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BJT biasing


Kevin Weddle
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Does low current biasing have more to do with transistor longevity? I don't think so, because I have seen transistors biased at low current but a high VCE. I plotted beta one time, and I think it goes like this : Increase the current from zero and the beta goes high, very high. Approach about half the current rating of the transistor, and the beta goes down from there. This area seems like logical bias.

If you bias a 200mA transistor at 5 mA collector current, you can only change the collector current by 5mA downward. So let's say the signal changes the current 1 mA, or .5 mA, or .1 mA, or 1pA. Is there any operating the transistor relative to magnitude? I wouldn't drive a 3A transistor with 1pA peak to peak base current.

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Hi Kevin,
Transistors last "forever" if you operate them within their ratings for voltage (including the max of only 5V to 7V for reversed biased emitter-base junction), max current and  min and max temperature (ambient and power dissipation).

All transistors have less current gain (beta) at currents near their max rating.

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A single transistor without negative feedback so its gain is about 170 distorts very much when its output level is high. An opamp has such a high amount of gain that a large amount of negative feedback can be used to reduce its distortion to nearly nothing, and still have high gain.

Here is a sim of a single transistor with no negative feedback. It is so distorted that you can't calculate its gain. To its right side is the same transistor circuit but with negative feedback added to reduce its distortion and gain. Its voltage gain is 9:

post-1706-14279143255143_thumb.png

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I don't see the feedback in the circuit.

The unbypassed emitter resistor provides negative feedback.

With the emitter resistor bypassed with the 100uF capacitor C2, the transistor's internal Re is 0.026 divided by its emitter currentof 0.43mA= 61.1 ohms. Therefore its voltage gain is 10k/61.1= 164 and its distortion at a high output level is very high.

Without C2, the voltage gain is 10k/(61.1 + 1k)= 9.4 and the distortion is also reduced. 
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The negative feedback in many transistor amplifier circuit is not always apparent. Take circuit A, for instance. It's an emitter follower configuration, which has a voltage gain of +1. There is 100% negative feedback present in this amplifier, as I will explain.

The transistor will conduct from collector to emitter when the base voltage is 0.7V or so above the emitter voltage. Increasing this base-emitter difference even slightly will cause the transistor to conduct heavily, and decreasing it only a tiny amount will cause the transistor to completely block current. Therefore, this transistor operates with a fairly constant voltage between base an emitter - about 0.7V.

If the base input voltage rises, then the voltage difference between base and emitter increases. The transistor becomes more conductive, increasing current flow from the collector down through the emitter and resistor. More current through the resistor means a greater voltage across it, meaning the emmitter voltage also rises.

When the base voltage falls, the base-emitter voltage difference is reduced, causing a drop in the transistor's conduction. So emitter current drops, and therefore also the voltage across the resistor.

In this way, the emitter voltage (output) rises and falls with the base voltage (input) in order to maintain a constant 0.7V between base and emitter - hence the name 'emitter follower'.

Even though we have placed no explicit feedback loop in this circuit, the very nature of the transistor and it's configuration in this circuit exhibits negative feedback, to ensure that the enmitter voltage follows the base. This is called "inherent negative feedback".

If we modify the emitter follower slightly by adding a resistor at the collector (circuit B), we can reduce the amount of feedback inherent in the circuit to something less than 100%. The voltage gain is thus controlable (-R1/R2), but still there is no visible negative feedback path! The feedback present in this circuit is still inherent, and due to the emitter resistor.

In the absence of an emitter resistor, there is no inherent negative feedback present, and so we must provide it explicitly (circuit C). Resistor R1 does this. The voltage gain is then controlable (-R1/R2), and the feedback greatly linearises the amplifier's response.

Sometimes we do not need any feedback at all, especially if we are switching something on or off. In switching applications we are not concerned with linearity, and we want as much gain as possible, and so we omit the emitter resistor, and provide no explicit feedback path. That's what's happening in circuit D.

Operational amplifiers (except in switching applications) are always connected with external components to provide feedback that determines the response of the system. Transistors, though, can be connected in such a way that negative feedback occurs even though no feedback path is provided explicitly.

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I did not realize that the internal re could be as high as 61 ohms. Also notice that when you remove the capacitor, the distortion is lower because the signal impedance is higher. Often you want to select the two base bias resistors low enough so that the impedance seen by the signal is not affected by changes in beta. You want the base bias resistors to carry more weight in determining the impedance.

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The internal emitter resistance of a transistor depends on its collector-emitter current. It could be very high.

A signal source usually has a lower impedance than a transistor circuit, then the input impedance of the transistor or its bias resistors don't matter.

The current in the base bias resistors should be from 5 times to 20 times the transistor's typical base current to allow for variations in beta and in temperature.

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Zeppelin pointed out to me in a nice and discrete PM:

A is a common-collector (emitter follower) & B is a common-emitter with emitter resistance.


That's a good point, and he asked me to clarify it. In my explanation I suggested that circuit B is a small modification to circuit A, insinuating that B is also a common collector setup, with a slight change. But, as Zeppelin says, the modification isn't really that trivial. It's quite a leap from A to B, with the two circuits operating in fundamentally different manners.

Given that the output is derived from the collector (in contrast to circuit A), circuit B is technically a common emitter configuration with a resistance present in the emitter for feedback. From this point of view, its behaviour can be described starting with a common emitter setup, introducing a resistance in the emitter to diminish and linearise the response to base voltage fluctuations.

I agree with Zeppelin in this sense, that classical transistor theory teaches this, and it is not wrong. I teach this too. But here (on this forum) I enjoy some license for deviation from the classical approach. Since the subject is inherent negative feedback in transistor amplifiers, I found it helpful to treat circuit B as a common collector system (an emitter follower), thus:

Given that for circuit B to work, the collector voltage must always be greater than the emitter, and the emitter voltage will still follow the base. Thus this configuration is still an emitter follower at heart. The introduction of a collector resistance capitalises on this behaviour. It merely develops the emitter current (resulting from the buffered base voltage appearing at the emitter) into a voltage which is proportionally larger (in most cases) than the emitter voltage.

Which approach is best? I think flexibity is more useful than any rigid textbook approach, but as Zeppelin notes, sometimes flexibility comes at the cost of technical correctness.

He is right. Circuit B is a common emitter configuation.

I would ask this though - take a phase spilitter, where RE and RC are identical, and outputs are derived from both emitter and collector. Is this a common emitter or common collector circuit?
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Practical Benefits of the Negative Feedback
Most practical amplifier circuits use negative feedback for the following practical benefits:

  • Stabilization of voltage gain
    Decreasing output impedance
    Increasing input impedance
    Decreasing distortion
    Increasing bandwidth



How NFB decreasing Zo, increasing Zin and bandwidth?
thanks
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How NFB decreasing Zo, increasing Zin and bandwidth?

The output impedance of a common-emitter transistor circuit without negative feedback is the collector resistor. Add a load and the signal across the load drops because of the current divider of the collector resistor and the load resistance. If there is negative feedback then when the output level tries to drop due to the current divider, then the negative feedback is decreased resulting in higher gain.

The same thing happens with bandwidth. When the output tries to drop at a high frequency then the negative feedback is also reduced which results in higher gain.

A bias resistor from the collector to the base of a common emmiter transistor circuit provides negative feedback but it decreases the input impedance. An unbypassed emitter resistor as negative feedback increases the input impedance because it is multiplied by the current gain of the transistor. 
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I've seen the constant current source in the opamp schematic. It's kind of a rudimentary one fashioned from a single transistor. Now, I thought it was just used as a general compliment to the design. I thought the high gain was due to driving the two or three bases simultaneusly. A constant current source at the collector of a transistor is not being driven by the signal.

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The signal at the collector of a transistor compresses half the signal because the collector resistor can't provide enough current. Therefore a constant current source provides a continuous current to stop the compression. The compression is distortion.

The voltage gain of a common emitter transistor is the collector load impedance divided by the internal plus external emitter resistance. Therefore since the current source is an extremely high impedance collector load then the voltage gain is very high.

Then the circuit has low distortion and high voltage gain.

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