Design a Simple Common Emitter Amplifier

The common emitter amplifier is a simple single BJT circuit that can provide a reasonably large open circuit voltage gain (output is inverted). Learn how to design (i.e., choose resistor values) for a common emitter amplifier in order to get the gain that you require

David Williams

[luisfranco]

To those watching confused about the 26mV, it is the thermal voltage of a bipolar junction transistor. It depends on temperature and is assumed to be 26mV at room temperature.

[EdwinFairchild]

Finally a video where you're not simply asked to find operating point and are asked a more real world problem of designing something as oppose to just finding currents and voltages

[ajaypun]

How did you get 26mV to find the value of R(e) at first?

[RexxSchneider]

A good exposition, but there are a couple of snags.
The value for re is both temperature dependant and non-linear, so the gain of the circuit (Rc/re) will depend on temperature and be non-linear. Also the bypass capacitor needs to have a reactance much smaller than re=6.5R for the frequencies we are interested in. That will be huge if we want to amplify down to 20Hz.
With a quiescent current of 4mA and a collector resistor of 650R, the quiescent collector bias point is just (4mA x 650R) = 2.6V below the positive rail, i.e. at 21.4V. That leaves very little headroom for output swings, a mere 5.2Vpk-pk compared with the 24V available from the supply.

Both of these issues can be addressed by splitting the emitter resistor and only by-passing part of it. That gives a larger effective emitter resistance and allows a larger collector resistance. The cost is a higher output impedance, but that is to be expected for a class-A amplification stage.

If you're willing to lose the 10V of headroom, then you can continue to use the same Rb1, Rb2. That leaves about 14V for the output signal, so you can increase Rc to (7V / 4ma) = 1.75K. To maintain an ac gain of 100, the effective Re now needs to be 17.5R, of which 6.5R can be assumed to be the base-emitter resistance, as before. So you split the 2.325K Re into 11R and 2.3K with the latter being bypassed by the emitter capacitor as before. Note that the bypass capacitor makes a high-pass RC-filter with the 17.5R effective emitter resistance (not the 2K3 emitter resistor), so it needs to be around 900μF to get a -3db cut-off frequent at 10Hz.

Personally, I'd not be so ambitious with the gain and work with a much lower base bias point. A gain of 20 using a collector current of 0.8mA could use a 12K collector resistor (setting the collector to 24.0 - 9.6V = 14.4V) and an unbypassed emitter resistor of 600R. That would swamp out the non-linear base-emitter resistance of about 30R, giving much more linearity because of the negative feedback. I'd suggest setting the emitter at around 4V meaning a total emitter resistor of (4V / 0.8ma) = 5K, made up of 600R unbypassed and 4.4K bypassed. The bypass capacitor would be 27μF for a cut-off of 10Hz. Working at the lower current would allow larger Rb1 and Rb2, so planning for HFE_min of 100, we would have Rb2 <= 100 x 5K / 10 = 50K. The current through Rb2 would then be (4.6V / 50K) = 92μA, compared with the maximum base current of (0.8mA/100) = 8μA. To be accurate we could calculate Rb1 to be somewhere between (24.0 - 4.6)V / 92μA and (24.0 - 4.6)V / (92 + 8)μA or 211K to 194K depending on transistor HFE. That's well within tolerances.

So I'd suggest Rb1=200K, Rb2=51K, Rc=12K, Re=560R + (4.3K || 22μF). Quiescent current ~ 0.8mA and collector biased at 14.4V giving a voltage swing of over 19Vpk-pk with a gain of 20 from 20Hz up. Input impedance would be 40K and output impedance 12K. If you need a lower output impedance, you can either sacrifice gain, or use more quiescent current and sacrifice input impedance. Or, of course, you could just add an emitter follower stage -- but that's another story.

[stephenj.wiener]

I think there is an error in the calculation for Zin. hie which is approximately 656Ω should be in parallel with 28K || 20K. This would make Zin closer to 621Ω not 4.17K as your calculation.

[zazugee]

where did 26mV come from?
i'm supposing it should be 24mV bc of the gain ?

[gurprikshakalra]

Awesome explanation...
This is what I have been searching for quite long.

Can you please explain how to rule out the value 26mV?

[maxp]

Hold on, how did you get your input impedance of 4.17k? If Rb1, Rb2, and 101*re are in parallel to each other, your equivalent resistance should be 622 ohms. It seems you multiplied re by 1001 and posted that number.

[fernandohood]

What is the little re on the diagram? Your talking I am looking at the diagram and suddenly your talking about little re. Where did the 26mv come from?

[jephthai]

Also, another question about the input impedance -- isn't the emitter capacitor a complex impedance, and so shouldn't the input impedance be 36.8+654.4j? If not, what would be the reason for considering it a resistance instead of a reactance? Also, is the choice of the correct value for that capacitor target frequency dependent?

[GrahamLappin]

I know I'm late in the day but... calculating the impedance (around 8 min 20 sec into video) from 28K ‖ 20K ‖ 101x6.5 (which is 656) comes to 621 Ohms? I calculated this using the standard method ie the reciprocal of the sum of reciprocals for each of the 3 resistors. Where am I going wrong in order to get the 4.17 K you calculated? (I see others have asked the same question but no reply, so I remain confused).

Having taught for many years (pharmacology, not electronics ... and that probably shows) I'd just say that it's worth joining the dots for individual steps. I see from the comments the 26mV caused confusion in this respect.

Otherwise I commend you in a very practical approach. So many videos on circuit analysis and yours is one of the few that actually starts with the design.

[CikaDraza]

Wait a minute, how on earth did you get input impedance of 4.17K?

[bearchow]

Way cool got into 26mv/ma for Re (aka Re', bulk emitter resistance, etc.) as a part of calculating voltage gain. I learned it in the early 70's from Tektronix app notes. In my later years a routine question to ask in an interview I drew the circuit and asked the candidate to tell me what the voltage gain was. As soon as they uttered something about beta I knew they weren't there.

[eugenepohjola]

Howdy again.
I was wrong. You and JohnAudioTech are correct. Rc is the output impedance. Then the output signal is ½ the unloaded and most power is transferred. I remembered JAT's reply wrongly.
Regards.

[Dannydawson]

I’m sorry please excuse my question but what exactly does a emitter do ?

[silvertongues]

hi there, why does your collector current (Ic) need to be 4ma, and thus your Ic(sat) be 8ma?

[robertdunn]

Very nice tutorial, much better at AC analysis/design than other common emitter design explanations that I've found. However, I'm a bit confused at the input impedance calculation of 4.17K. That calculation is 20K || 28K || 6.5 * 101 (or 656.5). That would be less than 656.5 and not nearly 4.17K. I did back out your result: I think that you used 6500 instead of 656.5 in the calculation. Am I right or did I misunderstand something?

[JohnLeski-npiy]

It would be helpful if you refer to the spec sheet for transistors. how do you know what I(sat) is?
spec sheet for a 2n2222 doesn't show I(sat). Otherwise it is a very good tutorial. makes me remember what I for got.

[michellenicholes]

when calculating re you used 26 mV. Where did you get this value from?

[demonsheadshot]

The outro with the hexbug, very nostalgic