Turning on a small load like an LED with a BJT in active mode.

Thread Starter

Rscott9399

Joined Jan 13, 2017
51
So was looking around at some circuits out there.

I have seen a few examples of people running simple things like an LED with a BJT just to light it up
I have stumbled upon several circuits where the bjt is in active mode not saturation.

These are examples from text books. Im surprised by this.
Is that ok to do? I know there are generally major concerns running a mosfet in linear region

Is this normally accepted in a BJT?

Essentially, my VB < VC but VB greater then VE only by the slightest bit, essentially the .7 drop to turn it on.

Any thoughts? Thanks
 

Thread Starter

Rscott9399

Joined Jan 13, 2017
51
lets forget about the mosfet for a second

and to answer your question, i dont know why i have concerns. I have just always been told a bjt as a switch should be fully saturated or in full cut off. one or the other
 

MrChips

Joined Oct 2, 2009
19,768
BJT and MOSFET can operate as switches.
BJT and MOSFET can also operate in the linear region. If there is any concern, perhaps it is the device power dissipation. This is true for any device that has resistance. The power dissipated can be calculated as I x I x R.
 

Thread Starter

Rscott9399

Joined Jan 13, 2017
51
BJT and MOSFET can operate as switches.
BJT and MOSFET can also operate in the linear region. If there is any concern, perhaps it is the device power dissipation. This is true for any device that has resistance. The power dissipated can be calculated as I x I x R.
yes i believe this is the main concern
So how do you calculate the power if you dont know R. Its easy with a mosfet because we know RDS on but what about a BJT
 

WBahn

Joined Mar 31, 2012
24,974
You mean the voltage drop across the BJT aka VCE??
Since the BJT is a three terminal device, there are three components to the power dissipation equation

Pce + Pbe + Pcb

Each term is the product of the current flowing between those two terminals and the voltage across those two terminals.

Ice·Vce + Ibe·Vbe + Icb·Vcb

The last term is usually negligible because Icb is usually small to nonexistent. But there are modes of operation where this is not the case.

For a transistor in the active region it is usually true (or at least assumed) that Ice >> Ibe and Vce >> Vbe, so the first term dominates.

But in the saturation region, Ib is often a significant fraction (or even perhaps larger) that Ic and Vce is usually significantly larger than Vbe, so it is often the middle term that dominates.
 

WBahn

Joined Mar 31, 2012
24,974
So was looking around at some circuits out there.

I have seen a few examples of people running simple things like an LED with a BJT just to light it up
I have stumbled upon several circuits where the bjt is in active mode not saturation.

These are examples from text books. Im surprised by this.
Is that ok to do? I know there are generally major concerns running a mosfet in linear region

Is this normally accepted in a BJT?

Essentially, my VB < VC but VB greater then VE only by the slightest bit, essentially the .7 drop to turn it on.

Any thoughts? Thanks
For light loads it usually doesn't matter depending on what your goals are. If you are making a circuit to establish a pretty constant current in the LED, then you would usually run it in the active region. Also, if you are trying to switch it on and off very quickly, you usually want to avoid both saturation and cutoff and keep it always in the active region.

For heavier loads where there is significant collector current, you generally want to drive it into pretty deep saturation to minimize the power dissipation (while keeping in mind that the base current starts contributing quite a bit to the power dissipation at some point).
 

Audioguru again

Joined Oct 21, 2019
543
We do not simple guess the value of a base resistor, instead we use a standard voltage divider for a linear transistor or a base resistor with negative feedback from the collector. Resistor value are calculated from the range of spec's shows on the transistor's datasheet.
 

MisterBill2

Joined Jan 23, 2018
4,564
For a long time transistors were mostly used in the linear mode and so there is a great deal of knowledge available in the huge amount of writing done about linear transistor operation. The switching mode is a special case of operation and also has a large amount of knowledge available. A whole lot has been presented in this thread, but there is a whole lot more available.
Aside from voltage and current ratings, power dissipation and heating are the things that can cause problems and are not so very well explained in many texts. But they are very real.
 

Audioguru again

Joined Oct 21, 2019
543
A linear transistor driving an LED can be in a constant current sink or constant current source circuit. Then the transistor replaces and heats like a series current-limiting resistor.
 

MisterBill2

Joined Jan 23, 2018
4,564
A linear transistor driving an LED can be in a constant current sink or constant current source circuit. Then the transistor replaces and heats like a series current-limiting resistor.
Totally correct. And also consider the linear operation of the transistors in all non-switching regulator circuits. That is why heat sinks are so common and so big. AND even in switching circuits, the transistor is in the linear mode for the entire time that it is moving between cutoff and saturation. I recall long integrals for calculating the power dissipated during that time period. It was far more critical back when transistor switching speeds were a lot slower.
 
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