Setting the operating point in a BJT amplifier

[@elwood.downey]

Best description of these options I've ever seen. Many thanks.

[@microled2012]

Very good video... although it seems like an "old topic" for the new generations, they are the base... the foundation of solid state electronics. Greetings from Colombia, South America

[@sc0or]

Not at all. Because there are no op amps that can be connected to an output stage directly. If we are talking about a sound amplifier, you can find an opamp only as an input differential stage, and zero monitor. The rest stages are made of discrete components. Of course if are not talking about LM-like all-in-one amplifiers PS D-class is another thing.

[@RexxSchneider]

You're missing a trick. The most stable circuit is when you use a collector-base feedback resistor divider _and_ an emitter resistor. That arrangement can easily improve both the temperature and β stability of the collector current by a factor of 4 or more even with as little as 250mV across the emitter resistor, compared to the circuit with fixed base bias divider resistors. That maximises both the gain and output swing at the cost of lowering the input impedance.

[@eugenepohjola258]

Howdy again. May I suggest a presentation of how emitter-bootstrapping increases the input impedance. Regards again.

[@peasant8246]

Excellent video, thank you.

[@t1d100]

Good information.

[@Mike-H_UK]

Whilst this is all electronically correct and good for illustrative purposes showing the benefit of negative feedback, it makes no sense to set the bias points without paying any attention to what the voltage gain of the amplifier needs to be. The voltage gain is determined by Ic, Rc and often to a lesser extent Re which in turn alter all of the component values that have been determined here. Regardless, the points about hfe sensitivity and temperature sensitivity are well made.

[@FesZElectronics]

Of course, it requires a lot of consideration to determine the intended bias point - as you said based on linearity, gain and so on. My objective however was to separate topics - I wanted to look at the step you take after figuring out an optimum bias point. I may treat the topic of bias point selection some other time.

[@RexxSchneider]

No, the maximum usable voltage gain depends entirely on the supply voltage and to a significant degree, nothing else. The most you can get out of a common emitter stage without distortion becoming significant is Vcc / 500mV. I'd be happy to show you the derivation if you don't already know it.

[@Mike-H_UK]

@@RexxSchneider For the general configuration of the common-emitter amplifier, the voltage gain can be closely approximated by -gm*RL for the degenerated case and -gm*RL/(1+gm*RE) for the non-degenerated case. My point was that if you choose RL and RE for an optimum bias point, as per this video, the resulting gain is very unlikely to be where you want it. So both operating point and gain need considering together. I take your point that for the specific case where gm*RE

[@RexxSchneider]

@@Mike-H_UK You'll notice that I stated "without distortion becoming significant" when I quoted Vcc / 500mV. For the general configuration of the common emitter amplifier, the voltage gain is indeed - Rc / (Re + re) where re=1/gm - in other words the slope of the V-I curve for the base-emitter voltage. That intrinsic resistance can be shown to be equal to 25mV / Ic. The operating point has little effect on gain as it is normally targeted at somewhere around half of the supply voltage. If the load resistance is Rc, and about Vcc/2 is dropped across it at no signal, then the collector current must be Vcc / (2 x Rc). If you have no emitter resistor, then the gain would be -Rc / re = - Rc / (25mV / Ic) = - Rc x Ic / 25mV which depends directly on the instantaneous collector current. That yields a higher gain for the negative-going part of the output signal than for the positive positive-going part which results in a nasty distortion of anything other than the smallest of signals. This would be the case for your calculation, where - Rc x Ic / 25mV = - Rc x (Vcc / (2 x Rc)) / 25mV = Vcc / 50mV, but that's not usable. You reduce the distortion to acceptable levels by having an emitter resistor of the order of ten times greater than the quiescent intrinsic emitter resistance (that's, of course, where my extra factor of 10 comes in). In that case, we set around 250mV across Re to make it around 10x bigger than re. The gain can then be approximated to - Rc / Re = - Rc / (10 x re) = - Rc / (10 x 25mV / Ic) = Rc x Ic / 250mV = Rc x (Vcc / 2 x Rc) / 250mV = Vcc / 500mV. Of course, you can increase that by a tiny amount by lowering the quiescent bias point, or by lowering the emitter resistor a little (at the cost of noticeably increased distortion), but I stand by my claim that you won't get a usable gain of more than about Vcc/500mV out of a single common emitter stage, and it's a good rule-of-thumb to design with. If you want a gain of 50, you're realistically going to need a supply voltage of at least 25V, and so on. Cheers!

[@Mike-H_UK]

@@RexxSchneider Thanks! What you say sounds fair enough and the calculation is the same as mine except you add ~10*Re for degeneration to reduce distortion. As a matter of interest (or because you seem genuinely interested in amplifier analysis), my circuits tend to have small signals (

[@lmwlmw4468]

Great video.

[@eugenepohjola258]

Howdy. Nice. I would suggest some modifications I have used with good results. 1. The basic base current biasing. To avoid thermal avalanching I use an extra reistor after Vcc. The base resistor and collector resistor connect to this extra Vcc resistor cold end. Then I use a suitable capacitor from this cold end to minus. I sacrifice 1 to 2 V of the Vcc in this extra resistor. If the total current tends to grow so will the voltage drop over the Vcc resistor and the base current will go down. 2. The base resistor to the collector version. The thermal regulation is good but much of the transistors raw gain is sacrificed. Also the neg. feedback tends to lower the input impedance. I suggest splitting the base-collector resistor into two parts in series. Then I suggest connecting a suitable capacitor from the split point to minus. This will eliminate the neg. feedback but not influence the thermal stabilization. In regards.

[@RexxSchneider]

It's not the case that a collector-base divider network sacrifices any significant gain. Try either building or simulating a circuit with a 10V supply, a 4.7K collector resistor, a 270R emitter resistor and a 100K - 10K divider to bias the base. You need a transistor with a β of at least 100. I've built and simulated it with a BC547B (β≈300) and BC547C (β≈500), You'll get a collector current around 1mA, an input impedance of 8K and an ac gain of 15.6. Now remove the 100K upper base bias resistor and put a 47K feedback resistor from collector to base. You'll find that collector current is roughly the same, the input impedance is 2.3K and the gain has dropped 1dB to 14. However, the collector quiescent voltage now only changes by about -7mV/°C instead of the -28mV/°C you get with the straight resistor divider circuit. There is also a drop in distortion by about a factor of 4 as well as a reduction in the output impedance. Now, of course, if you have a guaranteed high-β transistor, you can increase the values of the base-bias/feedback resistors and raise the input impedance if it's an issue. You lose a little of the extra temperature/β stability, but you're always better off than with the straight resistor divider base biasing.

[@eugenepohjola258]

@@RexxSchneider Howdy. Ok. Thanks for the thorough comment. Regards.

[@microled2012]

There are very good books that give us very good knowledge on this topic of transistors, their configurations and the calculation of their resistive and capacitive components, these books are both in their English and Spanish editions: "Electronics Devices by Thomas L. Floyd" ; "Electronic Principles by Albert Malvino", "Microelectronics Circuit Analysis and Design by Donald D. Neamen".

[@pluscrafter7117]

Will there be a series with Mosfets about amplifiers and biasing?

[@FesZElectronics]

Indeed! Mosfets is part 2 :D

[@stefano.a]

In the description are assumed some value (for Ic, Vce and voltage on Re) not explained. The problem in this kind of design is to keep the amount of stabilization feedback as high as possible without lowering too much the gain of the circuit and for avoiding thermal runaway.

[@RexxSchneider]

Small signal transistors don't suffer from thermal runaway as long as you design their collector current to something sensible. I'm not sure what kind of application requires 50mA collector current, but you really don't need that much current to design a single stage common emitter amplifier. Most of the time, you're far more likely to be working with currents around 1mA. With a 10V supply, that will allow a usable voltage gain up to about 20x and an output impedance of about 5K. The collector current determines the output impedance, so if you know what is needed to drive the next stage, you can work out the collector current you need to set the collector voltage somewhere about half of the supply voltage. The gain of the circuit is given by the voltage across the collector resistor divided by the voltage across the total emitter resistance. Because the transistor has an non-linear, internal emitter resistance of 25mV divided by the collector current, we normally want to set no less than 250mV across the emitter resistor to "swamp out" the changes in the transistor's internal resistance and reduce distortion. So the maximum gain you can reasonably use with a 10V supply is (half the supply voltage) divided by 250mV = 5V / 0/25V = 20. If you need less than 20 gain, you can use more voltage across the emitter resistor. Now you know the emitter current (= collector current) and the voltage across the emitter resistor, you can calculate its value. Since you now know the voltage across the emitter resistor, you then know that the voltage at the base is 0.65V higher. You know the collector current and the value of β for your transistor, so you know the base current. It is then just a matter of picking two resistors to bias the base to the voltage you calculated, while passing at least ten times the base current. So starting from a given supply voltage, a required output impedance and a required gain, you can calculate values for Ic, Vce, and Ve, as well as the values of all four resistors, assuming your transistor has a minimum specified β. All of those steps are probably too complex for a KZbin video aimed at covering the basics, but I'd be happy to run through a full practical design in the comments if you want to suggest the parameters (either input or output impedance, and gain) for the required circuit.

[@BIGRIP87]

👍 Nice

[@sanjikaneki6226]

great video as always + better and easier to understand then what was though at uni BTW in your linear power supply video you mentioned that you found a transformer but how did you test it it was good and how did you know how much power it could handle?

[@minahilashraf5617]

If the transistor has been properly biased in active region then which parameter defines the minimum alternating singnal input voltage or amplitude in datasheet?

[@josip1881]

Thanks for another interesting video. Could you build a mini wind turbine from one of those 12V DC motors and create chrager and BMS + maybe some low current inverter for it? Like real wind turbine system, but mini :D

[@bsuryasaradhi6816]

Use appcad software to automatically calculate the values

[@RussellTeapot]

8:47 "Negative feedback is good feedback" YOUR VIDEO SUCKS :D :D Just kidding, the only feedback I can provide here is absolutely positive, very informative video explained in a very easy to follow manner. Thank you very much!

[@genefulm]

I've got a simulation request you might be able to help with Fesz. I've been trying to create a circuit (like a B source) that when a voltage level is attained, linearly ramp a load current for a period of time up to a terminal level. This would be useful for simulating the soft-start turn on load of a buck regulator when the upstream capacitor get to full. I can't figure out how to do the time-dependent current ramp part.

[@FesZElectronics]

You can use "time" as a parameter in the behavioral source. For example in a transient simulation that takes more than 1 second, using "I=if(time

[@csoszadavid4803]

How can you choose the right Beta, If there is no problem everybody would choose 800, What is the method to choose "Beta"

[@FesZElectronics]

The component datasheet will usually give an interval; also the beta is current, temperature and Vce dependent; usually, to be safe, you should determine the realistic extreme values of beta (both maximum and minimum), and make sure the circuit works for both.

[@biswajit681]

Please make videos on control loop of switching converter..there is no good video on the KZbin

[@FesZElectronics]

At the moment I will try to finish the various series that I started - the amplifier classes and various small signal amplifier related topics; at some point I do intend to get back to SMPS topics like control loop.

[@biswajit681]

@@FesZElectronics Thanks 🙏

[@BIGRIP87]

right.

[@s_amoku]

kzbin.info/www/bejne/gmSwg6F7jsmcbsk

[@subramaniamchandrasekar1397]

Based on this video, if people who watched that some info on how to set the operating point in BJT amplifier will be learnt, will be disappointed. A dry talk professor from whom you do not understand anything. Sorry about this remark. Regards.

[@RealRobotZer0]

You are going trough simulations too fast, I think. Can you give us the .asc files?