transistor - limiting base current to bjt - electrical engineering - stack exchange

7/28/2019 Transistor - Limiting Base Current to BJT - Electrical Engineering - Stack Exchange

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I'm actually working with a darlington pair of BJTs.

My question is if I need to put a 1kohm resistor between my MCU pin and the BASE to use

it as a switch. Since the Hfe ratio for my transistor is quite large, and the collector currentwill be limited to a reasonable level, does that mean that the base current is dependent on

the collector current at all times and thus I can eliminate the usage of this current limiting

resistor?

transistor bjt

edited Jun 19 at 15:32 asked Jun 19 at 12:08

Steven Lu

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2 Answers

No, it's the collector current that is dependent on the base current, not the other way

around. No matter what the collector current is, the base current is .

Keep in mind that will be twice the value of another transistor, as there are twounctions between base and emitter.

VMCU VBE

R

VBE

But it's true that the collector current is what you want in the end. So to find the resistor

value (don't just pick 1k), you calculate . If you want = 2A and = 400, then

your will have to be . This is a value your microcontroller will be able to

deliver, but always check the datasheet.

=I BI C

H FEI C H FE

I B = 5mA2A

400

To put it all together, .R = ( )H FE

I CVMCU VBE

Olin is right about the resistor value being the maximum, i.e. the base current beingminimum. For many parameters in a datasheet you'll find more than one value, like typical

and maximum or minimum. You should always calculate for worst case conditions, and it

may require some logical thinking to find out whether worst case is minimum or maximum

for a particular parameter.

edit

Take . In my example I picked a value of 400. As higher is usually better datasheets

often mention a minimum value. What if it's higher? The base current won't be different, so

the collector current will be higher. If you drive the transistor in saturation will no longer

be determined by the transistor, but by the load's impedance will be a limiting factor. So,

while the transistor would very much like to draw a larger collector current, it won't change.So you think you're safe; the minimum specified is fine, higher is still OK. There's

something else to consider, however: is not constant, it varies with , and the

datasheet should have a graph for this. So check this for the wanted collector current.

. Two PN junctions, so that's 2 x 0.65V = 1.3V. Olin found that a 300 base resistor

H FE

I C

H FE

H FE I C

VBE

sistor - limiting base current to BJT - Electrical Engineering -... http://electronics.stackexchange.com/questions/15628/limiting-b...

ur 3 2011-11-10 20:50


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should be fine, in fact leaves some margin. But when I look at the datasheet for the it

says may be as high as 2.8V! That would result in a base current of ,

and that's too little to get the wanted of 2A: is only 670mA.

TIP110

VBE = 1.7mA3.3V2.8V

300

I C 400 1.7mA

ou're getting the idea. Don't simply use typical values, but make sure that your circuit still

works with components with extreme parameter values. This is not so much of a problem

with projects where you only build 1 device: you can see what's wrong and adjust. For

production you have no choice: always design for worst case.

edited Jun 19 at 16:51 answered Jun 19 at 12:57

stevenvh

31.2k 2 57 102

Vbe would actually be a single transistor voltage drop. regarding the use of the resistor, it could also depend on whether

the MCU IO pin is in push/pull mode or not. If it is, a resistor is probably not needed. dhsieh2 Jun 19 at 16:09

@dhsieh - The datasheet for the mentions a of maximum 2.8V. You could get such a high voltage

with just a single PN junction.

TIP110 VBE(ON) never

stevenvh Jun 19 at 17:08

2.8V looks extremely high, even for a double junction. Can you explain this high value somehow? I think I would try tomeasure it and if I can't, call my supplier's FAE for confirmation. Federico Russo Jun 19 at 17:20

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To go a little further, Stevenh's calculation shows you the base resistor. It's a good

idea to allow for some margin and provide a little more base current (a little lower base

resistor) than the absolute minimum required to get the desired collector current.

maximum

Let's expand the Stevenh's example and get some real numbers. Let's say the processor is

running from a 3.3V supply. A darlington has two B-E junctions between its base and

emitter, so let's say the overall B-E drop is 1.3V. That leaves 2.0V accross the base resistor.

2V / 5mA = 400 Ohms. If you're really sure about Hfe being 400 over your operating range

and that you don't need more than 2A collector current, then you can use only a little lower

resistor, like the common value of 360 Ohms. For more margin, use less, like 300 Ohms

maybe.

Now you need to go back and see what the load on the micro is. 2V / 300 Ohms = 6.7mA.

That will be OK for most micros, especially if its a PIC which tend to have particularly good

output current capability. However, I've seen some micros that are specified for less than6.7mA, so you have to check and possibly adjust things.

One thing to consider with darlingtons is that they are slow to turn off. You say this is for a

switching application, so turn off time could matter. If you're just driving a relay then this is

no problem, but if you're trying to do 10s of kHz PWM then this is probably not what you

want to use.

Another problem with darlingtons is the rather high on-state voltage. It is one B-E drop plus

one saturated C-E drop, maybe 900mV but could easily be more at high currents. At 1V the

transistor would dissipate 2W with 2A collector current. That will require some sort of heat

sinking or at least something like a TO-3 case mounted to the chassis or some metal.

There may very well be better ways to switch what you want without using a darlington.

answered Jun 19 at 16:07


ur 3 2011-11-10 20:50


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Olin Lathrop

16.3k 7 34

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ur 3 2011-11-10 20:50

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