What is the minimum input voltage needed to turn on an NPN transistor?

[TsAmE]

Homework Statement

What is the minimum input voltage for which the transistor will begin to turn on?

Homework Equations

None.

The Attempt at a Solution

I don't understand why the answer was 0.7V. I understand that yes by the time you reach the base there will be 0.7V left which will get dropped across the diode, but wouldn't Vin = 0.7 + voltage dropped across 100k?

 

[TsAmE]

Forgot to add the diagram

 

[berkeman]

Homework Statement

What is the minimum input voltage for which the transistor will begin to turn on?

Homework Equations

None.

The Attempt at a Solution

I don't understand why the answer was 0.7V. I understand that yes by the time you reach the base there will be 0.7V left which will get dropped across the diode, but wouldn't Vin = 0.7 + voltage dropped across 100k?

That's a very good question. They must be defining "Vin" as Vbe, not including the 100k resistor. But you are correct, you could just as easily call Vin the input voltage to the 100k resistor, so it would include Vbe and Vrb.

BTW, you also know that the transistor's Vbe is temperature dependent, right? So it's not generally just 0.7V or 0.6V, although for many problems we just assume a set Vbe value.

 

[vk6kro]

It is in the wording of the question, too.

for which the transistor will begin to turn on

This means before there is any current flowing in the resistor and therefore before there is any voltage across it. Once current begins to flow, then the resistor limits the amount of current, but before then, the resistor has no effect.

In this diagram (below), which I produced yesterday from LTSpice for another thread, we are talking about the region near the origin (550mV, 0 µA).
The drop across a 100 K resistor with even 1 µA flowing is 0.1 Volt and you can see that the current at 580 mV is a lot less than 1 µA.

[PLAIN]http://dl.dropbox.com/u/4222062/Base%20emitter%20curve.PNG

Base current of a 2N2222 NPN transistor vs base-emitter voltage

 

[berkeman]

Good point, vk6kro!