@Tech SUPPORT I didn't say transistors, I meant transistors specifically as an AC amplifier. And also said semiconductor electronics, where a synchronous generator is an AC machine

hi austin

hi helo g

I never understood why some people use two resistors and a capacitor until you explain it the way you just did Bravo I appreciate that bit if knowledge .. question what if somebody added a inductor what would that do?

@@SheikhN-bible-syndrome Adding an inductor in series (instead of the capacitor in parallel) would have the opposite effect, i.e. there would be a roll-off frequency above which the gain would decrease as the frequency increases. The roll-off frequency will be the frequency at which the reactance of the inductor is equal to the emitter resistance: f = R/(2πL).

@@RexxSchneider ah that makes since. What about leaving the resistor from ground to collector like normal and then bypassing it with a coupling transformer? That would let AC through while blocking DC , or would it have roll off as well? And is the 2 resistors and a cap the best approach ?

@@SheikhN-bible-syndrome Transformers are a more complicated subject. To DC they are of course just the resistance of the primary winding, which could be quite low. But their AC response depends on what is connected on the secondary winding. If that was purely restive, then at frequencies where its inductive reactance is large compared to its DC resistance, the primary would have an effective impedance of the resistance across the secondary multiplied by the square of the windings ratio. The 2 resistors and capacitor in the emitter is the preferred configuration when you want the operating point to be very stable across temperature variations. That means you make the DC gain a lot lower than the AC gain. Otherwise, a single unbypassed (i.e. no capacitor) emitter resistor, biased to drop around 250mV will usually provide a good compromise between maximising gain and output swing, against retaining a stable operating point over temperature swings.

@@RexxSchneider one thing that i dont get is that if there is a resistor or 2 on the emitter and a resistor on the collector and they are pretty large number. In order to keep the DC low to help with the over all efficiency my confusion is this How does it help efficiency ? Wouldn't it limit the amount of AC signal that they transistor can make because it can't get enough DC power to then convert into ac? So wouldnt it increase your AC current by lowering the value of the resistors allowing more DC in?

hi sandeep

To be honest, circuit design can be quite hard at times. In our previous video we have learned that the emitter resistance has an infuence on the allowed input resistance of the circuit and therefore depends on the actual gain of the transistor. Try to reduce the emitter resistance while keeping the gain ratio constant and you should be fine. I'd advise something like an R_E of 20 Ohms and a R_C of 200 Ohms and don't forget to attach a capacitor of about 1uF to the input. Also it's always good to use a simulation tool (like LTspice) before you design a circuit.

The minus sign indicates that the voltage is 180 degrees out of phase. If one were to apply an alternating voltage to the base, the output would appear to shift 180 degrees, i.e. the pattern of peaks and troughs reverse. In this type of circuit increasing the input voltage leads the output voltage to drop. Conversely reducing the input voltage causes the output voltage to rise. Remember the greater the voltage drop between VCC and the output, the smaller the output will appear when measured with reference to ground. As an example, assume the VCC is set to 12V. If the input signal causes a voltage drop across RC of 8V then the voltage measured between the output and ground would be 4V. If the input is then reduced, giving only a 7V drop across RC, the output would rise to 5V with respect to ground.

Simon Moring is right. There is an explanation at 3:36 just a few seconds before your question arises. The transistor is a current amplifier. A rising voltage at the base will cause the output current i_C to rise. A higher current i_C leads to a higher voltage (v_RC) at R_C. If the supply voltage V_CC is fixed, the output voltage v_OUT must therefore decrease, since V_CC = v_RC + v_OUT. In short: a rising input voltage leads to a falling output voltage. Input- and output-voltage are therefore 180° out of phase. We hope this helps!

@@imooxat Wow, that explanation was good

@@magical261 Thank you

No, the voltage change at the emitter resistor (v_E) is the same as the voltage change at the base (v_BE). The collector current i_C is the amplified base current and is nearly the same as the emitter current i_E. If i_C=i_E the voltage gain has to come from the ratio of R_C/R_E. Only if R_C is higher than R_E you will see a gain in voltage and this gain can only be seen at the collector. The voltage you see there will be negative though, compared to v_BE.

Yes, it is a small, time-varying signal, but it doesn't have to be sinusouidal.

This question is hard to answer, because such a value typically does not exist. In general, if you choose an appropriate opperating point (like it is shown starting at 10:08) and choose a very high gain (high collector resistor R_C and no emitter resistor R_E), you could amplify very small signals. But as it is discussed at 15:57, this comes at the cost of linearity. Also if the input signal is very small you could run into trouble with noise at the input. All in all your output signal might not look anything like the input signal, if the latter is very small and the gain is very high. In this case one might use other possibilities, like a second stage at the output [1] or other amplifier circuits, like an operational amplifier [2]. [1] wiki.analog.com/university/courses/electronics/text/chapter-10 [2] kzfaq.info/get/bejne/iamna91lvLLVkYk.html

Again we want to focus on the time-varying AC-signals written with small letters (v_out, v_RC and i_C), not on the DC-values written in big letters like V_CC.

Saturation effects happen when you choose the biasing point too high, like it can be seen at 11:28. This is only useful when you want to use your transistor as a switch, but not when you use it as an amplifier. You can find more information on using a bipolar transistor as a switch here: www.elprocus.com/using-transistor-as-a-switch/

Here, the phrase "saturation current" is used very differently from when a transistor is in saturation, and it's likely to cause the sort of confusion that the previous answer suffered from. It's in fact the transistor equivalent of the "leakage current" in a diode, and will typically have a value of around 0.01pA, although it will vary from device to device. It is used at 13:27 in the Ebers-Moll equation as I_cs. For example, if you set I_cs=0.01pA, V_be = 650mV and V_t = 25mV, you get 2mA. Although nobody seriously uses that sort of calculation for discrete design, when it's far easier to simply make use of an equivalent internal base resistance approximately equal to 25mV/Ic.

You probably mean the circuit at 17:39 with the bypass capacitor. A capacitor behaves as an open circuit for DC, which is why R_E can guaranty a stable DC bias point (in the large signal view of the circuit). For small-signals (AC-signals) the capacitor has a lower impedance, which in turn allows a high gain for these signals, since the total impedance of R_E in parallel to C_E becomes small.

"Delta" describes the CHANGE of the corresponding voltage. As can be seen in 2:24 the voltage change of V_be produces the same voltage change in V_e, since the input is already biased with a steady DC voltage V_BE.

This is a small-signal analysis. Small letters stand for alternating values without any DC-offset. In this case v_E really is v_B, because the 0.6-0.7V offset because of the transistor is a DC-offset. Just look at the graph at 1:12. The alternating voltage stays the same, as long as the DC-offset at the base is high enought not to cause clipping.

The internal resistance r_E depends on the transconductance gm, which varies widely and has a typical range of 1 to 400 millisiemens. If the external resistor R_E is small compared to the internal resistance r_E, the total gain of the circuit will change with collector current i_C. Therefore a larger external resistor R_E is often chosen even at the expense of gain to ensure that the internal resistance r_E does not influence the gain of the circuit too much.

Why r_E=Vt/Ic??

So you mean Re is chosen high ?

@@attarpattar V_T is the termal voltage and is a physical property of the transistor. You can find additional information in these handy notes, which might answer most of your questions: www.ee.ic.ac.uk/pcheung/teaching/aero2_signals&systems/transistor%20circuit%20notes.pdf

@@attarpattar R_E is chosen relatively high to r_E, yes, but at the sacrifice of gain.

I totally agree, but it's not easy to draw this line. If you are interested in more videos, we have a new channel and update our videos weekly. Ther also will be a video on differentlial amplifiers soon. kzfaq.info

@@ife.tugraz Ah, TU Graz's own channel! I hope you, or TU Graz staff or researchers talk about the so-called 'current mirror' and 'current source' which 'stabilise' lousy transistors and their unmatched-ness. My purpose is to give old school transistors chances because I had been playing with operational amplifiers too much that I become bored. There is nothing much to learn. I mean, there is too much about an op-amp that I yet to discover but the all-encompassing chip called 'integrated circuit' does whatever I need it to work. Heavy work had been lifted by the person or persons who designed the first integrated circuit so that successive generations would be more and more lazy and stupid (Anti-technology group such as my grandparents and parents always say!). I do not have extra money to spend on advanced chips and I do not have an income, so I think this is the best time for me to dig out components that I have left when I was active in electronics, and learn something new and advanced during this insane pandemic. I have already built three op-amp instrumentation amplifier. Believe it or not, I learnt so much just by building it! Stuff that I never learnt in school! I guess those are truly advanced and overly sophisticated for schools. For example : 1. The mysterious noise from resistors. Use small value resistors, as small as humanely possible while not creating a wire/short circuit. 2. Use massive capacitors whenever practical. High voltage rating. Not Farad. 3. All components are cunningly inductive AND resistive AND capacitive devices. Now I sound like a conspiracy theorist!

@@realchristopher4334 I agree, that you cannot learn electronics better than when you're building and repairing circuits. Your observations are good. It's true that very large transistor values often cause problems with noise in OPamp circuits (it also depends on the voltage noise and the current noise of the OPamp itself of cause). Also the issue of high frequency behavior of passive components will soon be discussed in our videos on electromagnetic compatability very soon.

@@ife.tugraz How very odd. I went to kzfaq.info as you provided but there is no one there to respond. That channel is less actively managed than this one? That is the first time I saw Nikolaus. After one week, I revisited that channel, I see that pretty boy again! Hahahahahaha! Sadly, the clock thing that he is explaining is beyond my ability to understand because not applicable for me and I do not own an oscilloscope. Having said that, one of my projects is making a lousy and simple oscilloscope with Arduino. The codes are already there, wiring should be easy (except buffer/voltage follower must be DIY), I hope not much power is sucked from my remaining little brain cells.

@@realchristopher4334 We did see your comment, but we didn't quite get the question. We are glad to answer any questions regarding the content of our videos though.

Yeah that bugs me too