Semiconductor Devices: Voltage Divider Bias

[@Laay_]

This is so well explained I wanna cry. Thank you!!!

[@anorthernsoul153]

Quite possibly the best explanation of BJT bias circuits that I've seen. Thank you so much prof! 😊

[@ElectronicswithProfessorFiore]

You're welcome.

[@davidluther3955]

THE BEST EXPLANATION I HAVE SEEN.

[@adambradley3284]

So good.... so clear...so concise. Also, I wish to add an extract from a Fiore response 2 years earlier, because I could not agree more; Fiore quote "I advocate returning to earlier material once later material is largely understood (because with the more advanced knowledge, you can better appreciate the subtleties of that earlier work that probably flew past you)." Lecturers in the 1980s attempted to impart this knowledge to Me, and although I left the institution with a Certificate (barely), watching Prof Fiore's videos on The Basics have had propound effect on my understanding of the more advanced circuits that I thought I already understood.

[@simonyoungglostog]

I'm enjoying your explanations. This is a subject I come back to every now and again. I remember the confusion when I first learnt it and I'm glad that you are going through the maths line-by-line. Getting a student to think about the maths too much makes them miss the point sometimes (in my experience). The Voltage divider starts with a Vth and Rth 'Unloaded". The load/unload part was what had me fooled in the beginning. Thanks again.

[@simonyoungglostog]

Still one of the best explanations around. As Beta increases, the difference between the Thevenin calculations (VB=VTh) and the actual numbers get closer. If we say VBE is 0.66 rather than 0.7 (Where we account for the RTh voltage drop), Ic works out to 1mA, same as the Thevenin calculation.

[@ElectronicswithProfessorFiore]

Certainly, Vbe=0.7 V is an approximation. It might very well be 0.66 V in some particular circuit.

[@whatever3041]

I created a voltage divider bias CE circuit and right next to it its Thevenin Equivalent circuit in Ltspice but both are giving different currents are voltages. So they aren’t really equivalent.

[@ElectronicswithProfessorFiore]

Assuming these are not very minor discrepancies due to rounding or finite precision, the chances that there is an error in the equivalent circuit or layout in LTspice are considerably greater than Thevenin having been wrong.

[@whatever3041]

​@@ElectronicswithProfessorFioreI figured out the mistake I made, thank you! You are very kind to reply so promptly.

[@arianagm2332]

​@@ElectronicswithProfessorFiore😃 ...I was actually just wondering what Thevenin must've been working on! I'll look it up!!!

[@torikenyon]

Mathematically I understand why a change in Beta won’t have much of a change in how the circuit behaves. In the equation for Ic, changing beta will only change a relatively small value. However I’m having trouble understanding intuitively why this is. I remember someone saying that the emitter resistor acts as negative feedback to changes in beta, but I didn’t really get what that meant.

[@ElectronicswithProfessorFiore]

The emitter resistor is sometimes referred to as "local feedback". Here is a quick example of how it works. Let's say that a voltage divider establishes a certain base voltage. That voltage, minus Vbe, drops across the emitter resistor, and that established the emitter current via Ohm's law. At this point, everything is stable, there being a certain base current established by beta (approx. Ie/beta). Now let's assume something causes beta to rise, for example, an increase in temperature. This increased beta, times the existing base current, should make the emitter current rise. As the current rises, the voltage drop across the emitter resistor must also increase (due to Ohm's law). This means that the emitter voltage is rising, BUT because the base voltage is fixed by the divider, that means that Vbe must be reduced, and any reduction in Vbe would force Ie back down. In reality, this happens on a continuous basis and there will be small shifts in both Vb and Ie, but those changes will be much smaller than the change in beta (in a well-designed circuit). It would not be weird for a doubling of beta to produce only a few percent change in Ie. What ends up happening is that most of the change is seen in Ib. That is, if you double beta, instead of Ie doubling, Ib cuts cut in half. This is why it is important that the divider current be much larger than the base current (otherwise, the changing Ib would load the divider excessively, causing Vb to shift).

[@torikenyon]

@@ElectronicswithProfessorFiore Well said, thank you!

[@lattehour]

16:00 voltage between 2 points is independent of the path .....

[@rudygomez6996]

cool

[@waynegram8907]

PROFESSOR FIORE, What is confusing is that Vth is 4vdc and Rth is 4K. If you take Vcc 12vdc/Rth 4k = 3vdc ( not 4vdc ), this is confusing because Rth is not the same as R total? Can you make another video about voltage divider biasing because there are 3 or more methods designers start with. They start with either #1 HFE beta , #2 datasheets IC specs or #3 The Resistor value gain ratios. By looking at the resistor values you know the gain ratio so the designers will use that as a starting point or they use the HFE beta as the starting point or the datasheets IC current specs. The Ib current formula changes for each method.

[@ElectronicswithProfessorFiore]

Dividing 12V by 4k does not yield 3V. Look at the units. It yields 3 mA. As far as design is concerned, we start with some desired performance characteristics of the amplifier, such as voltage gain, signal swing, and input impedance. That will give us an idea in terms power supply values, r'e, and the like. From there, we can set the desired bias levels and resistor values. Admittedly, this seems a little backwards because students have yet to study the AC analysis of these circuits. But you can't do the AC design and analysis until you can the DC part, so we study the DC part first. Thus, the student is left trusting that the instructor is not leading them astray. Not the best situation, admittedly. It is for reasons like this that I advocate returning to earlier material once later material is largely understood (because with the more advanced knowledge, you can better appreciate the subtleties of that earlier work that probably flew past you). But more to your point: Say you want a gain of 20 and you know the load impedance is 10k. A rough approximation then indicates that the total AC emitter resistance would be somewhere around 500 ohms (ignoring the effect of the collector bias resistor for a moment). That 500 is split between the transistor's r'e and whatever amount of unbypassed emitter resistance you have (which is partly dictated by the sort of distortion specs you're looking for). Once you have a target value for r'e, you can determine the required DC emitter/collector current. From there, you can determine the voltage divider that you need (given some knowledge of the required input impedance). If some of what I just wrote sounds confusing or unfamiliar, that would be normal as it's part of the AC analysis. Probably not the answer you wanted to see, but at this point you have to trust that the instructor is leading you to a place where everything comes into focus. "Wax on! Wax off!" ;-)

[@waynegram8907]

@@ElectronicswithProfessorFiore I meant to say that 12v/Rth4K = 3mA of Ib base current which doesn't make since if Ic is 1mA. What I'm saying is Rth is not the same as R total. You didn't show how to compute for Ib base current. You should make another video showing if the criteria with a load impedance and gain criteria how to design the correct resistor values. Example if the load impedance is 100 ohms and want a gain of 5, how to calculate the resistor values. The +VCC set at +9vdc and then at +12vdc would change the whole resistor values to get the same result.

[@ElectronicswithProfessorFiore]

@@waynegram8907 You're confusing the base current with the divider current, a common situation. Ib would not be 3 mA. Base current is always collector current divided by beta (or approximately emitter current divided by beta) assuming the circuit is not in saturation (in which case the effective beta collapses). In order to maintain a stiff voltage divider, the current down through the divider resistors needs to be much larger than the base current. In other words, the vast majority of the current flowing through the top divider resistor enters the bottom divider resistor with very little being shunted to the base. Don't forget KCL at that junction. Thus, if you have Ic=1mA and beta=100, then Ib must be 10uA (far smaller than the divider current). Calculations for voltage gain for a circuit like this can be found further along in the playlist.

[@waynegram8907]

@@ElectronicswithProfessorFiore The Thevenin Equivalent is Rth which when using the formula Vcc/Rth = Divider current? ( not the base current ). Only Ic/HFE beta = Base current. When you take the Collector Resistor value and use Vcc/Rc = Ic is not the collector current Ic? because the correct formula is Vcc/Rc+Re = Ic to get the collector current? Then once you have the Ic you compute for Ib base current.

[@ElectronicswithProfessorFiore]

@@waynegram8907 Vcc/Rc is not Ic. Vcc/(Rc+Re) is the collector saturation current (max Ic, not the ordinary bias current). The divider current is approximately equal to Vcc/(R1+R2), i.e., the sum of the divider resistors. I suggest that you rewatch the video, forgetting all of the assumptions you have mentioned.