Inverter using BJT

That's the basic idea. However, that 60 ohm resistor probably isn't the greatest idea. It shouldn't hurt anything, but you're pushing 66mA through the base of the BJT for no reason. Less than 1mA will do the job just fine. Try about 10k there. Also, 40mA is probably going to bake a 20mA LED. As LEDs get hot (which it will with almost twice it's rated current going through it), they draw more current. Now, the 100 ohm resistor would most likely keep it from blowing up, you're still going to at least reduce the life the LED for no apparent reason. Try 180-220 there.

Thanks for the suggestion, by the way how did you compute the values of resistor?

Ohm's law. It also helps to know the beta (aka gain or Hfe) of the transistor. This one has a default value of 100, which about right for a standard small signal transistor. What a beta of 100 means is that if you apply a current of X to the base (from base to emitter is more accurate), then the transistor will allow 100 times more current than X from collector to emitter. So if you have 1mA at the base, the transistor will flow 100mA from collector to emitter. We can do the same in reverse. If we're going to turn on a 20mA LED, we'll want to allow 20mA from collector to emitter. So we only need to apply .2mA (or 200uA) to the base for that to happen. Now, beta actually varies a little during operation. The amount of voltage and current applied will change it, temperature will change it, etc. So when we use a BJT as a switch like this, we commonly apply 5-10 times more current at the base than we need to ensure the transistor stays fully turned on in case the beta starts to drop. That's why I said 1mA, because that's about 5 times more than we need for 20mA through the collector. Now, how did I know what value resistors to use get that current? That's Ohm's law. The base-emitter junction of a BJT has a voltage drop of about .7V. Emitter is grounded in this circuit, so we want .7V and 1mA at the base. We have a supply voltage of 5V. So the resistor needs to drop 4.3V at 1mA. Ohm's law tells us that R=V/I (I is current). R=4.3/0.001, R=4300. So a 4.3k resistor would give us that 1mA at the base. Now let's see what current that 10k will give us. I=V/R I=4.3/10000 I=.000043, or about 430uA. About twice of what we need. That's a little on the low side. However beta shouldn't change because the supply voltage is not changing, and less than 1/8W shouldn't create much heat, we should be fine. You could always go with a 4.7k or even 2.2k resistor to be safe. If you're putting more than about 10 times the current you need to the base though, you're wasting power and creating heat for no reason. Again, you use ohms law to determine the value for the current limiting resistor in series with the LED. It's a 1V 20mA LED, with a 5V supply. We need to drop 4V at 20mA. That will require 200 ohms. 180 and 220 are standard resistor values near that.

Thanks