# Quick Question About Transistor Voltage Drop

Discussion in 'Homework Help' started by djwinger, Mar 12, 2012.

1. ### djwinger Thread Starter New Member

Jan 3, 2012
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In the above image, I was given everything in the diagram except for the two voltage drops I drew in red. All answers are correct, so there's no need to work out anything there.

What I'm having difficulty with is, to work out the voltages for Vrb and Vrc, I had to look up the answers to find out that Ve and Vce had no voltage dropped across them.

However, in the following Voltage Divider Bias example, both Ve and Vce have voltage drops.

So in the first image, how am I supposed to know straight away that there is no voltage drop for these values?

At first, I thought that since there was no component between the leg of the emitter and ground (Ve), there would be no voltage drop - fair enough; but why is the a voltage drop on Vc in that case? There is no component between the collector and Rc.

What confuses me most though, is understanding why there is no voltage drop between the collector and emitter (Vce), yet in the voltage divider example there is.

2. ### mlog Member

Feb 11, 2012
276
36
Wait a second. In the first figure, you say that Vce has no voltage drop. Do you mean that Vce=0? If so, then that's not right.

Look at the current, Ic. The Ic will produce a voltage drop (using Ohm's Law) across Rc. Since you know Vcc, you should be able to calculate Vc. Then you should be able to find Vce. What is it?

As for your question why does Ve have no voltage drop, do you mean why is Ve=0? I'm not sure if your confusing the term "voltage drop" with the more general term "voltage." Voltage is measured with respect to some reference, e.g. ground as a reference. Voltage drop is measured between two points, usually across a component like a resistor. Voltage drops are usually the result of a current through a resistance.

By the way, what purpose do you think the 560 Ω resistor has in the second figure?

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3. ### djwinger Thread Starter New Member

Jan 3, 2012
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This is where I was going wrong. I forgot that Vc was tied to ground...for some reason I was visualizing it as a drop between the collector and Rc. For this reason, I thought Vce would have to be 0V to satisfy the voltage loop (KVL). That's that sorted, thanks!

In relation to your question about the 560Ω resistor, I'm not too sure. I know it was specifically chosen to be 10 times smaller than the resistor to the left of it, to achieve some sort of rule, but I haven't got that far in the notes yet. Or did you mean something else by this question?

4. ### djwinger Thread Starter New Member

Jan 3, 2012
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Would I be right in assuming, since Vc=8.5V and Ve=0V; Vce=8.5V?

If Ve happened to be 1.5V (just for example) then Vce would have been 7V. Is that the correct way of working this out?

5. ### Audioguru AAC Fanatic!

Dec 20, 2007
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The first circuit is never ever used because the conduction of the transistor depends on the hFE of the transistor which is different for each one (even if they have the same part number) and the hFE changes with temperature, with the phase of the moon and with luck. Oh, maybe this horrible circuit is used in school.

The second circuit is normal and works with a transistor with minimum hFE or with maximum hFE and with a hot or cold transistor.

6. ### djwinger Thread Starter New Member

Jan 3, 2012
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Yeah this was the first circuit in the notes which was just used to introduce transistors, without making the schematic too confusing. We don't use this one for any lab tests. Only Voltage Divider, Base and Emitter Bias circuits for Year 1.

Good to know though, thanks!

7. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Voltage between collector and ground is equal to
Vc - Vgnd = Vc - 0V = Vc

And yes if emitter is connect to GND then Vc = Vce
But if not then Vce = Vc - Ve.

Thanks to this rule we can ignore the base current in our calculation.
Notice that base current load our voltage divider. So VB is not equal to
Vb = Vcc * R2/(R1 + R2)

So the base current Ib is equal to

$Ib=\frac{R2*Vcc-Vbe*(R1+R2)}{RE*(\beta+1)*(R1+R2)+R1*R2}=48\mu A$

$Vb=\frac{R2*(R1*Vbe+RE*Vcc*(\beta +1))}{RE*R2*(\beta+1)+R1*(R2+RE*(\beta+1))}=3.41V
$

And Ic = 100 * 48uA = 4.8mA; Ie = 101 * 48uA = 4.848mA

Last edited: Mar 13, 2012
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8. ### djwinger Thread Starter New Member

Jan 3, 2012
24
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Jony, thanks for taking the time to reply. I don't understand your equations though. Are you calculating "alpha" by any chance? We are working with beta for the moment.

Ib=Ic/β => 5.16mA/100 = 51.6μA

Vb is voltage divider rule (equation given on the sheet).

I was also just reading up this section about "the rule". It says that when (β*Re) > (10*R2), then Ie does not depend on β. Why do you say this rule allows us to ignore the base current?

Last edited: Mar 13, 2012
9. ### Audioguru AAC Fanatic!

Dec 20, 2007
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10. ### djwinger Thread Starter New Member

Jan 3, 2012
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Apologies, I was just reading it from my scribbles on the notes...I must have stuck the dot in the wrong place. What a noob mistake eh?

11. ### Audioguru AAC Fanatic!

Dec 20, 2007
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It doesn't matter because you might never see a transistor with a beta of 100. The beta might be 33, 300 or anything in between.

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12. ### jtrent New Member

Mar 11, 2012
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4
Ve does not refer to a voltage drop. It refers to the voltage at the emitter with respect to ground. Since the emitter is connected to ground then that voltage is 0 volts.

Vce is a voltage that you can calculate. This refers to the voltage at the collector with respect to the emitter therefore we use ce, collector to emitter. Read my full comments here.

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