In the below circuit, the circuit maintains a constant current on LED. My question is after the LED there is a voltage of 18V, still, need to be dropped. Will the NPN Transistor heat up while dropping the voltage? Also, the voltage source to led will vary from 12V to 20V. How can we modify the circuit? Or the existing circuit is fine?
Yes, it will work- it's a switched (not switching) crude linear regulator, but Q2 is operating in reverse mode and will regulate somewhat more poorly. It would be worse if your LED current was less since it has to sink about 3mA to throttle the base current of Q1, so that will consume about 1mA of base current in reverse active mode.
The transistor Q1 is dropping about 17.2V at 18.2mA so it will dissipate (ignoring base current dissipation, which is negligible) 313mW which is quite a bit for a TO-92 transistor. Assuming a 50°C maximum ambient and 200°C/W J-A thermal resistance, the junction will be at 112°C which is acceptable but not great for good reliability. If the '20V' was actually 30V it would be much worse, of course.
You can reduce the dissipation in the transistor by adding a resistor in series with the LED (based on the minimum the 20V supply could go such that you want to maintain regulation). For example, if it has to regulate from 15V to 20V you could add a resistor dropping about 5V, so 5/18.2 = 270\$\Omega\$.
If the 20V isn't going to change much you hardly need a regulator at all, and just switching a resistor with a single transistor (which will then run cold) would be better and simpler (and more reliable). For example, a 1K 1/2-W resistor.
It's not a very good regulator because the "reference" is the Vbe of Q2 and it will vary with temperature. In some cases that could be considered a feature but I'll leave that discussion for another time.
Here is an LTspice simulation, however the temperature curves apply to the junction temperature of the transistor. As you can see, temperature is much more of an effect than the voltage (ignoring self-heating).
What will happen in fact is that there will be some self heating so the current will start off higher and tail off as Q1 heats Q2.
If Q2 is connected correctly, the LED current will be a bit lower because the base does not need so much current. My simulation (using the LTspice models) shows a bit lower current than yours so the dissipation is proportionally less. The major factor may be how optimistic their number is for reverse beta.
For a little post-prandial edit, I've added plots of Q1 dissipation in relation to the LED supply and temperature (blue lines).
The circut will work and the main transistor will dissipate around 300mW and have a temperature of 55°C.
P = (Vc-Ve)*Ie
300mW = (17V-0.8V)*0.018A
(You could use Ic, the diffrence will be very small since Ie=Ic+Ib and base current is insignificant.)
You should use a transistor able to dissipate more than 0.6W (ideally 1W), such as 2N3904.
The temperature can be calculated this way:
Tj=P*Rthj-amb (thermal resistance)
This is hot to touch but will not damage the transistor, I suggest to put it away from the plastic case and capacitors.
Yes, the NPN transistor will heat up by exactly the same amount as a variable resistor if it were used to set the same LED current.
The circuit shown should work reasonably well down to a few volts but don’t just trust me on this - get a free sim tool and prove it yourself.
Constant current for LED ...the voltage source to led will vary from 12V to 20V... the existing circuit is fine?
No, the circuit as shown DOES NOT WORK
I assume that the circuit should be as shown here:
While you have not specified the devices used, I used a generic 2N3904 NPN here. The 2N3904, Q1 would dissipate about 288mW and it has a maximum rating of 625mW. At 25degC ambient you would expect the device chip to be at about 57degC and very hot to touch.
You could reduce the temperature of Q1 by adding a resistor for the LED, but all you are doing is spreading the dissipation between the resistor and Q1, you still have to dissipate 288mW.
A much better way to do this task is to use a single component such as the AL5809 constant current driver.
This is a single component, works from 2.5 - 60V and is cheap.
Update: It was suggested in comments that the circuit originally shown would work with the transistor in a reverse-active mode.
Maybe ….but the H(fe) would likely be in the single digit range (the 2N3904 is not exactly high gain) so trying to design the solution would be harder over the range of device parameters. See here.
Why anyone would suggest reverse active mode completely flummoxes me. There is no technical reason for choosing an unusual device operation mode for this type of simple application, so why even suggest it?
The feedback device (Q2 in my circuit) is never in Collector saturation, in addition it's never in Base/Emitter saturation, so there are no unusual configuration parameters.
The current does have a negative droop with temperature.