Much appreciated. Hopefully I will get better at this. Always looking to improve the material and how it is presented. I need a nice quiet shop!! Oh well. Thanks for the feedback.
Hello, I had to renumber my videos recently so you are probably looking for 27 at ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-Sq8EzHZgQvs.html 28 is at ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-rrp81VpFUd4.html. You can also go to my homepage and look under created playlists, Diodes and Transistors, and this will get you to the series. Thanks for watching.
I tested the base bias cyrcuit with two sources on multisym with BJT transistors 2N4315 and 2N3904, and when i vary Rc or Vcc β changes. So is this because of the model?
So the divider bias gives you something of a current source, which keeps Vc steady. Could you leave off the Emitter resistor? Or is the voltage feedback from that resistor critical?
OFFSET VOLT< when looking at transistor biasing resistor values how can you tell by the RATIO of the resistor values if its a gain of 1, 2 or 5 or 10? I think most transistors use 10mA for the Collector to emitter current as a standard?
When u calculate ve by substracting 0.7 from vb(voltage divider), u ignored the thevinin equivilent resistance voltage drop(tiny compared to voltage on re, if we multiply ie(beta by ib) by re and compare it to ib by the r thevinin and let assume re and r thevinin same, the voltage drop on re would be greater by beta value....and so we can then assume no drop on r thevinin, and analysis is very good correct! hope i am right...
The only way you could calculate Rb is by knowing what current you want for Ib. I don't think we knew that unless it's on the transistor charts. Please explain
You can calculate r'e with a fair degree of accuracy. You can also assume (wisely) the lowest hfe for the device. If you know emitter resistance, add it to r'e and multiply the sum by hfe. This will give you the lowest base resistance you should have. Because of the variability of hfe (Beta) it's hard to get an exact value of base resistance. Thanks for watching.
Yet another fantastic, informative video! (As always!) One question, however; in the base-bias circuit, If there are 14.3 volts on the base, how does it ever come OUT of saturation?
Hello, It is the current that counts in this circuit. The base resistor fixes the DC current on the base. This current is amplified by beta to get collector current (and really close to emitter current). The circuit could saturate or cutoff if the AC current that is applied increases of decreases the current to the point of saturation or cutoff.
Ah, I think I see - so, even though the voltage is that high, the current is low enough to bring it into the active region? Thanks, by the way, for the prompt response, and, of course, for all of these INCREDIBLE videos! I've learned SO much from them & I always look forward to the NEXT one!
The base bias circuit is erroneously drawn. The ground is connected towards the base which is wrong. It should be the posive terminal of the supply going to the base thro a resistor.
You mean with DC, only a single source, Vcc, such that the current flows from a value of Vcc greater than Vcc(min) but no current flows if Vcc is less than that value? Sure. You have to get Vbe < 0.7 to be in cut-off (the PN junction between the base and the emitter is then in cut-off) , and since, here, Ve = 0, that leads to Vb < 0.7. But Vb, by the tension divider, is equal to Vcc * Rb2/(Rb2+Rb1). Make that expression equals to 0.7 (the limit for the cut-off), compute the Vcc value by solving the equation (Rb2 and Rb1 are assumed known values) and here you have the "Vcc(min)" value for the limit cut-off/active. Sure, 0.7 is an approximation, can be anywhere between 0.55 and 0.75. And the resistors can be unprecise (10%? 1% …) as well as being themselves temperature dependent (decreasing as temperature increases for carbon and silicium and so, graphite based resistors, increasing as temperature increases for copper and most metal with good conductivity, including metallic based resistors).
In general though, it is more usual to have another signal, Vbb, acting as a command to the switching transistor, with either Rb1 = infinite, or a capacitance coupling (such as for amplification), if the switching has to be "fast" (short duration), or other techniques (such as starving the transistor to turn it off), than using "basically" a voltage divider as main tool for switching. And FET and MOSFET are easier to use, with less induced noise, if only switching is required.
Hello, There would be no current through RE which would mean no Ic. You would end up reading Vcc from collector to ground. Hope that helps and thanks for watching.
The beta of your transistors will vener ever cohere with the one you read bout in the specsheet. In other words. You gegerally have to calculate the correct Ice yourself. Wich is luckily very easy. Measute the Ib and divide the obtained Ice by that value and you have your gain.
Hello , Absolutely true. measuring each transistor in a production line is tough so the designed will use either the geometric average of the beta or the worst case beta from the spec sheet. this is what I went with. Thanks for your feedback and for watching.
I know, I did the entire video and caught the error at the end. Rather than shoot and edit everything again, there is a callout at the beginning telling everyone the polarities on the power supply are wrong and must be inverted. Thanks for watching.
