# Transistor current direction clarifications.

Discussion in 'General Electronics Chat' started by Thecomedian, Nov 18, 2013.

1. ### Thecomedian Thread Starter New Member

Oct 12, 2013
22
3
Ok so in voltage divider biasing of base of NPN BJT transistor, the voltage divider determines the quiescent voltage. 10k and 10k would put 4.5v across the base. 100 ohm and 100 ohm for R(L) and R(E).

Doesn't current flow from the bias resistors at the base to ground through the emitter? How come the voltage is still 4.5v (-the voltage drop)? Wouldn't the emitter be practically a short circuit compared to the bottom resistor in the divider/bias section?

Is there any voltage divider circuit that works without an emitter resistor causing negative feedback?

2. ### wayneh Expert

Sep 9, 2010
15,214
5,560
Yes.
It definitely is not.
Yes.

3. ### MikeML AAC Fanatic!

Oct 2, 2009
5,445
1,073
Look at the attached. Note that V(b) of 3.9V is only slightly lower than the expected unloaded voltage of 4.5V. The base current flows out of the tap between R3 and R4, and pulls the voltage a bit lower.

Note that this is not a very good bias circuit, because V(c) is a bit high.

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4. ### Thecomedian Thread Starter New Member

Oct 12, 2013
22
3
yeah, thanks. this is what I saw, too. What is that readout chart in LTspice, could you tell me how to get that to show?

edit: nevermind, I saw the .op thing and figured it might have something to do with selecting a different procedure. thanks for helping me learn about that awesome tool!

there is also still a large amount of current flow in R2, which seems to mean that only a bit of flow goes from R1 to transistor to R4, which doesnt make sense if R4 is like a short circuit in comparison to R2. Is there some principle of transistors or diodes that it allows such voltage dividers to work, even if the discrete components around the transistor should mean that R2 receives no current, and node R1-R2-Base should be almost no volts? As for the voltage drop, its 3.9v which is perfect if it is assumed there is 0.6v drop for the junction b-e, but otherwise the voltage divider controls the original voltage. That's why it doesn't make sense if the R(E) is parallel to R2 in the voltage divider network.

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Last edited: Nov 18, 2013
5. ### crutschow Expert

Mar 14, 2008
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Note that the base current is reduced by the beta gain of the transistor so the emitter resistor value basically looks like β*R4 from the standpoint of the base bias circuit. That's why the low value of R4 only has a slight effect on the bias voltage.

6. ### Thecomedian Thread Starter New Member

Oct 12, 2013
22
3
I never thought of it that way. It sounded from what I read about biasing transistors that β is an amplification of whatever base current is applied, not base current is reduced by 1/β from the max potential collector current.

7. ### LvW Well-Known Member

Jun 13, 2013
761
103
The base current is reduced ??

Thecomedian, I am afraid you misunderstood something.
* There is no reduction of the base current by a factor of 1/beta.
* And another misunderstanding could be caused by crutschow´s assertion that "the low value of R4 only has a slight effect on the bias voltage"

I don`t understand this sentence. In contrast, the resistor R4 has a major influence on the bias voltage Vbe. The voltage drop across R4 caused by the emitter current (according to the values as given in post#3) is

Ve=Ie*R4=31.9mA*100ohms=3.19 volts.

This gives a bias voltage of Vbe=Vb-Ve=3.94-3.19=0.75 volts.
That means, the value of R4 has a major influence on the bias voltage - and this is, of course, a desired effect. More than that, it is the only reason for using such an emitter resistor R4 because it provides negative feedback and stabilizes the operating point against temperature changes and parts tolerances.

8. ### crutschow Expert

Mar 14, 2008
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β is indeed the amplification factor between the collector current and the base current. My point was that, since the emitter current is equal to the collector current plus the base current, then the emitter resistance is multiplied by β as reflected into the base. That is why the emitter resistance has a much smaller effect on the base bias voltage then if there were no gain. (Note that from the base perspective, the emitter resistor is acting as an emitter follower load).

Last edited: Nov 19, 2013
9. ### crutschow Expert

Mar 14, 2008
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You are putting the cart before the horse. The emitter bias voltage is determined by the base bias, not vice versa (assuming there is an emitter resistor). Thus the value of R4 will affect the collector bias current but has a much smaller effect on the base bias voltage from base current loading as determined by the transistor gain (for reasonably low base bias resistor values). Of course the value of the emitter resistor has to be such as to provide the desired collector bias current with the given base bias voltage.

10. ### LvW Well-Known Member

Jun 13, 2013
761
103
I must confess I don`t understand the meaning and the contents of your reply. May be it is - again - only a linguistic problem.

*I think, everybody will agree that Vbe=Vb-Ve.
Having this in mind, how can you say that "the low value of R4 only has a slight effect on the bias voltage" ? As you know, Ve is developped across R4 !

* And what is the meaning of "The value of R4 ...has a much smaller effect on the base bias voltage from base current loading as determined by the transistor gain" ?

Without emitter resistor (R4=0) the base voltage (against ground) is, of course, much smaller than R4=100 ohms. What are you referring to?
What are you trying to explain? Please clarify.

11. ### WBahn Moderator

Mar 31, 2012
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Perhaps this will help clear up the confusion.

Remove the transistor and just place a resistor in parallel with the bottom 10kΩ resistor. What value would that resistor have to be in order to result in a voltage at the junction of the resistors of 3.9V? It works out to 32.5kΩ.

With the transistor and emitter resistor (and a collector resistor sized so that the transistor is in the active region) instead, what value does the emitter resistor need to be in order to result in the same 3.9V? From MikeML's simulation, it is 100Ω. Notice that the ratio of these two resistances is 325, which is essentially the β of the transistor.

Another way of saying it is that, looking into the base of the transistor, the Thevenin equivalent circuit is essentially a voltage source equal to Vbe in series with a resistor that is β times the emitter resistor.

12. ### crutschow Expert

Mar 14, 2008
20,494
5,803
True. But Vbe does not vary much. Thus the equation that applies is Ve = Vb-Vbe. The emitter voltage is then, for reasonable emitter resistor values, determined by the base voltage and not the emitter resistor, since the transistor is in the active region and acting as an emitter follower. For example, changing the emitter resistance from 100Ω to 150Ω will have only a small effect on the emitter voltage but a much larger effect on the emitter current.

The bias resistor network voltage is affected by the loading of the transistor base current. This current equals the emitter current divided by β. So a change in the emitter current changes the base current by a proportional amount but reduced by the value of the transistor gain.

Of course, in the limit, you can't vary the emitter resistance to zero. It has to be some value that gives the desired value of emitter current with the selected base bias voltage.

13. ### LvW Well-Known Member

Jun 13, 2013
761
103
Hi crutschow, thanks for clarification.
Now I know what you mean. Of course, as long as the BJT is in its active region, the voltage Vbe is approximately 0.65...0.7 volts - independent on the actual value of the emitter resistor. ***
I think, the misunderstanding was caused by the word "bias".
This can be (a) the voltage (against ground) developed by the resistive divider at the base node B or (b) the voltage developed between the nodes B and E.

***EDIT: Of course, this fact is obvious because during design and analysis of such an emitter stage we always assume a fixed value for Vbe in the region 0.65...0.7 volts (as we also did in the present example).

Last edited: Nov 20, 2013