Transistor questions

I'll assume you mean 'when calculating values for a transistor to be driven by an IC output...'

The chip outputs are fairly standard, but the outputs will probably not be at the power rails. These numbers are fairly close to the original TTL spec IIRC.

The output _capability_ is independent of the load. It means quite simply that the output of the IC, when high, can source 0.4mA into something less positive, and when low it can sink 8mA to it's negative supply (probably ground, but let's not assume anything). Just because those are the capabilities does not mean the chip _will_ source / sink those values - that depends on what is connected externally.

At the output point, it's merely a current path and current will 'split' according to effective loads if there are multiple paths. Two equal paths will split the current equally. Note that two transistors even of the same type aren't necessarily 'equal loads'.

Tell us a little mnore clearly what you are trying to do and we might be able to help.

Cheers

PeteS

Hi, Digit. I'm going to assume you're driving transistors from the outputs of an LSTTL gate -- that sounds the closest to your spec (although a standard LS gate is -0.4mA and 16mA).

Here's the drill. As you may know, the logic levels of TTL gates are supposed to be guaranteed to be at or over 2.8VDC when it's a logic high, and less than 0.4V when it's a logic low. Let's look at the datasheet:

On page 4 under "Recommended Operating Conditions" it says that I(OH)(max.)[High Level Output Current] is -0.4mA. That means the gate output voltage is guaranteed to be at least 2.8V if you are sourcing

0.4mA from the output. It also says that I(OL)(max.) [Low Level Output Current] is 16mA. That means that the gate output voltage is guaranteed to be less than 0.4V if you are sinking 16mA into the output when it's low.

Now if you're driving an NPN transistor, you want it to turn on when the gate output is high, like this (view in fixed font or M$ Notepad):

| | VCC | + | | | .-. | Load | | | | | | '-' | | | __ I(OH) | | -| \\ ---> ___ |/ | | )o-------|___|-o-| | -|__/ R | |>

| .-. | | R| | === | | | GND | '-' | | | === | GND | (created by AACircuit v1.28.6 beta 04/19/05

When the NAND output goes high, the NPN transistor will turn on. Now since the gate can only source 0.4mA, and is guaranteed to be only

2.8V, you'll have to size R such that 2.1V will drop across R when 0.4mA is going through it. Using Ohms Law, that means you have to make R at least 5250 ohms (choose 5600 ohms as the next greater standard value). This is pretty grim, because your transistor will only be able to switch 4mA to 10mA load current and stay in saturation.

Now let's look at the PNP case -- the transistor will turn on when the gate output goes low: | VCC | + | | | .-. VCC | | | + | | | | | __ I(OL) '-' | | -| \\

Hey thanks for the help, the PNP transistor is sounding like a much better idea. I run into one snag though, the PNP's I have can only support a max of 100mA at the collector and my load is six leds at 20mA each, so I need atleast +120mA. This is where I was heading asking about using 2 transistors on one output to split the load. I need to know how to calculate the resistor values need for the transistor.

IC specs: SN74LC148

There are bigger transistors, like 2N4403.

But with a 12 volt LED supply, are you willing to string 2 or 3 LEDs in series and run them on a single 20 milliamp current? That would make use of the supply a lot more efficiently.

But you can also provide base drive to two or more transistors from the same logic output. but even with the

2N3906, I doubt it is necessary for 120 mA.

If you add an NPN transistor (i.e. 2N3904) as a level shifter, you could tie its base to the +5 logic supply, its emitter to the logic output through a current limiting resistor of say, 470 ohms. Then its collector can tie directly to the PNP base. You should also provide a bleed resistor across the PNP base to emitter to get rid of any leakage, perhaps 10k. Of course, its emitter goes to the

+12 rail and the collector to the LED resistors.

Sorry I forgot to say I was running the leds in parrallel, thier voltage drop is about 4 to 4.5v so my only other option would be 3 rows of 2.

Yes.

I would ask in what configuration you plan to use the NPN or PNP external device btw. Common emiiter or emitter follower ?

No.

You need to use current sharing resistors. But then again you shouldn't connect a gate ( IC ) output directly to a transistor base anyway. There shoule be a series base resistor to set the desired Ib current.

It's more helpful if you think in terms of the IC's source ( current out ) and sink ( current in ) ratings btw.

Graham

Does anyone actually still use that stuff ?

Graham

Why not run the LEDs @ 16mA instead ( total 96mA ) ?

Graham

That would work fine, I think, and cut the switch current in half. The current would vary more if the supply is unregulated (resistors dropping less of the total voltage), but I don't know if that is a problem, in your case.

Yup. They're great for homework problems. ;-)

Doing basic interface with TTL is educational, because of all the gyrations you have to go through. At the end of the semester, they get let in on the secret that you'd just use a 74HC gate and a logic level MOSFET to solve this problem.

Cheers Chris

I'd hope the time might actually be spent learning some useful skills in place of the limitations of old technology.

Graham

TTL isn't difficult, and interfacing reinforces the basic lesson that all digital is analog, too.

Let 'em suffer. Myself, I'd make 'em all use slide rules and no calculators in the first semester, to improve the breed and give 'em the sense of the magnitude of numbers that all engineers and techs are supposed to have (but frequently don't, especially the newer grads).

Yup. We all used slipsticks back in the day, too. That's the way it was, and *we liked it*.

Cheers Chris

put in a 470 ohm R in series for each LED.

P.S. 4--4.5 volts seems high for a LED ?

--
"I am never wrong, once i thought i was, but i was mistaken"
Real Programmers Do things like this.
http://webpages.charter.net/jamie_5

I read this as the OP assuming common emitter. In that specific case, the answer would be yes. I think a blanket "no" to that question, while conditionally correct in the general case, will confuse the OP.

As you note, the OP must better define the question to recieve a meaningful/correct answer.

Digitmode:

This is correct for common emitter NPN or PNP transistors, for other configurations it is not necessarily true.

Martin

Hi, Digit. Calculating resistor values for transistors used as switches isn't too difficult -- I wonder why your teacher hasn't walked you through it.

OK. The information that we've gotten so far is that you have a TTL gate (which assumes a +5V supply) and a +12V supply. Your LEDs have a forward voltage of 4.5V to 5V (this indicates to me that they're probably logic level LEDs with built-in resistors). You haven't said whether you want the LEDs to turn on when the gate output is high or when it's low (this is useful to know, unless you happen to have spare inverters floating around). Let's assume to begin with that you want the output to be active-low (which was very common back in those thrilling days of yesteryear when men were men, women were glad of it, the Lone Ranger and Tonto ruled the West and TTL ruled the globe).

You would then drive a PNP transistor with the gate output, and that would then drive an NPN which would drive the load. Try a 2N3906 (PNP) and 2N3904 (NPN). They can handle 250mA, which should do the job. Your circuit should look something like this (view in fixed font or M$ Notepad):

| +12V | + | | | .---o---o-o-o---o---. | | | | | | | | | | | | | | | V~ V~ V~ V~ V~ V~ | +5V -~ -~ -~ -~ -~ -~ | + | | | | | | | | .-. .-. .-. .-. .-. .-. | .-. | | | | | | | | | | | |R3 X 6 | 22K| |+5VC | | | | | | | | | | | | | | | + '-' '-' '-' '-' '-' '-' | '-' | | | | | | | | |\\ ___ | |< | | | | | | | -| >-|___|-o-| 2N3906 o---o---o---o---o---' | |/ R1 |\\ | | | ___ |/ | o-|___|-| 2N3904 | | R2 |>

| .-. | | 10K| | === | | | GND | '-' | | | === | GND (created by AACircuit v1.28.6 beta 04/19/05

OK -- let's work backwards. You know your 2N3904 has to drive 120mA. To do that well, it will have to be saturated when it's on (Vce <

0.2V). The way to do that is to overdrive the base. If the Hfe of the transistor is typically 200, that means the average transistor will have 200 times the collector current as base current, assuming it's driving a current source. Of course, there's a wide range of Hfe for a given transistor, so your results may vary. And that's driving a current source, too. If you're driving a resistor-limited load, as Vce increases, the load current will go down. And high current times high Vce equals high power, which will smoke the transistor.

So, you use one of the two golden rules of thumb for driving a transistor as a switch. The first, most universal one is drive the base with 1/10th of the current you want to drive thru the collector. For every transistor with a typical Hfe over 100, that always does the trick. Another rule is to get the minimum Hfe, and drive the base with

10X as much current it would suggest. For example, if you've got a 2N3055 with a minimum Hfe at that current range of 40, you would drive the base with 1/4 of the current you need at the collector. This one is universally true. But coincidentally, since the minimum Hfe of a 2N3904 is 100, the answer is the same either way. And the hard saturation will mean transistor power dissipation will be small: Pd = V * I = .2V * .12A = 24mW. Won't even get warm.

Now if you're driving 120mA, you'll need to inject 12mA into the base. Now you can start sizing a resistor. Assuming the 2N3906 will be saturated (Vce < 0.2V), and since Vb of the 2N3904 will be about 0.7V (1 diode drop), 4.1V will be dropped across R2. Now you can use Ohms Law:

R2 = 4.1V / .012A = 341 ohms (choose 3330 ohms as standard value for R2).

Now we're getting somewhere. Since the 2N3906 is going to be driving

12mA, it needs about 1.2mA of base current by the same rules above. Since that's a lot less than the 8mA your LSTTL gate is guaranteed to sink, you can be comfortable that the low voltage will be about 0.3V. Also, there will be about 1 diode drop at the base (Vb = 5V - 0.7V = 4.3V). That means you'll need to choose an R1 such that 4V will be dropped with 1.2mA:

R1 = 4V / .0012A = 3333 ohms (choose 3.3K as standard value).

The 10K resistor is there to prevent leakage current from partially turning on the 2N3904. The 22K pullup is there to ensure that a logic "1" from the LSTTL output will be close to 5V. Without something of a pullup, it's typically around 3.5V, which might turn on the PNP.

Just for grins and giggles, let's say you're driving two NPN transistors which are splitting the load equally (60mA ea.). That means each transistor will require half the base current, so just use a resistor of double the value (6.8K as standard value). The PNP will still be doing the same job, so you don't have to change anything there.

I guess that pretty much covers sizing resistors for use with transistor switches driven by logic gates. You should know that, in the real world, people generally use 74HC-type outputs, which have a symmetrical drive capability of +/-4mA. That is, unless it's a bus driver-type gate, in which case it can be as much as +20mA/-64mA. Read The Fine Data Sheet for the details.

Also, if you run into a situation like this, you may want to use a darlington transistor. The 400uA I(OH) of output current should be enough if minimum current gain is 4000. That will mean you can save a transistor and two resistors. You may also choose to go for a logic level FET, in which case you can even skip the two resistors, and just have the 74HC output drive the gate directly (or use a 100 ohm series resistor between the output and your gate).

Hope you get an "A".

Cheers Chris

Elsewhere, he had written:

I think this means 9V. It was a different poster talking about 12V, I think.

Some of the blue LEDs (I'm looking at one from RS here) operate at 5V and 30mA. So 4.5 at 20mA wouldn't surprise me for this one. (300mcd at 30mA device, RS part number 276-311.)

Other discussions gave me to understand that:

I had missed the 9V statement, at that time. Anyway, there is the description you were looking for, I think.

Except that is not what he said. However, it is probably better that way, if he has the option to control that choice.

I'll leave it at that. The rest must await further discussions.

Jon

I went to the store the other day when there was a power outage, and the clerk, who was almost as old as I am, was making change by hand, the old way.

For some reason, that pleased me. :-)

Cheers! Rich

When I was teaching, I allowed calculators by default, but one day I managed to get up on the board (as an interim result) 10/2. Half the class reached for their calculators. That was it; no calculators for a month. Boy did they suffer.

Amusingly, one guy in the class was in his 40s and *had* a slidestick. So he brought that in instead ;)

Cheers

PeteS

Hmmm

Good point in a way, but I look at using older (SSI) technology in the same way as going through the basics in mathematics; they really need to understand the basics. I fully support getting them to use simple gate arrangements up to the point of a decent flip-flop for the same reasons. Such devices exist (within FPGAa etc) but how are they going to have a feel for setup and hold times if they haven;t struggled with making them operate in the first place?

Think about it; a lot of people use calculators as substitute brains nowadays - if you asked them to add two numbers together they'd be screwed.

Now translate that into the next generation of electrical engineers - I would prefer they understood the basics :)

Cheers

PeteS