using NPN as a switch - question

Hi,
If NPN is used as a switch, but the configuration is its emitter leg is not connected to gnd, but to other resistor, and then to ground (i mean it's not directly connected to gnd, but goes elswhere),
can I still assume the regular math for making it work like a switch?
Check the following picture.
If the emitter is connected to gnd, there's a voltage drop of 0.6v upon it. But if it doesn't, what is the voltage drop then? VBE is still 0.6v, but I need to find Vb and not Vbe in order to calculate Ibase.
How (or is it possible) to make the same math for calculating the Ibase needed for saturation in this case?
Thanks,
Tom

CHEECK THIS LINK: **broken link removed**

When there is a resistor connected to the emitter then you can estimate the base current using the following method:

suppose that you provide 5v to the base of the transistor, the Vbe voltage drop will be about 0.7v so this leaves 4.3v over the emitter resistor or base+emitter resistor if you are using both resistors.

for example if you use an emitter resistor 100 ohm then the base current will be 4.3v/100 ohm = 43mA
if you have a base resistor 100 ohm and an emitter resistor 100 ohm then the base current will be 4.3v / (100+100) = 21.5mA

All this is valid with no output load in the collector but note that Ie= Ib+Ic so as you start sinking current from the collector (your load) you will see the base current reducing because the emitter resistor will have to provide the output current too so the calculated current will be shared in the load and base.

Alex