29. Common Emitter Circuits

Common Emitter circuits can be configured in many ways. Three of these are covered in this video. CE's with no load, with a load, and with split emitter resistors are discussed. .I'll cover the AC parameters and how to calculate voltage gain, current gain, AC saturation and cut off, load lines, phase inversion and a few other things I can't remember.

[@TheZaratano]

1968 City College of New York; EE 104 ? Hard at that time; easy 52 years now with your online tutorial; congratulations, proffe !!! very clear and sharp.

[@robertallen6281]

This is one of the best explanation of the common emitter circuit. Now I have a better understanding of the AC path in the circuit which allow me to see a clearer picture on how the impedance in the circuit is arrived at. Thank you .

[@TheOffsetVolt]

I'm glad it helped!

[@RexxSchneider]

Well done. Very few videos on the CE circuit show any understanding of the problems inherent in fully bypassing the emitter resistor. I was pleased that you showed around 17:00 the changes in collector current when a large output swing occurs. You might have taken the opportunity to note that at 15mA instantaneous current, rej would be 1.67R, giving a gain of 336, and at 5mA instantaneous current, rej would be 5R, giving a gain of 112. When those are compared to the predicted small-signal gain of 224, it becomes obvious how much large output waveforms are distorted. I should correct one mistake on your part: when calculating the bypass capacitor, you want its reactance to be small compared to the _unbypassed_ resistance, because the gain depends on the total impedance at the emitter, which is rej + (Re || Xc). As long as Xc is significant compared to rej, it will produce a frequency dependant gain. So you find that even 100μF in series with the 2.5R will roll off the frequency response below 640Hz, which is why you had to show the gain at 5KHz. Of course, once you have a split emitter resistor with 100R unbypassed, you get a corresponding improvement in low frequency response, which will then be substantially flat down to around 16Hz. Note that if you use a 50R resistor as the unbypassed part, you will have to double the bypass capacitor, showing quite clearly that the response depends on the resistance that remains unbypassed, not the resistor that the bypass capacitor is connected across, which may seem a little unintuitive at first.

[@sbybill3271]

I had to rewatch the video to appreciate what you were saying. Pearls of wisdom!!! I wouldn't have known all this if these comments were not there. Thank you

[@darinmorgan3520]

You have a way of explaining complex topics. I have watched, read and practiced but for some reason your tutorials let this information, "click" and it all makes sense.

[@TheOffsetVolt]

Thank you Darin. Hope I can continue to help in the future.

[@Optinix-gz1qg]

wow great content, keep it up sir!

[@TheOffsetVolt]

Danke

[@stonedDawg]

very detailed explanation, but i still dont get it, if rej = 25mV/Ie, where 25mV come from? thanks for explaining!

[@TheOffsetVolt]

It is a value given from Boltzmann's constant and room temperature. It is thermal voltage, if you will. I discuss in an earlier video in the transistor series but can't remember which one :(

[@stonedDawg]

ok i will look for it! thanks for replying!

[@billwilliams6338]

THE OFFSET VOLT, Does the AC/DC load line mean the maximum output wattage power of the transistor or does the AC/DC load line mean the power dissipation of the transistor? Because the AC/DC load line is selected by the EE designer of the operating voltage and operating current for the transistor and the DC bias Q point which will determine the maximum output wattage power out of the transistor and the power dissipation of the transistor. I'm not sure what the LOAD means in the AC load line and DC load line what they are calling the LOAD.

[@billwilliams6338]

THE OFFSET VOLT, What is the difference between a DC load line compared to a AC load line? The +Vcc voltage and output impedance sets the DC load line and AC load line? but what is a GOOD DC load line and AC load line and what is a BAD DC load line and AC load line?

[@chenghaowang7610]

Fantastic video. Thank you very much. One question. For the third configuration which is not the 'real' common emitter circuit, should the saturation current be different from the second configuration with load because of that emitter end resistor which is not bypassed by the capacitor?

[@TheOffsetVolt]

The DC saturation should be the same (assuming the resistance on the emitter and the base voltage are the same) since the bypass cap only effects AC at that point.

[@kettelenegaspard685]

in a voltage divider transistor circuit can the input be bigger than the output, and also if i have an emmiter follower circuit with 2.5 v out what would be the input for a class A amp

[@TheOffsetVolt]

For an emitter follower, the input would be just over 2.5 since gain is less than 1. You can get you Av from the first stage and the EF wold provide Ai

[@minahilashraf5617]

How to define minimum signal level in voltage that can be amplified by common emitter amplifier?secondly which parameter on datasheet shows output resistance (ro)of transistor

[@kaustubhponkshe6186]

can you make video on single stage tuned amplifier for 27 Mhz

[@Sami-xc5pl]

Why do you use 15v as VCC while on your previous videos about transistor amplifiers commonly use 12v? Also what is the maximum wattage of your amplifier(s)? I've through your videos for a while and keep on replaying them :D

[@TheOffsetVolt]

Hello again, I used 15V just for the heck of it. However, depending on the biasing, a higher voltage gives you a larger swing for any AC output.

[@NoName-yy1jx]

23:20 how come Rc is parallel with RL and why did you assume that the current goes from the Emitter ground to RL ground can you explain this please

[@steevesmith1573]

That’s exactly what I was wondering

[@dextertech6570]

I didn't get how to calculate - r(ej). How did it come as 2.5 Ohm? I need help

[@neelamlashari1875]

Great sir

[@mohammedahmed-rd8xh]

in the first example, What is the value of current gain ? Is it equal to (Bac) or not ?

[@efox29]

Fantastic!

[@edbeckerich3737]

Where in the world did Beta ac = fT/fop come from??

[@emmetray9703]

How did you calculate a capacitors values?

[@TheOffsetVolt]

Hi Emmet, As a rule, you want bypass capacitors to have 1/10th, or lower, value of the resistance you are trying to bypass. If you know the minimum frequency that's is going yo be applied, just rearrange Xc=1/(2piFC) to C=1/(2piXcF). Hope that helps.

[@emmetray9703]

@@TheOffsetVolt Thank you for your reply. Please.. can you recommend me a books where from I can learn electronics , I have some knowledge ( I read Alexander and Sadiku's "Fundamentals of electric circuits" , but it's not enought) . I also know about the book "The art of electronics", but it's too hard for me. Thank You.

[@billyray1172]

@@emmetray9703 I know this old, but I'd recommend "Complete electronics, self teaching guide" it's a good book, somewhere between "Electronics all in one for dummies" and the "Art of electronics".

[@kettelenegaspard685]

What kind of diode control oscillator frequency

[@TheOffsetVolt]

Tunnel diodes and Gunn diodes can be used to control/create oscillator circuits.

[@kettelenegaspard685]

How will be the current through re if rc open

[@TheOffsetVolt]

Hello, There will be no current at the load but a small current will be present through re since the base emitter junction is forward biased

[@kettelenegaspard685]

Is the ac gain different than dc gain

[@snnwstt]

You will have to define what is the "dc gain". If you meant "beta", the answer is no, they are not necessary the same since beta varies with the frequency, for a given transistor. You could also consider that the "circuit" itself is not the same for the dc and for the ac analysis.

[@TheOffsetVolt]

Yes, the DC gain will vary with IC and Vce while the AC gain will vary with frequency. You can see this on a data sheet as DC Current Gain and Current Gain Bandwidth product for DC and AC respectively.

[@kettelenegaspard685]

So what if Rb2 is open, or Re is open

[@snnwstt]

If RB2 is removed, you get a Base Bias ( see kzbin.info/www/bejne/pXfCh3eqe5yjqsk as example ), highly dependent of the beta value. If RE is removed, you don't get a properly biased transistor ( IE = 0, for properly biased, IC =.= IE would also be 0, and you get a blocking transistor, a permanent "off-switch").

[@TheOffsetVolt]

Hello again. If RB2 opens, the circuit becomes base biased. Base current will go up which will cause Ic to increase - Ic would be RB1/(Vcc-0.7)XBeta.

[@snnwstt]

33:17 :-)