Understanding Transistor from the book "The art of electronics'

[khatus]

Hello guys I don't understand the meaning of a few lines here.
It is stated here that the base voltage Goes up to 0.6 volts when the switch is closed.
My question is how it is possible? Since we are not applying any voltage source across the base-emitter junction. So how it rises up to 0.6V??

[Benta]

No voltage source? What would you call the +10 V then?

[VooDust]

Yes we are? You are connecting Base to the 10V source by closing the switch. The voltage source is not drawn in symbol form, it is just implied by the "10V" label.

[Bud]

As indicated in the text, a base-emitter junction represents a diode in forward conduction, and when the switch is closed that diode becomes connected via the resistor in series to +10V voltage source. A silicon diode in such circuit will have ~0.6V voltage drop across it.

[Ian.M]

There's nothing 'magic' about 0.6V.  Its just an approximation for the typical forward voltage of the base-emitter diode junction (V_BE) for a small signal silicon BJT.  If you were working with very low currents, 0.5V V_BE might be more accurate and if you were switching high currents you'd probably see 0.7V or even more V_BE.   In other texts and as a rule of thumb, I believe 0.7V is (or at least used to be) more common.

* When the B-E junction is forward biased, V_BE is actually an inverse exponential function of (i.e. logarithmically proportional to) the  current into (for NPN), or out of (for PNP) the base.  AoE goes into this in section 2.3.

Edit: See VooDust's comments below.

[VooDust]

V_BE is actually inverse exponentially proportional to the  current

Eh, shouldn't that be logarithmically proportional? Because both dimensions are rising, and current rises exponentially with voltage, hence V_BE is logarithmical to current. "inversely" would imply one dimension diminishing, no?

[Ian.M]

Possibly, but the inverse function of an exponential *IS* a logarithmic function.  However unadorned 'the inverse of' is often colloquially used to refer to the inverse multiplicative function, and when used in the context of a number is commonly used for the reciprocal, so your query/clarification is justified.

[Ground_Loop]

Stick with response #3. It only gets more unnecessarily complicated from here.

[LoveLaika]

There's nothing 'magic' about 0.6V.  Its just an approximation for the typical forward voltage of the base-emitter diode junction (V_BE) for a small signal silicon BJT.  If you were working with very low currents, 0.5V V_BE might be more accurate and if you were switching high currents you'd probably see 0.7V or even more V_BE.   In other texts and as a rule of thumb, I believe 0.7V is (or at least used to be) more common.

* When the B-E junction is forward biased, V_BE is actually an inverse exponential function of (i.e. logarithmically proportional to) the  current into (for NPN), or out of (for PNP) the base.  AoE goes into this in section 2.3.

Edit: See VooDust's comments below.

If I may, it's been a while since I worked with transistors, so I wanted to ask. The base is forward-active, so the transistor conducts current, but the LED limits the current going through the collector? I'm a bit confused as to how the LED affects the current through the transistor. My initial thought was that the LED would limit the current to 100 mA, as that's what it's rated for. However, because of the BJT, there isn't a 10-volt drop across the LED (unless you live in an ideal world; might make more sense if you're using an NMOS). It took me a while to realize that, and then (just thinking it through), I would apply the constant-voltage drop model to solve it numerically. With that, and refreshing myself on how it works, the current through the collector is dependent on the base current and not on the LED. The 10-volt across the LED is just to ensure that the LED is biased to allow for current to flow. Is this correct?


https://inst.eecs.berkeley.edu/~ee105/fa14/lectures/Lecture10-BJT%20Physics.pdf

[TimFox]

The collector load in the textbook diagram is an incandescent lamp (resistor), not an LED.  Although the lamp resistance is not constant (positive temperature co-efficient), it will limit the current into an almost short-circuit (saturated C-E of the transistor).

[VooDust]

I don't see an LED in the top post. Do you mean the lamp?

[Benta]

Today, everything is LED   

[TimFox]

Of course, packaged LED indicators often include a series resistor to operate from a specified voltage, but the symbol in the schematic is the traditional incandescent bulb.  Applying a voltage from a low-resistance source to a roughly constant-voltage load (LED, forward-biased diode, or Zener diode) results in a destructive current through the load.

[LoveLaika]

Yeah, sorry. I just automatically assumed it was an LED.

[LoveLaika]

The collector load in the textbook diagram is an incandescent lamp (resistor), not an LED.  Although the lamp resistance is not constant (positive temperature co-efficient), it will limit the current into an almost short-circuit (saturated C-E of the transistor).

Thanks. I just assumed it was an LED. Reading the text, I guess I made an assumption of what operating region the transistor was in. That's something I always had trouble with. I just assumed the forward active region rather than saturation. When the transistor is on, because there's no current limiting resistor at the collector or the emitter, the transistor will want to pull as much current as possible. The only way it can do that is by having the collector voltage be as close to the emitter (ground) as much as possible. Ideally, the collector would reach ground (assuming a very ideal switch behavior), but it doesn't; we don't know the voltage of the collector, but we can assume that based on its assumed behavior, the collector voltage is low enough to make the BJT go into saturation. Is this correct?

This sort of behavior looks like it would blow up the lamp, though 100 mA doesn't seem like much.

[TimFox]

The saturated collector will reach a voltage (with respect to ground) much less than the 10 V supply voltage, so there will be almost 10 V across the lamp.  If it is rated for 10 V @ 0.1 A = 1 W, the bulb will be OK, except possibly for a short time when cold, since the cold resistance will be less than the hot resistance of 100 ohms.  However, that is no worse than applying 10 V through a mechanical switch to the cold lamp.

[rstofer]

The collector load in the textbook diagram is an incandescent lamp (resistor), not an LED.  Although the lamp resistance is not constant (positive temperature co-efficient), it will limit the current into an almost short-circuit (saturated C-E of the transistor).

Thanks. I just assumed it was an LED. Reading the text, I guess I made an assumption of what operating region the transistor was in. That's something I always had trouble with. I just assumed the forward active region rather than saturation. When the transistor is on, because there's no current limiting resistor at the collector or the emitter, the transistor will want to pull as much current as possible. The only way it can do that is by having the collector voltage be as close to the emitter (ground) as much as possible. Ideally, the collector would reach ground (assuming a very ideal switch behavior), but it doesn't; we don't know the voltage of the collector, but we can assume that based on its assumed behavior, the collector voltage is low enough to make the BJT go into saturation. Is this correct?

This sort of behavior looks like it would blow up the lamp, though 100 mA doesn't seem like much.

Very probably the base current is high enough to drive the transistor into saturation.  You can get the V_ceSat value from the datasheet but it usually on the order of 0.2V.  The same as V_be will be about 0.7V.  If you don't get a low V_ceSat, you might not have enough base current.  For most switches, I assume an H_FE of 10 so if I need 1A in the collector, I design to 0.1A of base current.  I certainly don't assume an H_FE of 100 or 200.

This is what experiments are made for.  Come up with a reasonable load for an appropriate transistor and measure I_b when you get V_ce into saturation.  Measure the collector current at the same time.  This is one of those cases where 3 meters come in handy.  I_b, V_ce and I_c.  V_be might be interesting as well.  Need another meter...  You don't seriously need 4 meters but it sure does help to match up all the numbers at the same time.

Look at page 2 of the 2N2222A datasheet and reflect on what is happening to V_ceSat when the collector current changes from 150 mA to 500 mA.  Note also that both values have the base current as 1/10th of the collector current.

Look at the base-emitter voltage (nominally 0.7V as I said above) is actually 0.2V to 0.6V at 150 mA collector current but as high as 2.0V at 500 mA.

There's a reason that all those numbers are given.  They define specifically where the 'corners' are when designing the device into a circuit.

https://www.onsemi.com/pdf/datasheet/p2n2222a-d.pdf

[TimFox]

The 1 k resistor from +10 V will supply 9.3 mA into the base (at +0.7 V) so with a conservative h_FE = 11, the collector current should achieve the 100 mA lamp current.