# Struggling to understand transistors

Discussion in 'General Electronics Chat' started by tjohnson, Mar 23, 2015.

1. ### tjohnson Thread Starter Active Member

Dec 23, 2014
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Recently I learned about NPN transistors in my physics class, and am having difficulty understanding them. I read about them in the AAC textbook, but some things still don't make sense to me very well.

If a transistor is used as a switch, is it true that there is an inverse relationship between the power going into the base leg and the power coming out of the emitter leg? An LDR has resistance that is inversely proportional to light, so it has the opposite of the behavior desired for a light that should only go on at night (like a streetlight). Would wiring it to the base leg of a transistor cause it to produce the desired behavior (so that the transistor would emit the most power when it is dark and the LDR is producing the most resistance)?

2. ### blocco a spirale AAC Fanatic!

Jun 18, 2008
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It is better to talk in terms of current rather than power in this case. The emitter current is equal to base current multiplied by hFE,

Ib x hFE = Ic = Ie

Remember this simple relationship and that Vbe = 0.65V and you will be able to understand the majority of transistor circuits, it's that simple.

Last edited: Mar 23, 2015
3. ### Papabravo Expert

Feb 24, 2006
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No it is not true. When used as a switch, all notions applicable to operating in the linear mode about a quiescent point go out the window. It is the external circuit that determines the relationship between base current and the other currents.

4. ### ScottWang Moderator

Aug 23, 2012
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Ic = hFE * Ib
Ie = Ic + Ib
Ib x hFE = Ic ≒ Ie

5. ### tjohnson Thread Starter Active Member

Dec 23, 2014
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Where does Vbe fit into that equation? I'm confused.

6. ### Papabravo Expert

Feb 24, 2006
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Vbe shows up in the so called Ebers-Moll equation. It is a necessary condition for the transistor to conduct. It is often referred to as a threshold voltage. Once the threshold is reached it is not possible to substantially increase the voltage across the base emitter junction.

7. ### crutschow Expert

Mar 14, 2008
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Not quite. Ie = Ic + Ib, otherwise Ib would be disappearing into thin air.

8. ### crutschow Expert

Mar 14, 2008
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It doesn't fit in directly.
That equation is used when treating the BJT as a current operated device where the collector current is proportional to the base current.
Vbe used only when you are calculating the base current through any resistor in series with the base.

If one end of an LDR is connected to V+ and the other end is connected to the transistor base, and the emitter is grounded, then the collector current will be maximum when the LDR sees light and off when the LDR is in the dark.

To reverse that behavior you could use a resistor to V+ in series with the LDR to ground and with the junction of those two going to the transistor base.
Then, when the LDR is in the dark and a high resistance, the current from the resistor will go through the base and turn on the transistor.
Otherwise when the LDR is in the light it will shunt the resistor current to ground and the transistor will be off.

9. ### tjohnson Thread Starter Active Member

Dec 23, 2014
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122
I suppose my question really is: What exactly is the relationship between the value of a resistor at the base junction and the current emitted by the transistor?

@crutschow: Thanks, I just mentioned LDRs as an example. I'm not trying to build a circuit with LDRs and transistors, but just to understand how NPN transistors work.

10. ### upand_at_them Active Member

May 15, 2010
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TJ, a key thing to learn is that when a transistor "acts like a switch" it doesn't have the same on/off behavior as a switch. If your intent is to switch something on based on the light level then you need a comparator like the LM311.

With a comparator you establish a trigger point and the output won't switch until then.

A single transistor will be varying degrees of "on" until saturated, and then it is "the most ON it can be".

11. ### crutschow Expert

Mar 14, 2008
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The transistor doesn't really "emit" current even though one terminal is called the emitter, it just controls the current flow from collector to emitter.
As previously noted the collector current in the active region (not saturated) equals the base current Ib x hFE (Beta).
The base current is normally determined by adding a resistor (Rb) in series with the base, connected to a voltage source (Vs).
Then the base current is approximately Ib = (Vs-0.7V) / Rb (with emitter connected to ground), where the typical base-emitter voltage (Vbe) is about 0.7V.
Thus from Ib and hFE you can calculate the collector (to emitter) current.

If you want to use the transistor as a switch then the usual design rule is to make the Ib at least 1/10th of the collector current to insure the transistor is fully saturated with minimum ON (Vce) voltage.

Make sense?

12. ### blocco a spirale AAC Fanatic!

Jun 18, 2008
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I know this but can think of few situations where it is relevant. Hence, for all practical purposes and for the sake of simplicity, Ie=Ic and Ib can almost always be ignored.

And.... I couldn't find the wavey "almost-equal-to" sign

Oh, here it is: Ie≈Ic, wasn't looking in the right place.

Last edited: Mar 23, 2015
13. ### crutschow Expert

Mar 14, 2008
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Maybe in practical applications, but it could be confusing to someone who is trying to understand the currents into and out of a transistor and might wonder what happened to Ib.

14. ### shteii01 AAC Fanatic!

Feb 19, 2010
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Vbe is used in two cases:
1) To determine that trasistor is On or Off.
2) To form KCL and KVL equations.

15. ### tjohnson Thread Starter Active Member

Dec 23, 2014
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Sorry, but I'm still struggling. I'm in over my head with this, since my knowledge of electronics is very basic and I still have a lot to learn, so those formulas don't make much sense to me. Right now, my questions are very fundamental, and I'm even having difficulty understanding articles about transistors written for beginners.

After looking at the AAC textbook some more, however, I think I'm finally beginning to understand the basics of NPN transistors better. Isn't it true that in terms of power (ignoring internal resistance), Base + Collecter = Emitter? Since a transistor used as a switch amplifies current, it must also decrease voltage, or else it would be violating the laws of science by generating power from nowhere. I still don't understand, though, what the exact relationship between the value of a resistor at the base leg and the current emitted by a transistor is.

The reason I started this thread was because I could not answer this question: Why does a larger resistor at the base leg cause an LED on the other side of a transistor to shine brighter? When I learned about this behavior, it surprised me, and led me to conclude that a transistor used in this way must reverse the effect of the resistor, since if the value of a resistor is increased in a circuit without a transistor, the LED would shine less (not more) brightly. If I can get this question answered, then I'll be happy with my knowledge about transistors for now.

Last edited: Mar 23, 2015
16. ### shteii01 AAC Fanatic!

Feb 19, 2010
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Lesson 1.
Forget power.

My advice is to go to the library and get a textbook that deals with diodes and transistors. There you will find description of what physically happens inside the transistor and the math that describes the physical behavior. My textbook was: Electronic Circuit Analysis and Design, Second Edition, by Donald A. Neamen.

tjohnson likes this.

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18. ### ronv AAC Fanatic!

Nov 12, 2008
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I think I know what is bothering you. I'm going to give you an ideal example and see if it helps.
Lets say you want to heat up a 12 ohm resistor that can take 1 amp of current and you have a 12 volt power supply.
You can put the resistor between +12 volts and the collector of the transistor. The emitter goes to ground. So when the transistor is on you have 12 volts across the resistor and 1 amp.
Now to control the transistor you need to apply a voltage to the base. Lets again use 12 volts. But you only need 1/10 the current into the base as you get from the collector. This is gain.
So you could put 120 ohms in series with a switch to +12. When the switch is closed 0.1 amps will flow in the base, (I=E/R) but 1 amp in the collector.

Mar 14, 2008
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