# BJT and FET circuits

Discussion in 'Homework Help' started by dmag0006, Dec 8, 2010.

1. ### dmag0006 Thread Starter New Member

Dec 8, 2010
17
0
Hi, I need to design two biasing circuits...one for a BJT and one for an FET. In my design, I need a certain output current, max signal swing and Q-point stability which I thought could be done by using a circuit similar to the one attached, and keeping VCE and IC constant. However I was told that my circuit will not provide stability when the transistor is changed, and that I need to alter my design. To be honest I don't understand what my problem is entirely, can anyone explain what my problem exactly is and maybe let me know how to alter my circuit in order to function the way I need it to?

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2. ### hobbyist Distinguished Member

Aug 10, 2008
761
56
Use only one supply voltage. Instead of the two shown.

Also, Your circuit using one resistor in the base loop, and no emitter resistor, makes it subject to instability, parameter wise and temperature wise.

You need to use a voltage divider to bias the base, that takes care of the device instability from one transistor to another ,different Beta's, and use a emitter resitor, for temp stability.

But all of that comes by knowing what your output current and voltages are, to have a way to start the design.

Last edited: Dec 8, 2010
3. ### mik3 Senior Member

Feb 4, 2008
4,846
63
Look for an emitter follower circuit or common collector.

4. ### dmag0006 Thread Starter New Member

Dec 8, 2010
17
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thanks for your help, but I need maximum signal swing, wouldn't an emmitter resistor change that or is that a misconception of mine?

5. ### mik3 Senior Member

Feb 4, 2008
4,846
63
If you need full signal swing then use an op amp to drive the base.

6. ### dmag0006 Thread Starter New Member

Dec 8, 2010
17
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So an operational amplifier across the base...and then a BJT as I showed in the circuit? I'm not that used to op amps that's why I'm making sure I understood well...

7. ### tyblu Member

Nov 29, 2010
199
16
Your schematic has a strong dependance on $\beta$, which varies considerably between discrete BJTs. $I_B = I_E/\beta$, so the drop across $R_B$ and thus the base voltage is highly variable. To stabilize $V_B$, use a voltage divider, as done here: [link]. If you will be amplifying DC voltages (as well as AC) you'll need a DC-coupled input -- a resistor $(R_B)$; if only AC inputs, you can use a capacitively-coupled input -- a capacitor. Note that the DC-coupled case will likely need a buffered input, or op amp.

8. ### hobbyist Distinguished Member

Aug 10, 2008
761
56

Ok, since you asked about the given circuit you posted, thinking that would do the job, it is very apparent, that you are at the basics of transistor circuit design.

So here is a more beginners approach to understand a very basic level of biasing your BJT.

Your instructor said, it is unstable with a change in transistors.

What is meant by that, is, every transistor has inherent parameters that vary from one transistor to another.

The parameter in question is the Beta, of the transistor.

First lets look at how you would design the circuit you posted.

In that circuit, you would need to know the beta of that particular transistor. Lets assume it is a beta of 100, for THIS transistor only.

So lets set some design parameters to design with.

VCC = 10v.
IC = 5mA.

So first you would design the collector circuit by making 1/2 of the supply voltage drop across RC.
That would be 5v. across RC which gives a current of 5ma and makes RC = 1K ohms.
Now you need to solve for base current needed to make this happen.
You use this formula, base current [(IB) = IC / Beta.] = 50uA.
Now to solve for RB, you would have to use this equation, [(VCC - Vbe) / IB] =~ 180K ohms.

Now that would put this transistor into whats called linear region for aplification, HOWEVER, when another transistor needs to be replaced in the circuit, the whole design would need to be redone all over again, just as the first was done, due to the natural inherent difference in Beta, from one transistor to another.

That is the reason a voltage divider is placed at the base of the transistor, so as to keep a constant voltage at the base so when other transistors are replaced in the circuit, the voltage will remain very much stable enough so as to be able to keep every transistor at its bias point, regardless of the difference in Beta's of each transistor.

Now concerning the emitter resistor, that is a negative feedback, so as to swamp out anby temperature changes that will always occur when a transistor is operating. The emiotter resistor in conjunction with the collector resistor, wiull also allow you to program to a certain extent the amnount of voltage gain for that transistor strage.

So the bias resistors are used to set the transistor in its linear region, without having to rely on the natural inherent transistor parameters, the resistors swamp out as much as possible the transistor variations. Be it beta, or temperature change, among other parameters.

So that way the designer controles the design rather than the transistor devices controling the design.

9. ### aliman1 Member

Mar 22, 2008
20
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I just learned something ! That was a Great explanation, even a novice like me could follow it. Super !

Aliman

10. ### dmag0006 Thread Starter New Member

Dec 8, 2010
17
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Thanks for clearing everything up (@ hobbyist) , it was all a big mess before that, and I asked my lecturer about the potential divider circuit and my lecturer said it was good to use it, so a big thanks to you aswell (@ tyblu)

11. ### dmag0006 Thread Starter New Member

Dec 8, 2010
17
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Now this design worked perfectly well for my BJT design, but when I apply a JFET and add capacitors to smoothen the output, the output current is changing with each different JFET I use...

12. ### dmag0006 Thread Starter New Member

Dec 8, 2010
17
0
The BJT worked perfectly, but will this work for JFETs aswell? Because I added two capacitors for smoothing purposes and the output current changed with every JFET I used...

13. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,916
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What typo of a JFET you use ?
And maybe you post the schematics.

14. ### dmag0006 Thread Starter New Member

Dec 8, 2010
17
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I used the 2N3819, however my circuit should be designed so that the output current would be 5mA whichever JFET I use. Umm I'm still a beginner to electrical engineering and I'm not exactly sure what do you mean by schematics...

15. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,916
1,084
Designing JFET amplifier is not so easy task.
2N3819 in PSpice has the following parameters
Vt = -3V; Idss = 11.7mA and you want Id = 5mA from Vdd=10V

$Id = Idss*[1 - \frac{Vgs}{Vt}]^2$

So we want to know Vgs for Id=5mA

$Vgs = Vt*[1 -\sqr{ \frac{Id}{Idss}}] = -3V*[1 -\sqr{ \frac{5mA}{11.7mA}}] = - 1.0388V$

So If we use this diagram

Then
Rs = Vgs/Id = 207Ω = 200Ω
Rd = 4.5V/5mA = 900 = 1KΩ
Rg = 100KΩ

gm = (2*Idss)/Vt * √(Id/Idss) = 5mS

Cs = 0.16/( 1/gm * Fc) = 47uF

So voltage gain

Av = gm* Rd = 5[V/V]

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16. ### dmag0006 Thread Starter New Member

Dec 8, 2010
17
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Yes but the same design with the same values had to give the same output current, so I doubt I can assume Idss is 11.7mA and Vt is -3V. Also, I think a potential devider is needed to improve stability...

17. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,280
479
You have to assume Idss and Vt. These two values are provided by the manufacturer. You do not get to pick and choose them. Sure, there will be some variation, but the variation comes from manufacturing process and you, the end user, has not control over it.