So the voltage at the base must accomodate the amplitude of the voltage swings and needs to be higher than the negative phase of the input AC by at least .7v or else it goes into cufoff.

With the simulator, I see the AC signal fluctuating above and below zero, but with the input capacitor, the AC signal shifted up. I dont unnderstand why.

Thanks,
Chris

 

"
So it seems the base voltage should be higher than just the sum of the emitter resistor voltage and Vbe.
You lost me here. ???"

Sorry, dont know how the quoting system works.

It may depend on whether you're viewing AAC in "blue mode" or "orange mode", and whether you're using the WYSIWYG editor or the other one; I've always used blue mode and the WYSIWYG editor, so YMMV.

There are several ways to do quoting. One is simply to hit the "Reply" button at the bottom of a post, which will cause the text of the post to appear in your edit window between quote and end-quote tags. You can then trim down the text to only that part(s) you want to reply to, or leave the whole thing intact.

The other way is to select the portion of text you want to reply to by click-and-drag; when you release the mouse button, a little balloon will pop up allowing you to select "+Quote" or "Reply" as shown in the following picture:



If you click on "Reply" in that balloon, the selected text will be copied into your edit window and you can go from there (that's what I did for this post).

 

With the simulator, I see the AC signal fluctuating above and below zero, but with the input capacitor, the AC signal shifted up. I dont unnderstand why.

It shifted up to whatever the voltage would have been if the capacitor were removed-- that is, the voltage established by the bias network.

 

So it seems the base voltage should be higher than just the sum of the emitter resistor voltage and Vbe.

I see where this can be a little confusing, but look at it this way, lets say you have your emitter resistor biased to 1V drop across it, and you design your voltage divider to have from ground to base voltage to be approximately 1.7V, then here is what would happen.

Normal bias no signal.



For demonstration purpose only, lets say you give your amp around 700mV input signal.
And consider the negative going half cycle.
To demonstrate this I just adjusted the resistor value to change base voltage.




Since the entire base circuit from emitter side is 1.7V then taking away the 700mV from the bias due to the neg. 700mV excursion, than that only leaves 1V left which is dropped across the emitter resistor, so the transitor is in cutoff, right?,,,,,Wrong,,,,,

Always think of your transistor base emitter as first in line to absorb the first 600mV-800mV of voltage in the loop, then whatever voltage is left gets taken by the remaining components in the loop.
Why?, because the resistor is a physical component of resistance, it works on a linear scale, the transistor Base Emitter is a nonlinear diode, so it works as a voltage regulator, it regulates any voltage above its threshold and keeps the voltage regulated at that, so the emitter resistor can have the rest of the voltage.

When this takes place you then have the amplification factor of the transistor, where the change in base voltage is reflected across the emitter resistor, this change across the emitter resitor then changes the emitter current flowing through it, 98% of this emitter current passes through the transistor out of the collector (electron current flow NPN) and drops voltage across the collector resistor, thus dropping the voltage at the collector which passes onto the next stage.

Now since input voltage at the base reflects change in the emitter voltage due to Vbe held regulated, the change in voltage across the Emitter Resistor, results in change of current flow through it, and this current flowing through it is in series with the collector resistor, then the voltage drop across the collector resistor is in fact the result of the change in voltage at the base.

Hence voltage gain. Now looking back at that, the voltage gain is the result of the change in current flow through the emitter resistor, now since this same current (98%) of it flows through the collector resistor resulting in voltage change, it goes to reason then, the Voltage Gain (Av) of a transistor stage can be directly related to the ratio of the collector resistor to the emitter resistor.

So you now have a way to design any DC, voltage gain within reason to meet specifications your circuit requires.

 

Usually an amplifier has negative feedback from its output to its inverting input so that the output level does not change much when it is loaded. Your transistor does not have its negative feedback from its collector.
Normally, the load on a transistor is capacitive-coupled to its collector and collector resistor so that the output DC voltage is not reduced. Since the output does not have any negative feedback then an added load causes the gain to be reduced.

Negative feedback allows a transistor circuit to use transistors that have different Vbe without fiddling with resistor values. Negative feedback also reduces distortion.

 

Let me see if I can summarize this.

Negative feedback is important because it allows for Vbb voltage stabilization ( in case Vbb supply decreases ), prevents thermal runaway, allows multiple transistors with different Vbe drops to function interchangably in the same circuit without having to fiddle with resistor values.

The mechanism for both Vbb voltage stabilization and prevention of thermal runaway has been previously discussed.

In the common emitter circuit, the negative feedback could occur either in the collector side by using a resistor from collector to base ( called shunt feedback ). That resistor shorts the CB junction which is reverse biased, allowing current to flow from collector to base, hence collector resistor voltage drops can influence the base via negative feedback. For example, a high Ic causes CE voltage contraction and collector voltage to decrease, decreasing Vb, decreasing Ib.

Another possibility is negative feedback using the emitter resistor. For example, a high Ic causes CE voltage contraction and emitter voltage to increase, reducing Vbe, decreasing Ib.

Negative feedback reduces gain because the output suppresses the input to a degree, wether the mechanism is shunt collector or series emitter.

Negative feedback also allows multiple transistors with different Vbe to function interchangeably. This is because of Vre. If Vb is 1.7 volts for example and we use a transistor with Vbe of .7, we still have 1 volt for Vre left. The .7v was used to forward bias Vbe while the remaining 1 volt will be dropped across Re. If a transistor with Vbe of .9 is used, we have 1.7v - 0.9 v or .8 volts dropped across Re. Whatever voltage remains in Re will function to reduce gain, stabilize Vbb, or stabilize Ic in the case that we have an increase in Hfe due to increasing transistor temperature.

The AC input at the base will have a positive and negative phase and it will add or subtract its voltage to Vb. It seems to stand to reason that I must increase Vb so that it is higher than the peak voltage of the AC input signal. For example, If the AC positive phase is 1 volt and the negative phase is -1 V. Then Vb should be at least 2.7 volts. That way, when the negative peak happens, Vb is at 1.7 volts and I have enough to forward bias BE, and enough for emitter negative feedback. The upward swing poses no issues. The transistor stays in FWD active with negative feedback working. So on the base side loop, we dont want to dip too much into the emitter voltage.

It also stands to reason that the emitter resistor voltage should be preserved on the collector loop. That is to say, we dont want Vce to dip into the voltage reserved for negative feedback during normal operation. That is because we need that feedback for extraordinary situations of Vbb dropping or thermal runaway. So the quiescent point must shift upwards at least by the amount of emitter voltage drop we desire for negative feedback. If for example, we want 1 volt at the emitter resistor for negative feedback, this leaves a maximum of 9 volts for the Vce swings with a Q point of 5.5 volts.

The capacitor at the input side of the amplifier following the AC signal is there to block DC signals from the previous amplifier block. The DC voltage would span the capacitor while the AC signal would go across R2 (the AC signal now alternating above and below 0 volts), adding or subtracting its value to Vb. Likewise, a capacitor on the output side would tend to block DC ( which was necessary on the input side in order to shift the voltages upward to allow the transistor to work in FWD active even in the negative phase of the AC), allowing only AC to pass on to the next amplifier block.

Is this correct?
Thanks,
Chris

 

Is this correct?
Thanks,
Chris

Too many words, didn't read all of them. But on a quick scan, I'd say OK. More or less.

I do think you focus unnecessarily on circuit conditions that tend to drive the transistor toward either saturation or cutoff, at the expense of what we refer to as small-signal behavior-- that is, with inputs that move the operating point up or down only a little bit; after all, that is the normal condition in an amplifier. A change of focus, there, might be instructive.

 

Is this correct?
Thanks,

You have a real good handle on the cause and effect of biasing transistor stages, next is to design and build the stages.
Now that you have this understanding of transistor action, you should be able to pick up just as quick the biasing methods for all transistor configurations. CE,CC,and CB.
When you learn biasing techniques for these, then you will be ready to start designing AC amplifiers.
Good job on your efforts to learn this.

 

You have a real good handle on the cause and effect of biasing transistor stages, next is to design and build the stages.

Maybe. Maybe not. It depends on what his ultimate goals are.

If he isn't aspiring to anything more than spending the rest of his days tinkering idly with simple (and largely obsolete) transistor amplifier circuits out of mere curiosity, then you're right: a more in-depth understanding of electronics isn't absolutely necessary for that. It certainly helps, but it isn't vital.

But if he wants to ever go beyond that level, and is presently using these circuits only as a vehicle for learning-- a stepping-stone to greater things-- he needs to realize that he's only begun to scratch the surface in understanding transistors, how they work, what can be done with them, and how to do those things. To tell him otherwise is doing him a BIG disservice because it encourages him to believe he doesn't need to learn any more.

 

To tell him otherwise is doing him a BIG disservice because it encourages him to believe he doesn't need to learn any more.

I will have to disagree with that, I believe this person has intellegence enough to know, that this is just the beginning of learning transistor circuit design. I believe he has the right approach to learn one step at a time the basics before advancing to more complex designs.

 

The AC input at the base will have a positive and negative phase and it will add or subtract its voltage to Vb. It seems to stand to reason that I must increase Vb so that it is higher than the peak voltage of the AC input signal. For example, If the AC positive phase is 1 volt and the negative phase is -1 V. Then Vb should be at least 2.7 volts. That way, when the negative peak happens, Vb is at 1.7 volts and I have enough to forward bias BE, and enough for emitter negative feedback. The upward swing poses no issues. The transistor stays in FWD active with negative feedback working. So on the base side loop, we dont want to dip too much into the emitter voltage.

Nearly.
Many (including some in this thread) say that the transistor is controlled by a current into its base. Actually, it is voltage controlled because that controls the electric fields in the transistor base-emitter junction. That might help you to consider how it works in a simple "general bias circuit" as shown in earlier posts where two resistors set the base voltage, the resistor in the emitter sets the current and the resistor in the collector circuit sets the voltage on the collector. When you say the base voltage has to be greater than the A.C. here is one way to think of it:

The base-emitter voltage of a transistor is biased at about 0.65V (0.7 perhaps). If the transistor conducts 1mA at that voltage, a 1k emitter resistor will set the emitter voltage to 1V. The base voltage will be 1.65V therefore. That is the voltage you need to define with the two base bias resistors. They need to conduct enough current that the current taken by the base does not materially change the voltage. If the transistor has a gain of 100 then it will need 10uA into the base to conduct 1mA in its collector. (The base current is also controlled by Vbe). So the current in the base bias resistors could be 100uA (though the top one will conduct 10uA more than the bottom). A large capacitor placed across the emitter resistor will hold the emitter voltage constant to A.C.

If another capacitor couples in some A.C voltage to the base, that changes the base-emitter voltage on the transistor. Because the collector current is exponentially dependent on the base-emitter voltage, a swing of about 18mV will double (for a positive swing) or halve (for a negative swing) the current. The important point here is that the A.C. cannot be too large, and is a lot smaller than D.C. value of Vbe so while you are partially right in thinking that the Vbe has to be greater than the A.C., the transistor can easily become fully on or fully off long before the D.C. value of Vbe is reached.

On the other hand if there is no capacitor on the emitter, then the emitter voltage will increase and decrease as the A.C. voltage is applied. In this case, the emitter resistor is providing a high degree of negative feedback. The input voltage could reach up to 1V - the voltage across the emitter resistor - plus the small incremental voltage across the base-emitter junction - before the transistor saturates or switches off, but that also depends on the collector resistor and how much voltage drop that generates as the current changes. You would normally select the collector resistor to set the collector voltage to the average of the emitter voltage and power supply voltage Vcc.

As an example: design a circuit to amplify a voltage with a voltage gain of 10. A power supply of 9V is to be used.
(1) Choose a suitable collector current. Say 1mA.
(2) Choose the emitter voltage to something that will stabilise it against variations in different transistor gains. As a minimum this could be a Vbe equivalent (0.65V) or 1V for better stability.
(3) Step (2) sets the emitter resistor to 1k.
(4) Set the bias chain up. The required base voltage is 1.65 (or thereabouts) as it is the emitter voltage plus Vbe.

 

Whoops- reply terminated before I'd finished.
The current in the lower bias resistor should be 100-10=90uA. The resistance should be 18k.
The current in the upper bias resistor should be 100uA. The voltage is 9-1.65=7.35V, Nearest preferred resistor value=75k
(5) Set the collector voltage to the average of the emitter and supply=5V. The required resistance is (9-5)=4V/1mA, nearest preferred value 3.9k
(6) Without a capacitor across the emitter resistor, a small voltage increase (say 0.1V) causes the current to rise to 1.1mA.
Therefore the collector voltage will FALL as the current increases from 9-3.9=5.1 to 9-(3.9x1.1)=4.71.
The ratio between the input voltage and collector voltage is (5.1-4.7):0.1=4. So the gain is set by the collector resistance divided by the emitter resistance (because the currents through these are more or less the same).
(7) For a gain of 10 we therefore need the emitter resistance to be 390 ohms. As there is already a resistor of 1k present, this needs a parallel resistor of 640 ohms, except that the transistor has an effective emitter resistance (this is internal to the transistor) of 25/Ie, where Ie is in mA, or 25 ohms. So reduce the resistance needed to 620 (nearest preferred value) but the 25 ohms needs to be subtracted from the 390, really. That would mean using 560 ohms rather than 620.
(8) This resistor cannot be connected directly across the 1k resistor or that upsets the D.C. voltage. Use a capacitor in series to separate the D.C. from the A.C. A value of 100 uF would be plenty.
(9) Now add a capacitor to the base to feed the signal in, say 10uF.
Then you have your amplifier stage that will accept input voltages to about 400mV but you should be able to simulate this circuit now.

 

@OBW0549 and @hobbyist

This is an important stepping stone for me. My objective at this time is to learn purely. I don't have an electronics background and I have a full time job in another field. The most immediate application is as a hobby, tinkering with devices such as microcrontrollers and sensors and motors. I could always find a howto on how to connect devices but it would be better to understand the reasoning for such circuits and connections. And I know I will learn to read faster than write. Even from this thread, I learned more than just transistor operation as it solidifies earlier concepts such as voltage division, linear devices, KVL calculations. I understand that this just scratches the surface. It is a fun adventure that I will pursue at a leisurely rate. Thank you for your time and attention. This will not be the last time that I post something that might seem misguided or very basic given my limited understanding. However, even telling me that I am wrong nudges me in the right direction.

@neonstrobe

If another capacitor couples in some A.C voltage to the base, that changes the base-emitter voltage on the transistor. Because the collector current is exponentially dependent on the base-emitter voltage, a swing of about 18mV will double (for a positive swing) or halve (for a negative swing) the current. The important point here is that the A.C. cannot be too large, and is a lot smaller than D.C. value of Vbe so while you are partially right in thinking that the Vbe has to be greater than the A.C., the transistor can easily become fully on or fully off long before the D.C. value of Vbe is reached.

On the other hand if there is no capacitor on the emitter, then the emitter voltage will increase and decrease as the A.C. voltage is applied. In this case, the emitter resistor is providing a high degree of negative feedback. The input voltage could reach up to 1V - the voltage across the emitter resistor - plus the small incremental voltage across the base-emitter junction - before the transistor saturates or switches off, but that also depends on the collector resistor and how much voltage drop that generates as the current changes.

So the swings in the base voltage, affects the swing in collector and emitter voltage. Applying the AC input to the base. An increase in Vbe -> increased Ib -> increased Ic -> decreased CE impedance -> decreased CE drop -> increased collector resistor drop -> collector voltage is more negative/ emitter voltage is more positive. This could make Vc < Vb putting the transistor in saturation or make Ve greater than Vb and put the transistor in cutoff with Ve > Vb. So these conditions can be hit before the input AC causes Vbe to fall below .7volts.

A bypass capacitor on the emitter reduces the effective emitter resistance for frequencies we wish to amplify, reducing the emitter voltage drop so that small positive swings of input AC in the frequencies we wish to amplify will result in higher Ic -> lower CE drop -> more negative collector/more positive emitter which will cause saturation/cutoff to happen sooner. In the absence of a bypass capacitor, the input voltage swings would be higher before saturation or cut off is reached.

So a bypass capacitor is really necessary and AC input must be small voltage amplitudes and I need to worry about the collector and emitter side as well as the base voltage when figuring the proper biasing. This is why multiple amplifier stages are serialized and amplifications summed up to achieve the total gain you want from a series of smaller gains.

(8) This resistor cannot be connected directly across the 1k resistor or that upsets the D.C. voltage. Use a capacitor in series to separate the D.C. from the A.C. A value of 100 uF would be plenty.

The resistor would only be connected when there is an AC signal of sufficient frequency that it shorts the capacitor. This seems like a good place for a potentiometer that will regulate the gain of the circuit.

Thanks,
Chris

 

So a bypass capacitor is really necessary and AC input must be small voltage amplitudes....

Then audio and video noise will be high and distortion will be high if the input level is increased even if the level is well below saturation and cutoff. The emitter resistor or another negative feedback method is always needed.

 

Ok, thanks. So a combination of both ( have feedback resistor in series with a capacitor bypassed resistor ) is probably the best as it allows for selective gain for frequencies you wish to amplify but there will always be a degree of negative feedback.

Been playing around with the biasing calculations. It seems one important consideration is to make sure that at the positive peak of the input signal, the resulting Vb must never go above the resulting Vc or else saturation happens. If it does, too big of voltage drop happened on RC. In this case, reduce the value of RC which also increases the maximum Ic and reduces the gain. The danger on the negative peak of the input signal is wether or not Vb falls below the minimum to forward bias the BE.

It would seem that it would be possible to set Qpoint up so that for a given Vcc you could maximally amplify only a half wave but more fully utilize the extents of the Vcc. For example, amplify an input signal of 1 volt to 10 volts using a Vcc of 12 volts. Then use a capacitor to remove the bias and a diode ( there will be some voltage drop) to isolate the amplified half wave. You will need two transistors operating at extremes high or low Q point, then join the two halfwaves to form one output. Or that might just be a really crazy idea. I wonder what the distortion would be like if it were possible.

Chris

 

Chris, your "crazy idea" will create lots if distortion by doubling the output voltage swing. But some "bridged" IC amplifiers are made by bridging two low distortion power amplifiers that are driven with opposing phase so that each amplifier drives one wire of a speaker producing almost double the output voltage swing then the output current swing is also almost doubled resulting in about 3.5 times more output power than one amplifier that uses the same speaker and the same supply voltage.

 

Not sure what you want to achieve. Old transistor radios using transformers had two transistors operating across the supply, one connected to each side of the primary, center tap going to battery. Push pull inputs allowed the transistors to conduct one after the other. Transformer combined signals back into one piece. Transformers not used for 30 years as too expensive (except perhaps for specialist applications and tube amps that can't work speakers directly).

 

I am just trying to learn. I don't have any specific projects.

Nope. Wrong.

Ic does not change because collector current is almost entirely independent of changes to Vce.

I'm trying to reconcile this with what I am seeing in LTSpice.

It seems changing the resistive load DOES change the current through R3. I thought current through R3 is Ic and only changes due to change in Ib.

In one case where R5 is 10k, the current through R3 is 1.72 mA and the voltage at the collector is 2.8 V. In the case where R5 is 50k, the current through R3 is 1.5 mA, and the voltage is 4.75 V.

This is a snapshot of what would happen if the base voltage is held at 2.1 Volts which is the top of the sine wave AC input. I'm not sure if this is an accurate simulation because I had to remove the coupling capacitor.

The only thing I could think of is that R3 current is technically not Ic anymore because of the resistive load. Ic is only the positive current entering the collector terminal of the transistor and going towards the emitter. That could still probably stay stable despite the load. This would happen if the current through R5 is being subtracted from current through R3 ( siphoning but technically not siphoning from Ic anymore because adding a node changes the definition of Ic).

 

This confirms it, if the simulation is correct. Ic remains the same, despite the load, even if current through R3 has changed. But how does one predict this voltage sag given a load?

 

This confirms it, if the simulation is correct. Ic remains the same, despite the load, even if current through R3 has changed. But how does one predict this voltage sag given a load?

Nodal analysis.