Design Article
Discrete audio amplifier basics - Part 1: Bipolar junction transistor circuits
John Linsley Hood
3/10/2010 2:00 PM EST
9.3 Stage Gain
The stage gain of a BJT, used as a simple amplifier, can be determined from the relationship:
Vout / Vin = hfeRL / (RS + ri)
where RS is the source resistance, RL is the collector load resistor, hfe is the small-signal (AC) current gain, and ri is the internal emitter-base resistance of the transistor. An alternative and somewhat simpler approach is similar to that used for a pentode valve gain stage in which
Vout / Vin = gmRL
where the gm of a typical modern planar epitaxial silicon transistor will be in the range of 25"40 mS/mA of collector current. Because the gm of the junction transistor is so high, high stage gains can be obtained with a relatively low value of load resistor.
For example, a small-signal transistor with a supply voltage of 15 V, a 4k7 collector load resistor, and a collector current of 2 mA will have a low frequency stage gain, for a relatively low source resistance, of some 300. If some way can be found for increasing the load impedance, without also increasing the voltage drop across the load, very high gains indeed can be achieved - up to 2500 with a junction FET acting as a high impedance constant current load.1
A predictable, but interesting aspect of stage gain is that the higher the gain, which can be obtained from a circuit module, the lower the distortion in this which will be due to the input device. This is so because if increasingly small segments are taken from any curve, they will progressively approach more closely to a straight line in their form. This allows a very low THD figure, much less than 0.01% at 2V rms output, over the frequency range 10 Hz"20 kHz, to be obtained from the simple NPN/PNP feedback pair shown in Figure 9.5 , which would have an open loop gain of several thousand.
Figure 9.5: NPN/PNP feedback pair.
The distortion contributed by Q2 will be relatively low because of the high effective source resistance seen by the Q2 base. A similar low level of distortion is given by the amplifier layout (bipolar transistor with constant current load) described earlier because of the very high stage gain of the amplifying transistor and the consequent utilization of only a very small portion of its Ic/Vb curve.
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abraxalito
3/10/2010 9:05 PM EST
I thought that Doug Self as editor would have caught the myth being propagated here that the BJT is a 'current operated device'. He makes plain in his own writings that its not, rather its voltage operated with an exponential law between VBE and IC.
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d-jeff
3/11/2010 3:32 AM EST
The bipolar IS voltage controlled with an exponential law (even though when a VBE is applied, some IB takes place). In this article on bipolar, this fact does not seem obvious, I agree with previous comment.
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bcarso
3/11/2010 12:17 PM EST
Hood was a delightful character, but wasn't always entirely clear on solid-state physics. I recall at one point in his description of JFET operation he refers to majority carriers "tunneling" through the pinched-off channel...
Although there are situations where a voltage-controlled analysis can be better suited to a given device, Barrie Gilbert makes the point that really both voltage and current are always involved (see his way-best-of-book articles in Toumazou's (et al., eds) collection Analogue IC Design: the current-mode approach). Another very powerful and seldom-referenced approach, analysis in terms of charge as the variable, is presented in Ed Cherry's Amplifying Devices and Low-pass Amplifier Design, which does an integrated presentation of tubes and transistors.
An aside: Barrie G. also mentioned once some experimental heterojunction ADI bipolars that had a beta of order 100k.
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Guru of Grounding
3/11/2010 7:32 PM EST
Arguing about whether a bipolar transistor is voltage or current "operated" is like arguing about how many angels can dance on the head of a pin! Notice that the context of his statement included the word "linearity". Many transistors have very linear hFE over many decades of current while Ic vs Vbe is anything but linear. Linearity is THE prime consideration in audio after all. Circuit designers simply want to characterize a transistor as a 3-terminal device and, for linear applications, the quantum physics involved inside are of little interest. I have disagreements with Mr. Hood, but they're about other issues. I think his explanation is quite appropriate in its context.
Bill Whitlock
president/chief engineer
Jensen Transformers
www.jensen-transformers.com
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bcarso
3/16/2010 4:06 PM EDT
Agreed Bill---although I would suggest that the expression should be "very constant beta" rather than "linear" beta, as a function of collector current. "Linear" beta for example would be where beta was given by an expression like m*Ic + b.
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c_abraham
6/12/2010 7:07 PM EDT
Then why does every semiconductor OEM call the bjt a current controlled device? Do you know more then them? How many semicond physics courses have you taken? Nothing personal, but if what you say is true, every OEM would concur, & every university semicond course would teach the same.
At the "black box" viewpoint, a bjt is classified as current controlled. For a deeper look involving internal charge distribution, depletion regions, etc., the charge control model is used, for the bjt & FET as well. Ultimately, at the atomic level, only quantum mechanics can describe what is happening. This is the universal model. Shockley stated this in the early 1950's.
Again, at the external viewpoint, current control is a good 1st order approximation. More in depth study produced the charge control model. Ultimately the device is best described by QM. Every OEM cannot be wrong. I took 1 undergrad course in semicond phy from the phy dept. Then I took 4 couses in semicond phyfrom the EE dept. in my MS & Ph.D. studies at 2 different universities. The bjt & FET are described as I just stated.
A bjt cannot be directly driven from a low impedance voltage source. Thermal runaway would take place. We always "current drive" a bjt. If the input signal source is constant voltage type, then a resistor must be used to drive the bjt. This is why bjt is "current driven". The Vbe is of course, absolutely needed, as are Ib & Ie. All 3 are needed for transisto action to occur. Without Ib, Vbe, & Ie, there can be no Ic. They are equally important.
Claude
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Kevin Aylward
6/2/2013 7:02 AM EDT
1 Semiconductor companies possibly just repeat the old wife’s tales that they know about. Possible its because good analog designers are hard to come by.
2 University semiconductor courses do indeed teach that the transistor is a voltage controlled device.
3 The Gummel-Poon charge controlled model expressly describes the transistor as a voltage controlled device. This is explained here:
http://www.kevinaylward.co.uk/ee/voltagecontrolledbipolar/voltagecontrolledbipolar.html
4 Driving a bipolar from a current source for biasing results in unpredictable biasing. This is explained here:
http://www.kevinaylward.co.uk/ee/bipolardesign3/bipolardesign3.html
5 Bipolors are easily driven from low impedance sources, e.g. Emitter resisters can be used when driving bipolars from large, voltage driven sources. Additionally, adding large resistors to the base circuit results in more noise, and lower bandwidth.
6 I could go on, but the original comments above are, essentially, incorrect.
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pcsalex
7/15/2011 9:47 AM EDT
by the way on figure 7B that is not a rearranged cascaded, but a differential amplifier!
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