by michael1978
Tags: amplifier, transistor
 P: 133 can somebody teach mm how to find desired voltage gain of ce a transistor amplifier wich is formula, i know the formula is rc:rl/re but i want for example a voltage gain of 50 for example input desired data: desired voltage gain 2.0 wich i want to know f_min = 10 HZ z_in = 50 K Z_OUT = 2k power suppply 12V now how is the formula to find r1 r2 i know to find but rc and re i dont know to find, that is my problem
P: 3,830
 Quote by michael1978 can somebody teach mm how to find desired voltage gain of ce a transistor amplifier wich is formula, i know the formula is rc:rl/re but i want for example a voltage gain of 50 for example input desired data: desired voltage gain 2.0 wich i want to know f_min = 10 HZ z_in = 50 K Z_OUT = 2k power suppply 12V now how is the formula to find r1 r2 i know to find but rc and re i dont know to find, that is my problem

You want output to be 2K, your rc has to be 2K. You want voltage gain of 2, so you want your re≈1K. Gain=rc/(re+r'e) where r'e= 1/gm≈Vt/Ic. Vt≈25mV at 25deg C. For Ie=1mA, r'e≈25Ω. So if you use 1mA bias current, re=1K is close enough.

If you choose a NPN with β>100, input impedance of the transistor is βXre≈100K, this is going to be a little tricky getting input impedance of 50K as the input impedance of the transistor is as low as 100K or higher if β is larger.

What is r1 and r2? is this a voltage divider to bias the base of the BJT? If so, you want Zin=50K, then you want r1//r2=100K so you get about 50K when parallel with the input impedance of the transistor. But this is not reliable.

You need to relax some requirement, if you can accept output impedance of say 20K, then re=10K and your input impedance of the transistor can by up to 1MΩ. Then you can ignore the input impedance of the transistor and make r1//r2= 50K.

For fmin, C=1/(2πRf) where R=50k input impedance, f=10Hz.
 P: 133 thnx for reply very good but can i ask you one question if you can explain in similary way what for example, i want to desigin a amplifier with gain of 50, how can i design? for example me i design a amplifier in the end i get other gain, and i want a gain of 50 can you teach me how to do it, to get desired voltage
HW Helper
P: 4,548

 Quote by michael1978 thnx for reply very good but can i ask you one question if you can explain in similary way what for example, i want to desigin a amplifier with gain of 50, how can i design? for example me i design a amplifier in the end i get other gain, and i want a gain of 50 can you teach me how to do it, to get desired voltage
For a high gain (e.g., 50) there are advantages in using two stages, each using one transistor. The first stage could have a voltage gain of x10, and the second a gain of x5.

But whatever voltage gain you need, for each stage you still use the equations that yungman provided, viz.,
 Gain AV=rc/(re+r'e) where r'e= 1/gm≈Vt/Ic. Vt≈25mV at 25deg C. For Ie=1mA, r'e≈25Ω.
P: 133
 Quote by NascentOxygen For a high gain (e.g., 50) there are advantages in using two stages, each using one transistor. The first stage could have a voltage gain of x10, and the second a gain of x5. But whatever voltage gain you need, for each stage you still use the equations that yungman provided, viz.,
yes i know, can you show the first stage gain for example to be x10, wich steps i have to take? WICH FORMULA? i know the gain is rcllrl/re , but i want for example x10 OF x5 ANY DESIRED VOLTAGE GAIN CAN YOU SHOW ME EXAMPLE PLEASE FOR DESIRED FIRST VOLTAGE GAIN
P: 133
 Quote by yungman You want output to be 2K, your rc has to be 2K. You want voltage gain of 2, so you want your re≈1K. Gain=rc/(re+r'e) where r'e= 1/gm≈Vt/Ic. Vt≈25mV at 25deg C. For Ie=1mA, r'e≈25Ω. So if you use 1mA bias current, re=1K is close enough. If you choose a NPN with β>100, input impedance of the transistor is βXre≈100K, this is going to be a little tricky getting input impedance of 50K as the input impedance of the transistor is as low as 100K or higher if β is larger. What is r1 and r2? is this a voltage divider to bias the base of the BJT? If so, you want Zin=50K, then you want r1//r2=100K so you get about 50K when parallel with the input impedance of the transistor. But this is not reliable. You need to relax some requirement, if you can accept output impedance of say 20K, then re=10K and your input impedance of the transistor can by up to 1MΩ. Then you can ignore the input impedance of the transistor and make r1//r2= 50K. For fmin, C=1/(2πRf) where R=50k input impedance, f=10Hz.
oh sorry i forget, if is rl load connected with rc, how you calculate can you teach like in first example please, show me example please thank you
 HW Helper P: 4,548 michael1978, you might find this thread useful, including the URL I provide of a reference for common emitter amplifier calculations : http://physicsforums.com/showthread.php?t=644679
P: 3,830
 Quote by michael1978 oh sorry i forget, if is rl load connected with rc, how you calculate can you teach like in first example please, show me example please thank you
If you connect rl to rc, then the gain will change like you said = (rl//rc)/re. The calculation I show was direct answer to your original request Zout=2K which is rc. Gain without load rl, equal 2.

For example, say if you keep Zout=2K, rc has to stay at 2K. Say if rl=2K, then rc//rl=1K. To get a gain of about 2, re=475Ω ( as for Ie=1mA, r'e=25Ω). But then, you have to worry about Zin as Zin=β(re+r'e)≈50K!!! That is not a reliable value.
P: 133
 Quote by yungman If you connect rl to rc, then the gain will change like you said = (rl//rc)/re. The calculation I show was direct answer to your original request Zout=2K which is rc. Gain without load rl, equal 2. For example, say if you keep Zout=2K, rc has to stay at 2K. Say if rl=2K, then rc//rl=1K. To get a gain of about 2, re=475Ω ( as for Ie=1mA, r'e=25Ω). But then, you have to worry about Zin as Zin=β(re+r'e)≈50K!!! That is not a reliable value.
sorry and have you get re = 475OHM? LIKE THIS 1000/2? is 500OHM
P: 3,830
 Quote by michael1978 sorry and have you get re = 475OHM? LIKE THIS 1000/2? is 500OHM
Remember gain=(rc//rl)/(re+r'e)

As I assume Ie=1mA, r'e=25Ω. rc//rl=1K. For gain of 2, re+r'e=500Ω so re=475Ω.
P: 133
 Quote by yungman Remember gain=(rc//rl)/(re+r'e) As I assume Ie=1mA, r'e=25Ω. rc//rl=1K. For gain of 2, re+r'e=500Ω so re=475Ω.
ah so r'e=25Ω 1ma and re 475 so in total 500, you mean like this? but how you get r'e=25Ω?
what i have to do to get r'e=25Ω, like this 25m:ie 1ma = 25ohm
P: 3,830
 Quote by michael1978 ah so r'e=25Ω 1ma and re 475 so in total 500, you mean like this? but how you get r'e=25Ω? what i have to do to get r'e=25Ω, like this 25m:ie 1ma = 25ohm
r'e=1/gm=Vt/Ic. Just trust me with Vt=25mV at 25 deg C. That gets into semi conductor physics. I am not expert in it and the approximation I gave is quite good already. That's all you need to know unless you really want to dig into it.

Remember this formula r'e=25mV/Ic. If you use 2mA, r'e become 12.5Ω. If you use 0.5mA, r'e become 50Ω. You get the drift?

Don't try to get precision gain using transistor, it drift a lot with temperature and different device even of the same name.( even using 2 different 2N2222 give you slightly different gain). That's the reason, using approximation is very good already. You want precision, you need closed loop feedback like an op-amp.
P: 133
 Quote by yungman r'e=1/gm=Vt/Ic. Just trust me with Vt=25mV at 25 deg C. That gets into semi conductor physics. I am not expert in it and the approximation I gave is quite good already. That's all you need to know unless you really want to dig into it. Remember this formula r'e=25mV/Ic. If you use 2mA, r'e become 12.5Ω. If you use 0.5mA, r'e become 50Ω. You get the drift? Don't try to get precision gain using transistor, it drift a lot with temperature and different device even of the same name.( even using 2 different 2N2222 give you slightly different gain). That's the reason, using approximation is very good already. You want precision, you need closed loop feedback like an op-amp.
is that 25mv/ie of 25mv/ic, i learn till now r'e=25mV/Ie not ic
P: 3,830
 Quote by michael1978 is that 25mv/ie of 25mv/ic, i learn till now r'e=25mV/Ie not ic
I believe it is supposed to be Ic, but then again, it is an approximation. Ie≈Ic in all respect if β>100. There are so many variables in these formulas you just use approx number. Here is the page explain in a lot more detail, actually they use Vt=26mV, it's close enough!!!

http://en.wikipedia.org/wiki/Bipolar...ion_transistor

Unless you really get into semi conductor physics, 25mV is good enough. I pretty much designing IC( actually all transistor circuits inside) using this approximation those days and it worked.

One thing I did not mention, you don't want to get gain of 50 out of one stage. Remember I show you how to calculate the Zin and Zout? You want Zout=2K, for gain of 50, the re+r'e=40Ω, that is low, then your Zin=βX(re+r'e)≈4000Ω. You cannot get Zin = 50K!!! Even if you can get the impedance you want, there are more limiting factor that you have not deal with, one namely Miller Effect that the circuit slow down as the gain goes up. These are a lot more important in real life than the Vt. If you want gain of 50, divide the gain into 2 separate stages of about 7 each.
 P: 358 If you want to be utterly pedantic about r'e $\Large re =\frac{dVbe}{dIe} = \frac{Vt}{Ie}$ And Vt we assume 25mV or 26mV http://en.wikipedia.org/wiki/Boltzma...hermal_voltage And transconductance gm $\Large gm= \frac{Ic}{Vt}=\frac{Ic}{Ie}*\frac{Ie}{Vt}= \frac{\beta}{\beta+1}*\frac{1}{re}$
 P: 133 thanx for help, is enough for begin do you know any good book about electronics transistor because now i am reading electronics principles by malvino
P: 3,830
 Quote by michael1978 thanx for help, is enough for begin do you know any good book about electronics transistor because now i am reading electronics principles by malvino
That's the best book....bar none!!!! Everything I posted is in the Malvino!!! I gone very far as an engineer using nothing more than Malvino. Yes, I got into BJT IC design with just the knowledge of Malvino, it's that good!!! Of cause I learn a whole lot more since then, but it's good enough to get you started as an engineer.