# Losses in BJT as current source

#### k1ng 1337

Joined Sep 11, 2020
23
Hi I'm investigating the efficiency of using a BJT as a current source to power an LED.

Let's say I have a base resistor who's gain provides 10ma to an LED, what are the losses in this circuit?

Aside from exactly matching an LED to a battery.. what is the most efficient way to power an LED with a forward voltage of 3v from a 5v source?

Thanks

#### ericgibbs

Joined Jan 29, 2010
11,551
hi 1337,

E

#### Ian0

Joined Aug 7, 2020
1,101
Matching source and load impedances is NOT the most efficient way to power the load. It is the way to extract the greatest power from the source.

#### DickCappels

Joined Aug 21, 2008
6,776
Is this homework or accidentally posted here?

#### BobTPH

Joined Jun 5, 2013
2,721
If there are no non-dissipative components, there is a maximum efficiency that can easily be calculated. And that efficiency is achieved by using only a resistor. A transistor can only make it worse.

With inductors and capacitors added, a more efficient circuit can be designed.

Bob

#### crutschow

Joined Mar 14, 2008
25,982
A resistor and transistor both limit the current to an LED in a similar manner, by dropping the excess voltage and dissipating the power from that.
The main reason a transistor instead of a resistor is used is that its current limit can be made relatively insensitive to the supply voltage (constant-current circuit).

#### Ian0

Joined Aug 7, 2020
1,101
The resistor has to be the most efficient as @BobTPH said. A transistor may regulate the current more accurately, but it has to be less efficient, because of the base current.
For a 100mA output a switched-mode circuit would be more efficient, but at 10mA it would be a challenge!

#### k1ng 1337

Joined Sep 11, 2020
23
Is this homework or accidentally posted here?
It was moved, I posted in power electronics and I am not taking any electronic courses in school..

The circuit I'm looking at is an npn common emitter with a single LED as a load with no series resistor

If I bias the base correctly I'll get 10ma current thru the LED, so I assume the losses are current X the voltage drop across the base resister + EB junction?

I understand that a SMPS is inherently better at larger currents but I am considering 5v input, with an LED forward voltage of 3v using an 2n3904.

My root question is, say my led has 10ma current and 3v across it, what happens to the remaining 2v? Is it dissipated as heat similar to a resistor in series with an LED or is it simply "unused".

I'll post a schematic later with my exact parameters.. maybe I'm missing something here

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#### Ian0

Joined Aug 7, 2020
1,101
I assume the losses are current X the voltage drop across the base resister + EB junction?
Plus the Collector-emitter voltage drop x the LED current. The collector-emitter voltage drop must be the difference between the power supply voltage and the LED voltage.

Good luck with that base bias circuit - you'll need it!
(Your best option is to put a resistor between emitter and negative supply, then use another transistor to divert current away from the base when the voltage across the new resistor is high enough to start to turn the second transistor on)

#### crutschow

Joined Mar 14, 2008
25,982
My root question is, say my led has 10ma current and 3v across it, what happens to the remaining 2v? Is it dissipated as heat similar to a resistor in series with an LED or is it simply "unused".
Whether you use a resistor or a transistor to limit the current, the power from the 2V times the current is dissipated as heat.
The transistor looks like a resistor.
There's no such thing as "unused" power.

#### BobTPH

Joined Jun 5, 2013
2,721
It is dissipated as heat, since the full current goes from the battery at all time.

The circuit you are describing is no better than a resistor in any way, and likely worse in both dissipation and current stability. To acieve better current control, you need feedback.

A transistor is more sensitive to temperature than a resistor.

Bob

#### k1ng 1337

Joined Sep 11, 2020
23
It is dissipated as heat, since the full current goes from the battery at all time.

The circuit you are describing is no better than a resistor in any way, and likely worse in both dissipation and current stability. To acieve better current control, you need feedback.

A transistor is more sensitive to temperature than a resistor.

Bob
I realized this after I posted once I considered KVL and KCL

I built a buck converter a few weeks ago and quickly realized the usefulness of SMPS, I achieved about 80% efficiency in my first attempt but it was much better than using a series resistor when compared.

Are there any other tricks like a SMPS that I should look into primarily concerning efficiency from a battery pack?

Thanks everyone

#### crutschow

Joined Mar 14, 2008
25,982
Are there any other tricks like a SMPS that I should look into primarily concerning efficiency from a battery pack?
Many forms of PWM using a transistor switch and an inductor should give higher efficiency than a linear approach such as a resistor or transistor.

#### Ian0

Joined Aug 7, 2020
1,101
There are probably dozens of SMPS ICs designed specially to run LEDs at the highest possible efficiencies.