# Common emitter npn Ib and Ic

Discussion in 'General Electronics Chat' started by cmisip, Oct 15, 2017.

1. ### cmisip Thread Starter Member

Sep 23, 2017
89
1
I am trying to understand transistor mechanics.

In a common emitter circuit ( where a resistor is connected to the collector ):
When the base voltage is HIGH, CE resistance is LOW (closed switch) and so C is connected to GND (LOW). When the base voltage is LOW, CE resistance is HIGH (open switch) and so the pull up resistor pulls C to Vcc (HIGH). In the active mode of the transistor, this results in an inverse phase between the input signal and the output signal voltages. I think the currents Ib and Ic would fluctuate in phase with the fluctuations of the voltages in their respective circuit loops. This would mean that when Ib is high, Ic will be low. If this is correct, then the relationship Ic=hfe*Ib does not seam to give a snapshot info of what is happening. That is to say, If I freeze the circuit at any particular time and take readings of Ic and Ib, won't I find that when Ib is up then Ic is down?

Thanks,
Chris

2. ### panic mode Senior Member

Oct 10, 2011
1,633
452
no, Ic is proportional to Ib as the equation shows.

you are not understanding voltage divider principle. i suggest to solve several examples where top resistor is fixed, say 1k. then see what would output voltage be if lower resistor has different value (0.1k, 0.5k, 1k, 5k, 10k). also check how voltage output relates with current through voltage divider.

transistor in your example is replacing that lower resistor of voltage divider

3. ### cmisip Thread Starter Member

Sep 23, 2017
89
1
When the base voltage is HIGH, CE resistance is LOW (closed switch) and so C is connected to GND (LOW). The collector resistor would be in series with a low resistance CE so their total resistance would be low leading to Ic being high. The voltage read at the collector would be low because a lot of the voltage drop has happened on the collector resistor.

When the base voltage is LOW, CE resistance is HIGH (open switch) and so the pull up resistor pulls C to Vcc (HIGH). The collector resistor would be in series with a high resistance CE so their total resistance would be high leading to low Ic. The voltage read at the collector would be high because the collector resistor would be less than the resistance of CE it is in series with causing minimal voltage drop on the resistor and maximal voltage drop on CE.

Is that better?

Thanks,
Chris

4. ### Audioguru Expert

Dec 20, 2007
10,605
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hFE is not used when a transistor is a switch. hFE is used when the transistor always has plenty of collector to emitter voltage so it is a linear amplifier, not a switch. The datasheet for the transistor shows that the base current is usually 1/10th the collector current when the transistor is a switch, even if the hFE is much higher.

5. ### MrChips Moderator

Oct 2, 2009
17,317
5,333
Are you trying to understand why the voltage Vc on the collector is out of phase with the bias voltage on the base VB?

Let's go back to the load line analysis.

We superimpose the load line on top of the I-V characteristic curve of the transistor for fixed values of IB.

The Q-point, or quiescent point, is the DC (or static) condition of the transistor for a given base current IB.
In this example, we are given,

VCC = 12V
RL = 1.2kΩ

ICmax should be 10mA (well, never mind. Let's assume that RL is 1.3kΩ).

If we were to bias the transistor with IB = 46μA, the transistor's Q-point, taken from the graph, is

IC = 4.8mA
VCE = 5.8V

Now let's modulate the Q-point by superimposing a sinewave voltage on the base.
If we increase IB from 50μA to 80μA, IC increases from 4.8mA to 7.7mA.
VCE decreases from 5.8V to 2.0V.
Since the emitter is at GND, the collector voltage is also 5.8V to 2.0V.

Can you see now that the collector voltage is out of phase with the base voltage?
In other words, the common emitter amplifier is a voltage inverter.

Note that all of this analysis is done with no reference to beta.
(from the graph we can see that beta is 100 approx.)

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6. ### dl324 AAC Fanatic!

Mar 30, 2015
7,421
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No.

Stop thinking in terms of HIGH and LOW and resistance between the collector and emitter.

When a transistor is used as a switch, you alternate between cutoff and saturation modes.

7. ### cmisip Thread Starter Member

Sep 23, 2017
89
1
Thanks guys, this is really helpful.
Sorry to make the question confusing with the use of the word switch. I was thinking about amplifier circuitry and forward active mode of transistor operation specifically in the common emitter configuration. The switch analogy was just conceptualizing what is happening at the CE junction but staying in the active region. The way I see it now, the base current sets the impedance of the CE junction and depending on that impedance will be the voltage drop across it. Since the resistor is on the collector side connected to Vcc and it is in series with CE, the current going through both the collector and CE will fluctuate up or down depending on the base current ( as it controls CE impedance). Now the higher the base current, the lower the CE impedance, the higher Ic and Ie will be. Now I can see the relationship of Ib=100*Ic.

In the graph above, an Ibmax of 80 uA corresponds to a Vce of 2 V (Ic of 7.7mA) and an Ibmin of 20 uA to a Vce of 9.34 V (Ic of 2 mA). The hfe for both conditions is approximately 100.

On the CE side of the transistor, Vce fluctuates inversely to Ic. And Ib fluctuates directly with Ic.

On the base side of the transistor, there is not such a "variable resistor" mechanism ( like the one provided by the CE junction) so I would assume the base current will fluctuate in direct proportion to the base voltage.

I have a question though. The graph above I also found from electronics tutorials:

http://www.electronics-tutorials.ws/amplifier/amp_2.html

It is very helpful but I dont understand a couple of things.
1. There is a resistor on the emitter side. Why would this still be a common emitter configuration?

2. If there was no resistor on the emitter side, Vbe would be just .7 volts. This means the input AC signal would subtract its negative phase from the base voltage and hence pull BE into reverse bias, turning off the active mode.

3. R2 provides a parallel path for current to get to GND. The other path is through the base, through BE, to GND. KVL equation was written so that R2 voltage equals Vbe (+ the emitter resistor voltage drop if there is such a resistor). In a loop where there is a battery source, the battery source internal is the "opposite path" to the circuit voltage drops. In this case, it seems I must imagine that R2 is internal to the battery and therefore provides a current opposing Vbe ( or the other way around). After all R2 or the BE junction is between Vbe and GND which I imagine to be the battery terminals. Is that the way to conceptualize KVL in a scenario where there is not a discrete source like a battery?

Thanks,
Chris

8. ### Audioguru Expert

Dec 20, 2007
10,605
1,182
Without the emitter resistor and if the transistor is biased from a 0.7VDC source to the base then some transistors would be saturated and others would be cutoff (because they are all not the same even if they have the same part number). Also a higher temperature would cause some transistors to be saturated. Also the distortion for an audio or video circuit would be awful. You cannot simply bias the base with a fixed current instead of 0.7V because the beta of each transistor is different and it varies with temperature.

The emitter resistor adds DC negative feedback to decrease the bad DC problems and if it does not have a parallel capacitor the resulting AC negative feedback reduces the voltage gain and reduces the distortion.

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9. ### cmisip Thread Starter Member

Sep 23, 2017
89
1
So the emitter resistor raises Vb so that it is above the threshold that is minimally required to turn on the diode of the BE junction. If the BE junction happens to have a forward voltage of .9v for example, if Vb was only .7v, then the BE junction will be nonconducting. If the Ve is at 1 volt, then that 1 volt needs to be maintained and it will be common to two loops (BE loop and CE loop). There will be three resistances in the CE side and the middle resistance will have a fluctuating voltage drop between 2 V to 9.34 V and the other two would have to proportionately split the remainder.

I have to read up on the negative feedback thing.

Thanks,
Chris

10. ### MrChips Moderator

Oct 2, 2009
17,317
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Look at it from a different perspective.
You are not trying to maintain VE at 1V. That is a target to determine the Q-point.

In the example given, we will assume

VCC = 12V
R1 = 20kΩ
R2 = 3600Ω
RC = 1200Ω
RE = 220Ω

I am not going to contest the values shown in the diagram but simply give my own analysis.

For VE = 1V
IE = 1V/220Ω = 4.54mA

From this, we can extract the Q-point from the graph shown before.
VB = 0.7V + VE = 1.7V
IB = 46μA

R1 and R2 creates a simple voltage divider to set VB. If we make the current through R1 and R2 much greater than IB then we can ignore IB.

R2/(R1 + R2) = 1.7/12

We have infinite choices that we can make for R1 and R2 with the above constraint. Our choice will determine the input impedance of the amplifier.
You can make a "stiff" voltage divider by selecting low values of R1 and R2 but your voltage gain will suffer. Relax the voltage divider and the circuit becomes sensitive to beta and temperatures. Choose a compromise.

Yes. The purpose of RE is to provide negative feedback which helps to stabilize the circuit. The purpose of CE is to restore the AC gain and hence will influence the frequency response of the circuit.

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Feb 17, 2009
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12. ### Audioguru Expert

Dec 20, 2007
10,605
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Negative feedback reduces the voltage gain so that the collector operating points of transistors with a Vbe of 0.6V and 0.7V are almost the same instead if being amplified so they do not work because one transistor is saturated and the other transistor is cutoff.
The reduction in voltage gain also allows a transistor with an hFE of 100 to perform almost the same as one with an hFE of 300 or more.
The same for temperature changes.
The same for distortion reduction.

An emitter resistor without a bypass capacitor also increases the input impedance of a transistor.

Negative feedback can also be made when the base is biased with a resistor from its collector, but then the input impedance of the transistor is reduced.

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13. ### cmisip Thread Starter Member

Sep 23, 2017
89
1
Trying to grasp this negative feedback thing. So I saw from a circuit simulator that the voltage on the emitter resistor does fluctuate up and down along with the fluctuations of the collector resistor and Vce. I have to guess that their combined voltage drops equal a total of Vcc. So when Vce is down, then the voltage on the collector resistor and emitter resistor are high but distributed according to the value of their resistances. In the setup I saw on falstad, the emitter resistor voltage varied from .5 to 1.5 volts.
The base bias voltage is calculated with static value of 1 volt for the emitter resistor. So when there is no AC input, the voltage at the base is ~1.7 volts. With an AC input, the base voltage would start to fluctuate up and down but so will the emitter resistor voltage. When the voltage drop on the emitter is high, this reduces Vbe (1.7 v -1 v = .7 v ) causing a decrease in base current and hence a reduction of the emitter potential drop due to a reduction of Ie. When the voltage drop on the emitter is low, this increases Vbe (1.7v -.5 v = 1.2) causing an increase in base current and hence an increase in emitter voltage drop due to an increase in Ie. This happens because the base voltage is held steady by the bias resistor ( but it will fluctuate with the AC source on that side of the transistor).
Is this correct?

Thanks,
Chris

14. ### MrChips Moderator

Oct 2, 2009
17,317
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Another way of stating it:

When base current increases, collector current increases and voltage at emitter increases. This reduces base-emitter voltage which reduces base current. The effect is the amplifier gain is lower than what it would be without negative feedback.

15. ### cmisip Thread Starter Member

Sep 23, 2017
89
1
So this is another difference from the transistor switch configuration. In the amplifier case, the voltage drop across BE changes and is not at a fixed .7 v like the switch case. The .7v volt BE drop is only used when figuring out the bias voltage when no AC signal is applied when setting up the quiescent point.

Sorry to digress, but:
In the transistor switch circuit, there is a resistor in front of the base to limit the current and the voltage. Without that resistor, the base current and voltage would be high (high Vbe > .7 v and provided it does not fry the transistor), it would still turn it on ( as long as the collector voltage is less than the base and the base voltage is greater than the emitter). I guess my point is that Vbe can be any value above the minimum to turn it on and that is what we mean when we say the base current determines the collector voltage because what we are manipulating is really the magnitude of the Vbe ( not the absolute value of the the base voltage, but the difference between the base voltage and emitter voltage).

I just read up on common base configuration and it looks just like a common emitter except the input signal is applied on the emitter resistor. I think the same thing would be happening as with the common emitter. Input signal would fluctuate the emitter resistor voltage, which in turn fluctuates the base voltage, which in turn results in fluctuating Ic on the collector.

Is this analysis correct?

Thanks,
Chris

16. ### MrChips Moderator

Oct 2, 2009
17,317
5,333
I would not fixate too much on the base voltage. The base-emitter junction is a P-N junction and like an LED or a zener diode, the current rises rapidly over a very narrow VBE range, i.e. there is a sharp rise in the I-V characteristic curve. Hence VBE = 0.7V holds true for most circuit applications. If you do not limit the base current you will destroy the device, in the same way you would destroy an LED.

Yes, we are modulating VBE but in reality we are modulating IB.

17. ### Audioguru Expert

Dec 20, 2007
10,605
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Usually when you design a transistor circuit you want it to work with any passing transistor of one part number since there are weak ones, typical ones and sensitive ones. The Vbe and other characteristics are shown as minimum, typical and maximum ranges in a datasheet.

When I bias a transistor I look at its datasheet to see what its Vbe is at the temperature and current I want. For a 2N3904 a graph shows the "typical" Vbe when it is saturated and another graph shows when it is ON (as an amplifier) when the collector current is from 0.1mA to 100mA and when it is very cold, at room temperature and very hot. At room temperature the "typical" Vbe goes from 0.6V at 0.1mA to 0.8V at 100mA but some transistors have a higher voltage and others have a lower voltage (a range of voltages with the maximum voltage shown in the text of the datasheet at 0.95V). So I select the "typical" voltage at the temperature and current that my 2N3904 transistor is operating at then I use enough input signal level or add enough negative feedback so that all transistors work well.

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18. ### hobbyist AAC Fanatic!

Aug 10, 2008
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85
You have the right understanding of that, I think where you were getting a little confused about was biasing the base with base voltage, and Vbe inherent in the transistor to turn it on.

Lets do a few examples to show how you can't just make the Vbe voltage at the base any value greater than 0.7V and think it will turn on the transistor no matter what.

Assuming you are using no emitter resistor, to save on parts to build a simple amplifier.

Just as an example of clarification, lets use a low supply of 5V. Now if we decide to put around half the supply voltage at the base terminal, we know we have all the voltage we need to turn on the transistor. (its well above the 700mV) So we should have no problem turning on the transistor.

So we can use 100ohms for the collector load and use the same values for the base voltage divider.
We may get something like this.

We know the b-e voltage is going to drop the voltage across the lower resistor, but look at the base current and collector current, and look at the collector to ground voltage. The transistor is saturated, no good for an amplifier. The excessive base current just puts it out of commission as an amp.

Ok what can we do to make it work as an amp with no emitter resistor.
Here is how its done.

You assume a Vbe of around 700mV.
Now you design your base divider to have 700mV. dropped across the ground resistor, calculate that resistor value, then calculate the resistor value for the pullup supply resistor, to drop the remainder of the supply voltage across it.

Then you breadboard the circuit, (just don't rely on computer simulators) and you check the voltage at the collector from ground ref. and adjust either the ground or pullup resistor to get close to your half supply voltage dropped across the transistor at the collector terminal.

I built the circuit and had to adjust the pullup resistor from 153 ohms to 160 ohms and got around 2.4V at the collector. Then I tried 3 more transistors in place and the voltage varried around 2.2V to 2.7V at the collector.

This is what the base to collector current ratio should be more like.

Do this experiment than read up on how designers will incorporate emitter resistors into there amp circuits, so as to controle gain and parameter independancies.
When you design using emitter resistors you just take your emitter voltage, and add the Vbe to it to bias the base divider.

Hope this helps clarify, that you need to calculate base current by using Vbe assumptions, rather than just make the base voltage large enough above 700mV. to turn on the transistor.

There's a limit to how large the base voltage needs to be for linear operation of the transistor.

I am not negating what you came to in conclusion, I agree with what you said comparing the Vbe to emitter voltage, I'm just sharing this to help clarify in visual example of what takes place under these conditions.

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19. ### Audioguru Expert

Dec 20, 2007
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It is crazy to use such high currents for a little transistor.
The distortion will be extremely high without any negative feedback.
Without an emitter resistor or bias from the collector then the a base resistor value must be adjusted for each transistor (and for each temperature).

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20. ### cmisip Thread Starter Member

Sep 23, 2017
89
1
@hobbyist

I tried the configuration on a circuit simulator ( I know you suggested a bread board but I needed to wrap my head around the concept first ). The way I see it, there is no way to drop the voltage at the collector to 2.617 volts. If I trace a path from the 5 volt supply to ground on the collector side, I only see one resistor to drop the voltage so the voltage drop there must be close to 5 volts leaving only a few millivolts on the collector. We cant satisfy Vc > Vb. The only solution I found was to add an emitter resistor which will increase the collector voltage or a base resistor which will reduce the base voltage.

If I disconnect the collector from the 5v source, then tracing from Vcc to the first resistor to the base of the transistor yields the base voltage of ~.7 volts as 4.3 will drop on the first resistor and the BE drop is .7 volts. Its when I connect the collector loop that I see the transistor going into saturation onstead of forward active.

I am finding out I hope correctly that the way to quickly figure out voltages via KVL is to simply expect that whatever Vcc we apply, it should always drop to zero by the time we get to GND, regardless of which path we take.

@MrChips

You said Vbe=.7v holds true for most circuit applications. Do you mean when there is no AC input because I see it fluctuating in the simulator, granted that the fluctuation is minimal ( a few millivolts only ) when there is an AC source. It is however not fluctuating like the emitter resistor voltage which swings from .5 volts to 1.5 volts. Simply adding the emitter resistor voltage to the bias at the base does not yield the base voltage I expected. Could it be that this is because base voltage is pulled down by the emitter voltage when emitter voltage is going down and base voltage is pushed up when the emitter voltage is going up so that the difference is always around .7v? I mean, the opposing forces ( the bias resistors holding the base voltage steady, and the emitter resistor voltage pulling it up and down) find a point of equilibrium at any moment close to .7 volts).

@Audioguru
I am still trying to wrap my head around the concept of using a emitter resistor to ensure that variance in different transistor's Vbe (under different operating temperatures) could be negated with a single value of emitter resistor. But I see on the simulator that increasing the emitter resistor seems to decrease the swings of Vbe so a lower value of emitter resistor results in a wider amplitude of Vbe, hence increased likelihood of meeting the Vbe of different transistors. I just need to find the conceptual explanation for this.

Thanks,
Chris