Basic electronics: Transistor switch

Question

I am new to electronics. I am reading this book called Practical Electronics for Inventors.

This is a portion about bipolar transistors from the book:

There are two things I don't get from this:

1. When the switch is thrown to the "on" position, I think \$I_{B}\$ should be:

$$I_B = I_{R_1} = \frac{V_{cc} - V_B}{R_1} = \frac{V_{cc} - 0.6}{R_1}$$

Supposing \$V_{BE}\$ = 0.6V. Why does the author have \$I_B = 0.6/R_1\$ ?

2. When the switch is thrown to the "off" position, why do we need R2? How about this:

If the author has to have R2, I still have a question: How does the current can goes from the collector to the base to R2 to ground? Look at this:

I think the current would have trouble if it wants to follow from collector to base because there is a NP junction between the collector and the base. How does the current flow when the switch is "off" with R2 in the circuit?

Answer 1

1. You are correct. The book is wrong, strike one.

2. Again, you make a good point - current can't flow out of the base (OK, it can, but it's picoamps, nanoamps at worst, just leakage current). However, while it's unlikely that leaving a bipolar transistor's base unconnected will cause any great current to flow (either into the base, or from collector to emitter), it is possible. The base is a relatively high impedance, and it's easy to couple current into it capacitively (from your own body in its vicinity), or electromagnetically (like an antenna). If the load being driven were less heavy, such as an LED, you would probably be able switch the LED on by just touching the base with your finger, injecting current into the base (thereby also raising its potential), current which is coupled capacitively to your body from the mains wiring around you. I've had fun making touch sensors this way. The author uses R2 to prevent that from happening, but actually that's an extra, unnecessary component. Connecting the base directly to ground would have the same effect, holding the base potential at zero. I think that's strike two for this book.

3. I know there's no question three, but I'm going to answer it anyway. The author states that the relationship Ic = hfe × Ib holds true unless Vc drops below 0.6V above Ve. This is also incorrect. It holds true right up until saturation, at which point the collector voltage is only 0.2V higher than the emitter (or somewhere in that vicinity, depending on the transistor - some transistors are better than others in this respect). That's strike three. It is most definitely out.

Answer 2

For the point 1) the current loop is VCC, R1, and the VBE junction which is more-or-less a 0.6V diode for most practical purposes.

So you have VCC=I*R1+VBE and the book has a serious mistake…

For the second point, you current diagram is wrong: you can't pull collector current from the base (in the usual biasing mode!). When you switch an NPN transistor you put current thru the VBE junction and it gives collector gain.

Turning it off empties the VBE junction from whatever charge remains and stop the collector flow. In practice with an open base you would only have the collector cut-off current (about 100nA, usually trascurable) but from parasitics some current could leak into the base and turn on the transistor.

If you actually switched the transistor with a mechanical switch like in figure you could ground the base without issues. In practice using a throw switch is not useful and the 'usual' transistor setup is like this

^{simulate this circuit – Schematic created using CircuitLab}

(resistor values and voltages are not correct, it's just to show the shape of the circuit)

This is usually known as a common emitter switch. Common emitter since the control current flows from base to emitter (base current) and the load current from collector to emitter (collector current).

In this circuit you can see why the base-to-ground resistor should be somewhat big: otherwise all the control current would go thru it!

The calculation of the two resistor values is called biasing the transistor: you would substantially want some mA of base current when the switch is closed and less than 0.6V of VBE with the switch open. Hopefully your book will explain that a couple of pages later.

Answer 3

1. Yes, you are correct, the book has an error.

2. Taking into account leakage, the collector current Ic = \$\beta I_B +(\beta+1)I_{CBO}\$ Adding the resistor in the absence of Ib reduces the collector current to \$I_{CBO}\$. The resistor needs to be low enough that the highest vaLue of \$I_{CBO}\$ only presents a few hundred mV at most at the base. A short is an acceptable value, though you may prefer a resistor in some cases when possible failure modes are considered (beyond the scope of this answer).

The difference between \$I_{CBO}\$ and \$(\beta+1)I_{CBO}\$ may not be all that great under benign conditions (room temperature) since \$\beta\$ is very low at nA collector current (it's not a constant as the simplified view holds- it decreases at both high and low collector currents).

Especially at high junction temperatures (or with leaky transistors such as the old germanium types) \$I_{CBO}\$ is exponentially higher and the total collector leakage can rise to objectionable levels. For example, an LED might appear visible illuminated with only a few uA of current. Or a battery supply could be drained prematurely in standby state.

Answer 4

1. It should be a typo. I = (Vdd - 0.6)/R1.

2. If we don't have R2, the base could be floating. We need something to pull down. There can be some leakage current happening in the BJT.