Basic doubt...kindly clarify...

[Bhanumurthy]

Hi All,
My basic doubt is "Will the input impedance of a Common Emitter Amplifier be less than kΩ?"
Regards
Bhanumurthy.

 

[Kral]

Bhanumurthy,
I don't know what you mean by "less than KOhms". The input impedance of a common emitter amplifier depends on the load impedance, the Beta (Hfe) and the collector current. Do you need the quation for the input impedance?
Regards,
Kral

 

[rikie_rizza]

Yea, Karl is right.
In fact, mostly is in several hundred K.

 

[LvW]

No, it is not always as high as 100 kOhms (mostly depending on Ce)

Normally, for a common emitter stage, the input resistance is some kOhms.
In this context, it is good to know that the output load has only minor influence, but the most important role plays the emitter path with Re (perhaps shunted by Ce)

Here is the formula for the input resistance at the base pin:
Rin=h11 + g*Re
(g=transconductance, Re=emitter resistance for ac)
In addition, you have to consider the bias circuitry in parallel to Rin.

 

[cydi]

generaklly its less than Kohms only

 

[LvW]

Hi Bhanumurthy,

sorry, but there was a silly typing error.
Here is the corrected input resistance at the base of an BJT:

Rin=h11*(1+g*Re)=h11+h21*Re

 

[hellovinay]

sir this vinay i found that ac resistance is around less than 100ohms

 

[LvW]

sir this vinay i found that ac resistance is around less than 100ohms

Usually, the parameter h11 alone has already a value of at least 1.5...2 kOhms.
More than that, since in most cases negative signal feedback is used, at least a part of the emitter resistor in NOT bypassed by a capacitor - resulting in an effective ac emitter resistance Re,ac. Thus, a part (h21*Re,ac) is added.
Example: h21*Re,ac=150*100=15 kOhms.

 

[FvM]

If internal emitter resistance can be neglected (and no external resistor exists), it's simply the differential resistance of base-emitter junction:

rbe = ib/26 mV (silicon diode @25°C)

 

[Audioguru]

Look at the input impedance of a 2N3904 vs collector current on the datasheet from Fairchild.

 

[laktronics]

Hi,
Obviously an R = I/V is dimentionally incorrect.
The input impedance of common emitter transistor is
Rin = Rb + hfe re

If an input Vb is applied across base and ground, it practically appears across re and produces an ie = hfe*ib. So,

Vb = re*ie = re*hfe*ib
So,
Vb/ib = Input inpedance (assuming zero Rb) Rin =hfe*re.
If any Rb, then,
Rin = Rb + hfe*re

Regards,
Laktronics

 

[LvW]

For my opinion, the above "calculation" looks somewhat strange. It is a "good" example for two errors within a calculation which cancel each other.

Quote: If an input Vb is applied across base and ground, it practically appears across re and produces an ie = hfe*ib.

If there is an emitter resistance Re (Re instead of "re", because we should discriminate between static/ohmic and dynamic resistances), the base voltage Vb will certainly NOT appear across Re.

Quote: Vb = re*ie = re*hfe*ib

Therefore, this equation is simply false, because it forgets the voltage across the base-emitter path (which has the resistance h11 or rbe, but don´t write Rb).

And now the second error which, however, corrects the first error resulting in a formula which is correct:

Quote:If any Rb, then, Rin = Rb + hfe*re

There is always any Rb (should read: rbe) - and it is not correct simply to add this resistance to the first part hfe*Re , as you are not allowed to add two resistances which have different currents.

 

[laktronics]

Hi,
Thank you for your comments.
If there is an external resistance Re, then the emitter resistance will be sum of re plus Re and the base input resistance will be hfe(re+Re). Also the value of rbe will be much less than hfe*re as other wise transistor will not be of any use in amplification.The applied vb in this case will appear across re+Re, after drop in Rb effective.
It is ok to add rb or (rb+Rb) to hfe*re since the term hfe*re is the resistance referred to base and not emitter.

Regards,
Laktronics

 

[LvW]

Hi,
Thank you for your comments.
If there is an external resistance Re, then the emitter resistance will be sum of re plus Re and the base input resistance will be hfe(re+Re). Also the value of rbe will be much less than hfe*re as other wise transistor will not be of any use in amplification.The applied vb in this case will appear across re+Re, after drop in Rb effective.
It is ok to add rb or (rb+Rb) to hfe*re since the term hfe*re is the resistance referred to base and not emitter.
Regards,
Laktronics

1.) Please, can you explain to me the physical meaning of "re" ?
2.) Also the value of rbe will be much less than hfe*re as other wise transistor will not be of any use in amplification.
I cannot follow these arguments; please explain. Do you mean "re" or "Re" ? (because you are using both symbols)
Regards
LvW

 

[jony130]

In my country re is equal
re=1/gm=Vt/Ic=26mV/Ic in room temperature. And represent the small-signal impedance locking into the emitter, for the base.
So Rin=(hfe+1)*(re+Re)≈hfe*(re+RE)
And
Ku=Rc/(re+RE)
RE-emitter resistor

 

[LvW]

OK, I see. Thank you for clarification.
I can agree to this way of calculation. However, for my opinion it it is somewhat uncommon to use the emitter input resistance 1/gm (which you call re) for the calculation at the base input. Furthermore, for my opinion it is not quite logical to explain why the sum (re+Re) has to be multiplied by (1+hfe).
Instead, I always explain the derivation of the correct formula starting with the sum of two voltages between base and ground (Vbe+Vem). After substitution by ib*rbe resp. iem*Re and with iem=(1+hfe)*ib we arrive at the correct result.

 

[bheema rajulu]

hello sir, this is bheema
i want to know how sreen printing is done in final stage of PCB.what chemical methods are there to do manualy & what chemicals are used and what are the tools are used.can u give me a linksite for this pls..