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In my book I've seen many assumption done for to analyse BJT circuit. Like; Thumbs rule (when we assume Vce~ Vcc/3, Ic*Rc ~ Vcc/3) or assumed Ve is one tenth of supply voltage in another analysis (where it says that it is in the range from one fourth to one tenth ). This assumption are taken in the process of circuit design where Rc and Re along with other many values are to be determined. Again in this example one answer is given using assumption that under quiescent condition collector current voltage will be half of Vcc.

Q1: how do I know When I do which kind of assumption? Is there any rules?

Q2: My friend said that I can make assumption during circuit design but when I'm solving one I can't do assumption. Is it write?

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2 Answers 2

I have never used any of the assumptions you mention, so the answer is "never". There are some simplifying assumptions that make designing with and analyzing BJT circuits easier, but the ones you mention are very dependent on the circuit topology, and probably wrong a lot of times even when the applicable circuit is used.

As always, there is no substitute for actually understanding the components you are using. Designing by assumptions and rules of thumb will get you into trouble, exactly because you then don't know when they are applicable and when not.

Here are some simplifying assumptions you can make for BJTs much of the time:

  1. The B-E voltage is around 600-750 mV when the transistor is on. Basically, the B-E junction looks like a diode to the circuit. Which end of this range to use depends on what the transistor is doing and how hard you are driving it. A small signal high-gain transistor can start to come on at 600 mV or even lower in very low current applications. Figure the 750 mV when driving the transistor into saturation at a good fraction of its maximum current.

    Of course this is just a simplifying assumption and not a rule. Often a 100 mV difference doesn't matter, so this is close enough. When it does matter, you need to be more careful and this simplifying assumption isn't useful.

  2. The gain is infinite. The gain of real transistors can vary quite widely even within the same model from the same manufacturer. Note that datasheets generally only guarantee the minimum gain at a few operating points. Actual gain being 10x higher in some cases for some parts is not out of line. Good BJT circuits therefore work with the transistor having the minimum guaranteed gain for the operating point, up to infinite gain. Analyzing a circuit at infinite gain is often a good starting point, since it has to work at that point anyway. Then you can see how much margin there is and look at what happens incrementally as the gain is changed from infinite to the minimum guaranteed.

    Note that infinite gain means the base current is zero.

Again, these are simplifications that let you do a quick analysis. Often the insight gained from that analysis is sufficient for the purpose. When not, it is still often useful to start with the operating point from the simple analysis and change it incrementally as a result of making the parameters more realistic. This is essentially a iterative solution technique.

However, in other cases these simplifying assumptions are too simple and therefore not useful. If you understand how the devices work, then you should be able to see for yourself when the simple analysis is good enough. Eventually you'll gain some intuition that shortcuts the process of knowing what can be applied when.

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I'm assuming a common emitter amp with emitter resistor and potential divider on the base to bias it. My two-penneth in addition to Olin's notes: -

  1. If the output is at the collector it's reasonable to assume (in many cases) that the quiescent operating (DC) point of the collector is half Vcc - this allows maximum peak to peak swing before clipping occurs. This "rule" is slightly tempered by "rule 2"
  2. For decent linear amplification you benefit from an emitter resistor and this of course eats into the maximum p-p swing mentioned in "rule 1". As a guideline I aim for an emitter resistor volt drop of about 0.75V - this can vary and be pushed lower if the design requirements are such.
  3. Output impedance from the collector IS the collector resistance so if you want an output impedance of 100 ohms then you better use a 100 ohm resistor - the output load (external circuits) will also be "seen" in parallel with this resistance and that means that a heavy load reduces the gain... this leads on to the next rule of thumb
  4. You have your collector resistor value to suit your load and you choose your collector current to approximately sit the collector voltage at halfway between the positive rail and 0V (this assumes NPN transistors) but, due to (2) it should sit a bit higher (say) half of 0.75V above mid rail. Now you calculate your collector current to make the quiescent collector voltage match (\$I = \dfrac{V_S+0.75}{2\cdot R_C}\$). If your supply is 10V and Rc is 100 ohms then Ic = 54mA. Emitter resistance that caused the modification of 0.75V to the above is 0.75/0.054 = 13.9 ohms - note that I've assumed collector and emitter currents are equal - this is not unreasonable and matches Olin's assumption about the gain being presumed to be infinite.
  5. DC voltage on base - assume about 0.7V for the base-emitter volt-drop and then add on the 0.75V from (2). Assume the current through the biasing resistors is between ten and fifty times lower than Ic - let's say 25 times smaller - this sets the bias current through the potential divider that feeds the base at about 2mA - biasing from a 10V rail will total up to about 5000 ohms and this needs to be two resistors providing a potential divider ratio of \$\dfrac{0.7+0.75}{10}\$ = 0.145 - then calculate the two resistor values that feed the base - you know the pot divider ratio and you know they must add-up to 5000 ohms so the lower value will be 14.5% of 5000 = 725 ohms and the upper will be 4275 ohms.
  6. Inputs and outputs will need capacitors and knowing the collector load resistance and the base biasing resistor values allows you to choose cap values that are suitable for the lowest frequency you wish to pass through the circuit.

Voltage Gain of the circuit is approximately Rc/Re and this will be about 7.2. If you need more AC gain then use a capacitor in parallel with Re.

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