what you want to do: +12V---bulb | | R1 | | C TTL ----+-----B | E R2 | | | +------+ V Ground Choose R1 & R2 such that (12*R2)/(R1+R2) is less than 5V but high enough to saturate the transistor. Going higher than 5V could damage the TTL output driver; probably you want to design for 4V to be safe, or maybe drive the base bias from the +5V TTL supply instead of the +12V supply. When the TTL output is low, it will drag the base bias voltage down. Check the TTL data sheets to see how much current your particular part will sink and choose R1 accordingly -- too low an R1 will result in damage to the TTL output drivers again, or you simply might not drag the base down low enough to get the transistor into cutoff. You might want to use a 74HXX (high power) part for this gate; they can sink more current. Depending on how much current you need to light your lamp, you might also want a darlington. Uncle Ralph would be proud of me. -- Roy Smith, Public Health Research Institute 455 First Avenue, New York, NY 10016 {att,philabs,cmcl2,rutgers,hombre}!phri!roy -or- roy@alanine.phri.nyu.edu "The connector is the network"
ISW@cup.portal.com (Isaac S Wingfield) (07/15/89)
Try this: connect TTL output directly to transistor base connect 1K resistor from TTL output to +5V Now, when TTL is low, it sinks 5mA, no problem when TTL is HI, it does nothing and 1K provides about 4-4.5 ma of base drive be sure to use a transistor with high enough beta; 200 or so would be good Good luck, Isaac isw@cup.portal.com
john@stiatl.UUCP (John DeArmond) (07/16/89)
In article <3868@phri.UUCP> roy@phri.UUCP (Roy Smith) writes: > > The problem is that TTL can sink current, but not source it. With >TTL you really have to think current sinks instead of voltage sources. For >example, the typical circuit to light an LED: To clear up a small point. Only open collector devices do not supply current in the high state. All other TTL supplies relatively high current through a totem-pole output arrangement - it must in order to charge circuit capacitance and achieve high slew rates. My TTL book from National shows that a typical gate (7404 hex inverter to be exact) can supply 35 ma in the high state. Not a lot but still enough to work with. -- John De Armond, WD4OQC | Manual? ... What manual ?!? Sales Technologies, Inc. Atlanta, GA | This is Unix, My son, You ...!gatech!stiatl!john **I am the NRA** | just GOTTA Know!!!
abali@parts.eng.ohio-state.edu (Bulent Abali) (07/17/89)
In article <5841@stiatl.UUCP> john@stiatl.UUCP (John DeArmond) writes: >In article <3868@phri.UUCP> roy@phri.UUCP (Roy Smith) writes: >> The problem is that TTL can sink current, but not source it. With >>TTL you really have to think current sinks instead of voltage sources. For >>example, the typical circuit to light an LED: > >To clear up a small point. Only open collector devices do not supply >current in the high state. All other TTL supplies relatively high >current through a totem-pole output arrangement - it must in order to >charge circuit capacitance and achieve high slew rates. My TTL book from >National shows that a typical gate (7404 hex inverter to be exact) can >supply 35 ma in the high state. Not a lot but still enough to work with. >John De Armond, WD4OQC | Manual? ... What manual ?!? There is either a typo in your databook, or you are mistakenly reading the ICCH rating. This is the current supplied through Vcc to the gate. This is NOT the TTL output current. The correct place to look at is the I_OL and I_OH ratings. As the previous poster stated, TTL outputs can sink current, but cannot source as much. My databook from TI specifies recommended maximum LOW output current as 16 ma (sink), and HIGH output current as 0.4 ma (source). Back to the original question: 12 volts : : BULB :--------- 2.2K C-----: : TTL------<><><>----B C < E----B > R E < : : GND GND A darlington transistor pair with a collective beta of 500 or more is sufficient enough to drive a 200ma bulb. R is a warmup resistor for the light bulb to prevent cold rush current (Just for the sake of completeness. I don't think it is really necessary.) -=- Bulent Abali Ohio State Univ., Dept.of Electrical Eng. 2015 Neil Av. Columbus, Ohio 43210 abali@baloo.eng.ohio-state.edu
geller@bnlux0.bnl.gov (joseph geller) (07/17/89)
In article <4363@merlin.usc.edu> cyamamot@nunki.usc.edu (Cliff Yamamoto) writes: >Greetings, > I know this is a simple circuit, but I can't seem to get the bias >right or something. The circuit is as follows : > > C----bulb---> + 12 volts >TTL---1 Kohm----B NPN transistor > E---+ > | > GND > >The problem is when the TTL is high, the bulb barely lights up. This circuit >is for automotive use so the bulb is just an instrument panel bulb (not a high >powered spot light). Isn't 2N2222 or 2N3904 good enough for this? How does >one go about selecting a transistor for this? And lastly, how does one figure >out the bias needed to fully turn it on? > >Thanks for all your EE101 advice! :) >Cliff Yamamoto To interface from anything (such as TTL) to the outside world (as a lamp) you must consider two questions first. 1) What is the drive signal ? 2) What are the current and voltage requirements of what is being controlled ? 1) Your drive signal is the output of a TTL gate. There are many TTL families (S, LS, F, etc.) but for the moment lets assume the original TTL, as in 7400, etc. TTL devices are specified by fan out. A typical logic element, as the 7400, has a fan out of 10. When one TTL gate drives another it supplies a one or a zero to the input of the following gate. Since TTL is made of transistor stages you must consider voltage and current at this point. It is not good enough to simply sat zero is zero volts and one is 5 volts. When you design a TTL circuit that does not exceed the fan out limit, the voltages are well defined. A zero may vary from 0 volts to .8 volts and a one may vary from 1.8 volts to 5 volts (the .8 and 1.8 may not be exact for standard TTL, but there are spec'd numbers such as these). If the TTL output of the driving stage is a one, it is actually back biasing the input transistor on the following stage. Since the load is actually off, with a one input, the driver supplies very little current, or about 40 microamps. With a fan out of 10 a standard TTL chip is rated to source only 400 microamps. The voltage is typically on the order of 4 volts. When the driver is sending a zero, it really turns on the input stage of the following gate. Now the current is much higher, on the order of one and half milliamps. So a typical TTL gate can sink over 15 mA. So ... your drive signal can supply 400 uA at about 4 volts. The limit in your circuit is not the transistor or the resistor. It is that you simply cannot draw 4 mA from the standard TTL gate at a one output. There are many many ways to interface to TTL. For typical LED or small incandescent lights drawing less than 30 mA, many engineers opt for the standard 7406 or 7407 open collector outputs. To use the output you wire the drive power supply to your load (if it is an LED you MUST use a series resistor) and the other side of the load to the chip output pin. For higher current loads the Motorola ULN chips or the Sprague chips, as the UDN series are fine. You can also build a power FET stage, or a darlington stage, but, since this sounds like a first TTL interface project I would recommend one of the packaged drivers. In a pinch, you can use the 7407 or 7406 to sink the current from a resistor tied to Vcc when its output is on, and when its output is off use the same current to bias your 2N2222. Since the 06 and 07 have opposite logic, one or the other will light your light at the right time. I don't want this to be too long, so just quickly lets look at your load, the light. If it draws 500 mA at 12V, the control transistor must have a maximum collector current of at least 500 mA (and usually more for engineering margin). The 2N2222 is about 800 mA, so it is close but, OK. If the 2N2222 has a voltage drop of about .2 V when it is full on, you are asking for .2 * .5 or about 100 milliwatts of power dissipation, also OK.
dk@thumper.bellcore.com (Dan Kahn) (07/18/89)
cyamamot@nunki asks why he can't fully light an instrument-panel bulb in the collector of a 2N2222 driven by TTL. john@stiatl suggests > Your ciruit would work if the load was not a bulb. The problem is the > characteristics of a tungsten bulb. The resistance of tungsten varys widely > between ambient and operating temperature. The net effect is that the bulb > draws much more current cold than after it is lit. While your dashboard bulb > may only be a 6 watt bulb (indicating about 500 ma), it will probably > momentarily draw 10 amps or so when you apply power. > > Normally this is not a problem. Most circuits have enough reserve, either > from adequate capacity, stray capacitance or inductance to provide this boost 1. I can't think of a practical case where the additional inrush current would be provided by "stray capacitance or inductance," but that is not relevant to the main point. 2. The steady-state (rated) current of a 12-volt instrument panel bulb is not likely to be anywhere near 500 ma, but let's assume it is. 3. It's true that when such a bulb is placed across a stiff 12-volt source, it will at first draw much more than 500 ma. The crucial error in John's explanation, however, is the assumption that a bulb _must_ go thru this high-current inrush stage in order to reach its rated-current steady state. That this is not necessary can be established with either a real or a "thought" experiment. Connect the bulb rated at 12 volts and 500 ma to 112 volts thru thru a 200-ohm resistor. Note that I've chosen the values such that if 500 ma flows thru this series circuit, there will be 100 volts across the resistor and 12 volts across the bulb, for a total of 112 volts, = the source. With 12 volts and 500 ma, the lamp would be at normal brilliance. Will this actually happen? According to John's posting it will not, since when the bulb is placed across the 112-V/200-ohm series combination, the most current that could possibly flow thru it would be 112 V / 200 ohms = 560 ma (obtained by assuming that the cold resistance of the bulb is 0), only a bit more than its steady-state current, while John's claim is that several amps is required to get the bulb going. But without even hooking up any wires, think about the implication of John's claim: if the bulb doesn't light up, its resistance is very low, thus close to 560 ma is flowing thru it. Alongside of it we have an identical bulb connected to a stiff 12-volt battery, brilliantly lit, with only 500 ma flowing thru it. The bulb with _less_ current is the bright one under this claim. The light output of incandescent bulbs is far from linear with current, but it sure is monotonic! And for those unimpressed with my "thought" experiment, try it in your lab. Precise voltages and resistors are not required. For ex., you can use 120 VAC, a 6- or 12-volt bulb, and whatever resistance you need to get close to the bulb's rated current. Not only will the bulb light up, it'll be a lot happier not to have been put thru the shock of the inrush surge that it experiences when placed across a stiff source at its rated voltage. 4. I've tried to illustrate the irrelevance of the "inrush" phenomenon to cyamamot's problem without the complicating factor of the transistor. But the transistor doesn't change anything. The bulb should light if enough current is supplied to the base. Several postings have addressed the issue of the ability of TTL to source adequate current. Dan Kahn Bell Communications Research Morristown, NJ
storkamp@sjs.sj.ate.slb.com (Mark Storkamp, 408-998-0123 (ex 2079)) (07/19/89)
In article <5841@stiatl.UUCP>, john@stiatl.UUCP (John DeArmond) writes: > In article <3868@phri.UUCP> roy@phri.UUCP (Roy Smith) writes: >> >> The problem is that TTL can sink current, but not source it. With >>TTL you really have to think current sinks instead of voltage sources. For >>example, the typical circuit to light an LED: > > To clear up a small point. Only open collector devices do not supply > current in the high state. All other TTL supplies relatively high > current through a totem-pole output arrangement - it must in order to > charge circuit capacitance and achieve high slew rates. My TTL book from > National shows that a typical gate (7404 hex inverter to be exact) can > supply 35 ma in the high state. Not a lot but still enough to work with. My TI data book shows, for the 7404, maximum high-level output current as -400uA. Somebody has sugested using a 74Hxx part, this has only -500uA. If you realy must source current, the 74S will give you -1mA. A pull-up on the base of the transistor to 5V should give you more drive, but the maximum low-level output current on a 7404 is 16mA, so don't use anything smaller than a 330. A 2N2222 has a minimum hfe of 100 @ 150mA, so this should give you at most 160mA for the bulb. For anything larger, try a darlington configuration. > > -- > John De Armond, WD4OQC | Manual? ... What manual ?!? > Sales Technologies, Inc. Atlanta, GA | This is Unix, My son, You > ....!gatech!stiatl!john **I am the NRA** | just GOTTA Know!!! Mark Storkamp Schlumberger Technologies
john@stiatl.UUCP (John DeArmond) (07/21/89)
In article <561@sjs.sj.ate.slb.com> storkamp@sjs.sj.ate.slb.com (Mark Storkamp, 408-998-0123 (ex 2079)) writes: > >My TI data book shows, for the 7404, maximum high-level output current as >-400uA. Somebody has sugested using a 74Hxx part, this has only -500uA. Don't confuse the current specification an output can supply and stay within the TTL voltage specs with the maximum available output current. TI may not even specify this value. When driving a transistor, we are interested in the current drive and not the voltage as long as the driver can supply at least 0.6 volts or so. To moot this question, I actually measured the output while driving a diode (same as B-E junction). The non-descript gate I pulled from a grab-bag supplied 25 ma into a 0.6v load. Not spec but not bad. John -- John De Armond, WD4OQC | Manual? ... What manual ?!? Sales Technologies, Inc. Atlanta, GA | This is Unix, My son, You ...!gatech!stiatl!john **I am the NRA** | just GOTTA Know!!!
tomb@hplsla.HP.COM (Tom Bruhns) (07/22/89)
john@stiatl.UUCP (John DeArmond) writes: >> >Your ciruit would work if the load was not a bulb. The problem is the >characteristics of a tungsten bulb. The resistance of tungsten varys widely >between ambient and operating temperature. The net effect is that the bulb >draws much more current cold than after it is lit. While your dashboard bulb >may only be a 6 watt bulb (indicating about 500 ma), it will probably >momentarily draw 10 amps or so when you apply power. > ... >If you want to actually measure the cold resistance, you must use a low current >ohmmeter. No, your Simpson 260 won't cut it. The problem with VOMs is that the >current from the meter will heat the filament. The temperature vs resistance >curve for tungsten is fairly steep so no heating is tolerated. A DVM on the >low current ranges (ranges designed not to turn a diode junction on) will >do pretty well. I've played around with this a bit. I've measured the >effect of self-heating by carefully breaking a bulb and measuring the >resistance of the filament with a DVM while the filament is immersed in >oil as a cooling media. The difference is not great but is measurable. > >Hope this helps.. > >John These two paragraphs, separated by several others in the original, seem to contradict eachother. If "my" Simpson 260 will heat the filament enough to significantly change its resistance, then I'm quite sure the 2n2222 won't have much problem overcoming the cold- resistance effect.
tomb@hplsla.HP.COM (Tom Bruhns) (07/22/89)
In the nine responses I've seen here, the closest I've seen to my favorite solution to interfacing TTL to bipolar transistors is reference to the concept that TTL generally sinks current much better than it sources it. First, I would have used a 2N4401 or 2N4403 instead of a 2N2222/2N3904. The 4400-4403 series is designed to handle currents much higher than the 2222 et al. As I recall, the 4400 series betas peak at about 150 mA collector current. They are about as cheap as 3904s. Second, I would look for a way to turn the transistors on when the TTL output is low: to do this, consider TTL output to R to base of PNP. Emitter of PNP to +5. Resistor shunting base to emitter. Proper choice of resistors can guarantee turnoff and substantial base current to the PNP in the on state. The PNP can then drive a load directly, or drive the base of an NPN power transistor, at in excess of 100 mA easily, and that's assuming very low betas. I'd generally try to use an integrated power driver in a commercial application, but for hobbiest hacking where it's what's in the junkbox that counts, I've become quite fond of the ttl --> pnp solution; it uses the TTL the way it was intended w.r.t. output currents. Another note: several folk suggested a Darlington connection. Be aware that the drop across a Darlington won't get much lower than a volt for silicon devices at room temp and moderate current. A cleaner output (lower drop) results from running the first of the two transistors from, say, +5 on its collector; ya gotta limit current somehow, probably most easily with a resistor from first emitter to second base. But the problem with all this sort of thing is that the output voltage vs current for TTL in the high state is not well-controlled. They guarantee some minimums, but your circuit should be able to handle TTL output anywhere between Voh(min) and Vcc when the current out is no more than the Ioh rating. By contrast, the Vol range is very limited, and the available current is much higher. Thus, my TTL -- pnp -- npn solution has the same nominal available net current gain as a ttl -- darlington, but has much more available output current, since it's "on" when the TTL output is "low". Especially if this is going to be used in an automotive environment, you should design for the extremes you will likely encounter.
geller@bnlux0.bnl.gov (joseph geller) (07/22/89)
The article from tomb@hplsla.HP.COM offers a real nice solution
to the original TTL - light bulb question. I'd just like to correct an
error I made in my reply.
I stated that the maximum high level output for a standard TTL
output is 400 uA. This is the maximum 1 level steady state output current
you should design for, but as others have mentioned, the totem pole output
is capable of much higher 1 level currents.
The higher currents are specifically intended to charge circuit
capacitance to give the 10 - 20 nano second TTL transitions. To design a
circuit which uses the 1 level output current (the totem pole sourcing I)
to directly drive a transistor is iffy at best.
The TTL totem pole output stage looks like this :
+5
|---| The voltage drop across Ric goes up
Rib Ric as the one level current (Ioh) does.
| | The power dissipation, or chip heating,
| C also goes up with Ioh.
internal chip-- | -B NPN At Ioh max, 400 uA, the power
E dissipated by Ric may be as low as
| micro watts at very low switching speed.
\/ As the frequency of the switching goes up
-- so does the average power, because of the
| load circuit C's.
|--(TTL OUTPUT PIN)
| The real iffy part is to try to say
C what an acceptable Ioh above the 400 uA
internal chip------B NPN might be. If a circuit draws 10 mA from
E the output (Voh would be about 2.2 V), the
| power dissipation in Ric is up to on the
common order of 10 milli watts steady state.
Remember that Ric and the NPN's are
integrated components on the chip.
Any way my rough calculations for that original circuit show that
several mA might have been available; my comment that 400 uA was the limit was
incorrect. Thanks to brianr@sbs.uucp for his correction by email.
Perhaps integrated drivers are not the best answer when a junk box
full of transistors is available. I would recommend using the PNP solution
first. It's certainly got the big advantage that when the PNP driver is off
you are not dumping power into a resistor. If your not on battery power
a pull up resistor from totem pole or open collector TTL, which then biases
an output NPN when the TTL output is high, also works fine.
Component failure due to heating is usually one of the first failure
modes in a circuit were ratings were exceeded. I suppose there are cases where
it might be OK to exceed mfgr max values but, my experience is that there are
plenty of "gremlins" to deal with even when those max values are not exceeded.john@stiatl.UUCP (John DeArmond) (07/24/89)
In article <5170048@hplsla.HP.COM> tomb@hplsla.HP.COM (Tom Bruhns) writes: > >These two paragraphs, separated by several others in the original, >seem to contradict eachother. If "my" Simpson 260 will heat the >filament enough to significantly change its resistance, then I'm >quite sure the 2n2222 won't have much problem overcoming the cold- >resistance effect. Well try it and see. I'm sure you'll see the light (or lack thereof). The simpson WILL provide enough current to alter the "cold" reading. Remember the objective was to obtain the cold resistance, right? The ammount of current necessary to heat a filament a couple hundred degrees and the ammount needed to reach incadescence (sp) is vastly different. When you are trying to mentally analyze this, remember to account for radiative energy losses which go up exponentially with temperature. (or is it geometrically? anyway, a lot :-) -- John De Armond, WD4OQC | Manual? ... What manual ?!? Sales Technologies, Inc. Atlanta, GA | This is Unix, My son, You ...!gatech!stiatl!john **I am the NRA** | just GOTTA Know!!!
kline@tuna.cso.uiuc.edu (Charley Kline) (07/26/89)
In article <1383@bnlux0.bnl.gov>, geller@bnlux0.bnl.gov (joseph geller)
and others speak in detail of how to build slightly higher-power drivers
for TTL outputs.
I just finished a project, which works fine, where a 74198 shift register is
driving the LED side of an optoisolator through the following circuit:
[sorry about the words, i don't have the patience for ASCII graphics]
TTL output of 74198 to a 10k resistor to B of a 2N2222-style transistor. E
of transistor to ground. C of transistor through 15 feet of ribbon cable
to another enclosure, to a 270 ohm resistor, to the cathode of the LED.
Anode of LED to +5.
Now as I mentioned, this all seems to be working perfectly. My only
reservation is that in operation, the 74198 gets a little warm. Not
alarmingly so, not even warm enough to bother me really, but when I had this
thing breadboarded on my bench and was just driving an LED through a 270 ohm
resistor from the 74198, with none of this transistor driver jazz, the 74198
ran with no detectable (to my finger) increase in temperature.
I would have thought that the 10k resistor between the driver and the 74198
would be limiting the high output current to something much less than even
the simple LED load would be presenting. Am I wrong? What else could be
causing this thing to heat up?
_____
Charley Kline, Assistant TD, Celebration Company at the Station Theatre
c-kline@uiuc.edu
uunet!uiucuxc!kline
"Just another useless dead thing, I've been killed by love."tomb@hplsla.HP.COM (Tom Bruhns) (08/01/89)
dk@thumper.bellcore.com (Dan Kahn) writes: > > ... > >need to get close to the bulb's rated current. Not only will the bulb >light up, it'll be a lot happier not to have been put thru the shock ^^^^^^^ Actually, I don't think anyone has yet proven that light bulbs experience happiness :-) :-) >of the inrush surge that it experiences when placed across a stiff >source at its rated voltage. > > ... > >Dan Kahn Bell Communications Research Morristown, NJ >---------- But seriously, thanks for the explanation -- it's a better way to communicate the same thought I had when I read the posting on the ohmmeter changing the resistance of the bulb significantly while the transistor (capable, presumably, of passing a lot more current than the ohmmeter sources) wouldn't, in the poster's opinion, be able to turn on the bulb because of it's cold resistance!
davidc@vlsisj.VLSI.COM (David Chapman) (08/01/89)
In article <1647@thumper.bellcore.com> dk@thumper.bellcore.com (Dan Kahn) writes: >cyamamot@nunki asks why he can't fully light an instrument-panel bulb >in the collector of a 2N2222 driven by TTL. john@stiatl suggests > >> Your ciruit would work if the load was not a bulb. The problem is the >> characteristics of a tungsten bulb. The resistance of tungsten varys widely >> between ambient and operating temperature. >> >3. It's true that when such a bulb is placed across a stiff 12-volt >source, it will at first draw much more than 500 ma. The crucial >error in John's explanation, however, is the assumption that a bulb >_must_ go thru this high-current inrush stage in order to reach its >rated-current steady state. > >[sensible explanations deleted] > >And for those unimpressed with my "thought" experiment, try it in >your lab. Precise voltages and resistors are not required. For ex., >you can use 120 VAC, a 6- or 12-volt bulb, and whatever resistance you >need to get close to the bulb's rated current. Not only will the bulb >light up, it'll be a lot happier not to have been put thru the shock >of the inrush surge that it experiences when placed across a stiff >source at its rated voltage. Oh, yes, why didn't _I_ think of that? I've been watching this discussion from a distance (I personally wouldn't use incandescent lights for anything - strictly semiconductors :-) and never once did I remember: There's a market for "light bulb life extenders" that do precisely that - limit the inrush current while the bulb warms up. A surge is clearly not necessary to light the bulb. Personally, I'd use some sort of Darlington arrangement with much more gain than necessary. TTL chips, even from the same manufacturer, vary widely in their output current capability. If it says 400 uA, don't try to use any more than 400 uA even though you "know" that number is conservative. (Of course, I expect to light an LED with the output of a 4049 - CMOS buffer - in series with a 330 ohm resistor :-) >Dan Kahn Bell Communications Research Morristown, NJ (space filler) So who was the idiot who wrote inews? Why should there be more of "my" text than what I quoted if the article is short? They should be _glad_ that I'm not contributing _more_ to net.noise! Grrr..... -- David Chapman {known world}!decwrl!vlsisj!fndry!davidc vlsisj!fndry!davidc@decwrl.dec.com