Clarification with common collector...

Thread Starter

clickman

Joined Dec 9, 2016
28
Hello everyone, i'm new in this electronics thing and in this forum, but eager to learn.

I've been learning and experimenting with the common collector amplifier. I've been able to design some and test them with a simulator successfully. I've also learned that they are called emitter followers because they act as a buffer and can "copy" the voltage applied to them (with a Little loss due to a voltage drop of the transistor).

What i can't understand is something about them having the ability of providing or amplifying currents for big loads. does this means that if for example I connect a 8 ohm speaker will it be able to work? In the simulations i've made the current indeen amplifies but the voltage drops sometimes even lower than the input.

What could i be not understanding?

thanks in advance to everyone.
 

#12

Joined Nov 30, 2010
18,222
Consider this: If you want to start with some sort of input voltage, but you don't need any more voltage or any more current, you can do that with a piece of wire. Now why would anybody install a transistor configuration that doesn't have any voltage gain? It must be the larger current it can provide. Whether you can drive an 8 ohm speaker depends on how much you started with and how much current gain your transistors can provide.
 

Thread Starter

clickman

Joined Dec 9, 2016
28
first of all thanks for your quick reply.

Maybe my mistake is thinking that the emitter follower Works like an op amp, a (constant?) voltage, that can provide a current according to the needs of the load (am i confusing the concept of load?)
 

#12

Joined Nov 30, 2010
18,222
I can see a lot of difference between an emitter follower and an op-amp. The emitter follower can't switch the polarity of its output compared to its input and has only one input. You can configure an op-amp like that, but you can't configure a single transistor like a dozen other ways an op-amp can be used. You should look at, "Class A" amplifier. That's what you're talking about when you talk about an emitter follower driving a speaker. Yes, it can be done. Back in the vacuum tube days, it was considered practical to do a class A amplifier. Now that transistors exist, a two transistor output stage is much more efficient.
 

crutschow

Joined Mar 14, 2008
27,395
What i can't understand is something about them having the ability of providing or amplifying currents for big loads. does this means that if for example I connect a 8 ohm speaker will it be able to work? In the simulations i've made the current indeen amplifies but the voltage drops sometimes even lower than the input.
A bipolar transistor has current gain (typically about 50 to a 150).
Thus the base-emitter current is typically about 50 to 150 times less than the collector-emitter current (the emitter carries both the base and collector currents combined).
Thus a load on the emitter can have much higher current than the base current so it provides current amplification. For a low impedance load this can mean there is a big power gain (power = V*I).
Since the base-emitter voltage is typically about 0.6-0.7V, the emitter voltage is the same as the base voltage minus the base-emitter drop.
And since the base-emitter voltage increases slightly with increasing current, this gives the added drop you saw in your simulations with an 8 ohm emitter load as compared to a higher impedance load.
(There is also an intrinsic emitter ohmic resistance that adds to the drop).
 

Thread Starter

clickman

Joined Dec 9, 2016
28
thanks for your reply

The drop I see in simulations is very sharp. For example an input of 1V shows drops to maybe 27mV. this is seriously confusing.

I should ask you if an emitter follower should/could be used in audio applications or should I use an op amp.
 

Thread Starter

clickman

Joined Dec 9, 2016
28
Could it be that i'm attempting to use only 1 transistor in a task in which 2 transistors paired would be better?
 

Thread Starter

clickman

Joined Dec 9, 2016
28
e_f.JPG

This is one of the circuits i'm experimenting with. I'm one of those that learns only by making.

I've "designed" this according to the method taught by a Carnegie Mellon professor that posted a video on YouTube about emitter follwer setup

The sine voltage provides a 1 volt input and the 8 ohm resistor simulates a speaker.

It is a very simple setup
 

AnalogKid

Joined Aug 1, 2013
9,283
Vbe, the base-emitter forward voltage of a transistor, is not a constant. As the emitter current increases, Vbe increases - even though most of the emitter current comes from the collector. Your circuit attempts to place the emitter at 1/2 the supply voltage by making the two base bias resistors unequal, and that works up to a point. For a small signal transistor, Vbe can vary over 300 mV from the start of conduction to fully saturated at peak rated current.

Also, this variation in Vbe is instantaneous. That is, it tracks the instantaneous emitter current throughout the sine wave cycle. So Vbe is less down around the negative peaks where the transistor is almost cut off, and greater up around the positive peaks where the transistor is near saturation. This translates into harmonic distortion in the output signal.

ak
 

#12

Joined Nov 30, 2010
18,222
The drop I see in simulations is very sharp. For example an input of 1V shows drops to maybe 27mV. this is seriously confusing.
So is your wording. Are you saying a 1 volt sine wave input produces a 27 mv sine wave out put or are you saying that when the input is at the instant of 1 volt above ground the base to emitter voltage is 27 mv?
 

Bernard

Joined Aug 7, 2008
5,788
I just pulled up this old print for another post & may be helpful here ?
As AK said Vbe changes with loading as shown here starting at input 2.325 V Vbe = 1.205 & @ 12.04 V, Vbe = 1.33 V. Not a problem when driving a motor or controlling LED brightness.Brightness Control 00000.jpg
 

WBahn

Joined Mar 31, 2012
26,398
View attachment 116615

This is one of the circuits i'm experimenting with. I'm one of those that learns only by making.

I've "designed" this according to the method taught by a Carnegie Mellon professor that posted a video on YouTube about emitter follwer setup

The sine voltage provides a 1 volt input and the 8 ohm resistor simulates a speaker.

It is a very simple setup
Your output has a time constant of R·C = 160 μF · 8 Ω = 1280 μs. This corresponds to a cutoff frequency of about 125 Hz. This is a high pass filter, so signals below that cutoff frequency are going to be attenuated. It's a first-order filter so the skirt is -20 dB/decade. Going from 125 Hz to 40 Hz is about half a decade so that's a 10 dB loss, so your voltage attenuation should be about a factor of 3.2 or so. That would be in the 300 mV range if you are starting at 1 V. But that's just the output. You also have a high pass filter on the input. The effective resistance in that filter is approximately the parallel combination of the bias resistors, or 900 Ω || 1 kΩ, which is ~475 Ω. With a capacitance of 16 μF, the cutoff frequency is about 21 Hz, so that shouldn't be killing of the input signal too much. So your 27 mV output voltage (assuming you are measuring it across the 8 Ω resistor) is about an order of magnitude below what would be expected (based on a rather ideal look at the input/output attenuations). What we haven't looked at yet is the output impedance of the amplifier. It would not have to be very large before you picked up quite a bit of attenuation into an 8 Ω load. Assuming that the input source resistance is negligible, the contribution of the transistor input resistance would be something like 2 Ω, so that might be very noticeable, but shouldn't account for an order of magnitude.
 

#12

Joined Nov 30, 2010
18,222
Another way to look at the behavior: Your emitter resistor has too many ohms to discharge the capacitor when the transistor is nearly off. The DC level on the emitter will climb until very little signal can get through. That makes the transistor act kind of like a rectifier which is charging up the output capacitor faster than the emitter resistor can empty it.
 

Thread Starter

clickman

Joined Dec 9, 2016
28
Your output has a time constant of R·C = 160 μF · 8 Ω = 1280 μs. This corresponds to a cutoff frequency of about 125 Hz. This is a high pass filter, so signals below that cutoff frequency are going to be attenuated. It's a first-order filter so the skirt is -20 dB/decade. Going from 125 Hz to 40 Hz is about half a decade so that's a 10 dB loss, so your voltage attenuation should be about a factor of 3.2 or so. That would be in the 300 mV range if you are starting at 1 V. But that's just the output. You also have a high pass filter on the input. The effective resistance in that filter is approximately the parallel combination of the bias resistors, or 900 Ω || 1 kΩ, which is ~475 Ω. With a capacitance of 16 μF, the cutoff frequency is about 21 Hz, so that shouldn't be killing of the input signal too much. So your 27 mV output voltage (assuming you are measuring it across the 8 Ω resistor) is about an order of magnitude below what would be expected (based on a rather ideal look at the input/output attenuations). What we haven't looked at yet is the output impedance of the amplifier. It would not have to be very large before you picked up quite a bit of attenuation into an 8 Ω load. Assuming that the input source resistance is negligible, the contribution of the transistor input resistance would be something like 2 Ω, so that might be very noticeable, but shouldn't account for an order of magnitude.
I've been simulating with 40 Hz, could that be a problem?
 
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