Common emitter amplifier, emitter capacitor, ac voltage gain and pwr supply caps

[ericohman]

Hi!

I'm trying to understand this very basic amplifier circuit.
There's a few things I haven't understood.

1. What network will modify the amplitude of the output ac signal?

2. The emitter capacitor, is that used to make sure all frequencies are amplified equal?

3. I noticed there's no caps from the power supply, like in the rangemaster circuit there's a electrolytic cap to ground....
a) What's that cap for?
b) What is the audible difference between having a, say 100 uf compare to a 1uf connected to +9V to ground.

http://www.electronics-tutorials.ws/amplifier/amp5.gif


I thought opamps were harder but I have read a lot about them, and I actually find these transistor amplifier circuits harder to understand.
All the best,
Eric

[R.G.]

1. What network will modify the amplitude of the output ac signal?

I don't understand your question. Every part in there will modify the amplitude of the output AC signal. So i think you meant something else, like maybe "what parts set the gain?" or similar. Can you restate the question?

2. The emitter capacitor, is that used to make sure all frequencies are amplified equal?

No. It's there to increase gain.The emitter resistor stabilizes the transistor's bias point for DC conditions, but it does this by introducing negative feedback at all frequencies. The capacitor "shorts out" the emitter resistor for AC signals above some frequency, and restores the lost AC gain above the "shorts out" frequency.

3. I noticed there's no caps from the power supply, like in the rangemaster circuit there's a electrolytic cap to ground....
a) What's that cap for?

This is an example circuit, where everything not shown is assumed to be perfect; in this case, the power supply is assumed to be perfect. Real power supplies are not perfect. The use of capacitors across the power supply is a technique used in real world circuits to make up for the imperfections of real power supplies. That capacitor lowers the impedance of the power supply, and reduces power-supply-borne noise and interference.

b) What is the audible difference between having a, say 100 uf compare to a 1uf connected to +9V to ground.

Since this is a part needed for possible real-world situations which are somewhat unpredictable, it is impossible to say. Theoretically, there is no difference at all. In the real world, 1uF may or may not be big enough to make up for the shortcomings of the real power supply, and 100uF - or 1000uF - might be. It depends on the rest of the world that the circuit is connected to. You're asking about the difference between theory and practice.

We learn theory so we can understand practice. Theory is a logical abstraction, like a line in geometry is a thing with zero width and infinite length. Lines do not and can not exist in the real world because in the real world, all real objects have non-zero width and less-than infinite length. All we have are approximations to a theoretical line. Same thing with theoretical circuits. We study the perfect to understand the imperfect.

I thought opamps were harder but I have read a lot about them, and I actually find these transistor amplifier circuits harder to understand.

Actually, opamps were invented to make up for the imperfections of real tubes, transistors, etc. An opamp uses the high gain of the circuit to "hide" the imperfections of the real circuit. So yes, transistor amplifier circuits require much more learning to really understand.

[ericohman]

Thank you!

Yes, question 1:
I meant what parts sets the gain.

3b:
Is there a big possibility that I will here a difference if I do NOT use any electrolytic to ground? If powered with batteries.

I've also noticed sometimes the "signal" goes out from the emitter instead of the collector. But, before I get too confused. Let's keep talking about the picture above, and how to set the gain.

Found a link: A lot to read...might be relative to this thread:
http://www.opamp-electronics.com/tutorials/feedback_3_04_10.htm

[GibsonGM]

RL and RE set the gain, with RE having a 'noticeable' effect if you stick a pot there.  The cap 'bypasses' AC at frequencies relating to its reactance, and will cause the gain to vary at a first-order rate.  Try building it with, and without, the cap and playing!  Won't cause any harm.  Experimentation is the best way to "get" what parts are doing. 

RE and the bypass cap make a shelving filter.  If the cap is large (1uF or larger, say) all freq's we care about pass, and you hear 'normal' guitar.  If it is small (<1uF) you start to get selective treble boost.   R.G. could tell you ALLLLL about this, at great length!!! But that's the gist of it.  The -3dB point on the line would be at 1/2piRC; it is easier to set up LT Spice or another simulator to take a look at the curves for that cap - or just change its value til it sounds like what you want!

You can play with RL, too, but be aware that that resistor sets the current thru the device, which may terminate if such current becomes too large (too small a value for RL).  LOL.

[ericohman]

I have a rangemaster on a breadboard with pots for changing the emitter resistor as well as the voltage divider which sets the base. I have been trying some things and there's a significent difference.
Will have a go with LTSpice tonight.

If I want to breadboard a npn common emitter amp, will a BC108A TO-18 NPN 20V 100mA transistor work good?
Those are easily available here in Sweden so that I don't need to wait for the shipping
Want to do this weekend

[R.G.]

Yes, question 1:
I meant what parts sets the gain.

In this case, an invisible part. Bipolar transistors are complicated intermally. They're sufficiently complicated that an entire generation of electrical engineers that were skilled with tubes was lost because they could not make the transition to transistors right.

A bipolar transistor like this has a current gain and a transconductance, and it can be described either way, both being a different way of looking at what's going on inside.

Let's remove the emitter capacitor from the circuit for a start. One very important thing to remember about bipolar transistors is that if you want them to amplify, the collector-base junction must be reverse biased and the base-emitter junction must be forward biased. The reverse biased collector-base junction stops any current flow. But current fed into the base-emitter "poisons" the ability of the collector-base junction to hold off current flow. The amount of this "poisoning" is described as the current gain, how much current is let flow in the collector divided by how much flows into the base. It's also called hfe, Hfe, beta, and other things. It's current gain. It's nonlinear, but approximately described by datasheet numbers.

In a transistor that's set up like the example, resistor on the collector, resistor on the emitter, bias on the base, the base is pulled up to some voltage by the left hand resistors. Since this elevates it above the emitter, current flows into the base.

The base-emitter junction has a "turn on" voltage of about 0.5V, below which little or no current flows, and above which an exponentially-larger current flows as voltage increases. For all practical purposes, you can't pull the base more than about 0.7V above the emitter because you can't supply enough current.

So the base is pulled up to some voltage, and the emitter is pulled to ground by that resistor. That lets the base be more than 0.5V more positive than the emitter, and base current starts to flow. But that also "poisons" the ability of the collector to hold off current from the power supply, and the collector lets current through too. Both the base current and the collector current combine and flow out the emitter. The current let through the collector is bigger than the base current, many times bigger. For good transistors, hundreds of times bigger.

What that does is to cause the emitter voltage to rise, because the current through the emitter resistor causes a voltage drop across the emitter resistor by Ohm's law, V = I * R. This voltage raises the emitter toward the base, which decreases the relative difference between the base and emitter, and therefore acts to reduce the base-emitter voltage. This decreases the current going into the base, cutting down the collector - and emitter! - current.  It's negative feedback.

What this does is to cause the circuit to balance where there is just enough base-emitter voltage to cause just enough base current to flow to make just enough emitter current to flow to keep the emitter where the base-emitter voltage just keeps that amount of current flowing. That statement is confusing because it is circular. Negative feedback is circular in nature. What's important to realize is that the emitter voltage balances where the emitter is between 0.5V and 0.7V lower than the base, and comes to rest there. We don't know exactly the base emitter voltage, because it varies from transistor to transistor, but it's in that range, and that's close enough for us to use.

So we set the base *voltage* by setting the base divider network, the left hand two resistors. That forces the emitter to rise to between 0.5 to 0.7V below the base (as long as there's enough current available in the base circuit and also from the collector). And it balances there. What lets the emitter rise in voltage to balance is the emitter resistor.

The emitter resistor sets the emitter current by Ohm's law, I = V/R where the V is the emitter voltage which is Vbase minus 0.5 to 0.7V. Since the base current is less than 1% of the collector current for transistors with gains of 100 or more, we'll ignore the base current and say that the collector current is equal to the emitter current.

The collector voltage is equal to the power supply voltage minus the collector current times the collector resistor.

What happens if we apply a signal at the base and wiggle it around?

The base voltage moves up and down. This increases/decreases the base current to match, and the emitter follows the base by negative feedback. So the signal on the emitter follows the base with a gain of 1.00000 minus the little bit lost to creating the base current changes. The emitter follows the base, and that's where we get signals coming out of emitters - it's the "emitter follower" circuit.

Meanwhile, back at the ranch... uh, I mean collector!    The emitter current wiggles around in an amount determined by the base voltage signal size divided by the emitter resistor. The collector current is equal to that, minus the gnat's eyelash change of the base current. The amount of change of the voltage at the collector is the change in collector current times the collector resistor. This is equal to the collector current change (and that's substantially equal to the emitter current change, remember) times the collector resistor.

OK - emitter voltage moves substantially equal to the base movement. Emitter current equal to emitter voltage divided by the emitter resistor. Collector current equal to the emitter current. Collector voltage changes by power supply minus the collector current flowing in the collector resistor. The voltage change at the collector goes *down* when the collector current goes up, so the voltage change on the collector is inverted compared to the base and emitter because more base current raises the emitter and lowers the collector. And since the amount of current in the emitter and collector  is the same **the size of the change in voltage on the collector is equal to the ratio of the collector resistor and the emitter resistor**.

>>The voltage gain is equal to the ratio of the collector and emitter resistors when there is no emitter bypass cap.<<

PHeww!

But wait - there is an emitter bypass cap. That has the effect of "shorting out" the emitter resistor for AC signals. How can there be an emitter voltage to make feedback?

If you now put the emitter bypass cap back in, we come to the "invisible component" that started all this.

There is an internal emitter resistor in all bipolar transistors. It's called the Shockley resistance, and it's actually a side effect of the internal physics, not a real resistor. It is quite small compared to real external resistors and changes with the changing collector current.

Adding the bypass cap to the emitter has the effect of removing the external feedback loop for AC signals while leaving the DC feedback in place. So the AC gain is now much larger. How much larger depends on the value of the internal Shockley resistor, which depends on the internal current gain, and the amount of change of base current (and collector current, therefore) with changes in the base voltage. This setup is best described not by current gain but instead by transconductance - the change in collector current with changes in base-emitter voltage from 0.5V turn on to 0.7V max.

As you're starting to suspect now, putting a number on the exact gain of a transistor with a completely bypassed emitter is hard. It's bigger, sometimes a lot bigger, hard to determine exactly, and varies with different transistors and the phase of the moon. I made up that last thing about the phase of the moon.   

Answering your question, what sets the gain for the bypassed-emitter setup is 1. choice of transistor, 2. the size of the collector resistor. It goes up with bigger collector resistors, but if that gets too big or too small, the DC bias won't be in the linear region any more and you'll get distortion.

3b:
Is there a big possibility that I will here a difference if I do NOT use any electrolytic to ground? If powered with batteries.

Probably not. Batteries are generally pretty good approximations to "perfect" power supplies if they're freshly charged and have a lot more current capability than the circuit needs.

[GibsonGM]

Like he said, LOLOL. Copy and save that one!   I am stuck in the world of tubes, where a bypass cap is the holy grail of tone shaping. You learn something new every day )