Active loads for BJTs

Thread Starter

nn_in

Joined Apr 14, 2009
14
How using constant current or active loads used in audio amps in diff amps and common emitter mode enables higher gain?

How is it better than using a resistor?

Any videos or actual experiments folks can share to explain this better.

Thanks.
 

LvW

Joined Jun 13, 2013
1,020
The key word is: Dynamic (differential) resistance.
* For an ohmic resistor the STATIC resistance R is equal to the DYNAMIC resistance r (effective for small voltage/current changes only).
This property results from the linear current-voltage relation.
Therefore, in order not to have a very large DC drop across the collector resistance, the ohmic value is limited to some kOhms (50kOhm would produce 50 volts for Ic=1mA).
* However, for an active load with a non-linear characteristic (like the C-E path of another transistor) the DC drop Vce can be limited to some volts only and - at the same time - the DYNAMIC resistance rce (1/h22), which is responsible for the voltage gain of the stage can be made much larger (for example rce=50 kohms).
 

Thread Starter

nn_in

Joined Apr 14, 2009
14
Is this correct . For a CE gain is Rc/Re or Rc/Re' (if no Re is used) .Hence the gain is higher with constant current source?
So the Bjt rce adds up with the external constant current source rce?
 

LvW

Joined Jun 13, 2013
1,020
Is this correct . For a CE gain is Rc/Re or Rc/Re' (if no Re is used) .Hence the gain is higher with constant current source?
So the Bjt rce adds up with the external constant current source rce?
No - that is not "correct". Please note, that in electronics no formula is really "correct". Everything is just an approximation because of some simplifications.
That means: We are using formulas and values which are good enough with respect to other uncertainties (tolerances, temperatur influences, parasitics,...).
For a common emitter stage (with RE feedback for signals to be amplified), the following gain expression is normally used:

G= - (Rc||rce1)/[(1/gm) + RE]
with coll. resistor Rc, output resistance of the main transistor rce1, transconductance gm.

The ohmic coll. resistor can be replaced by rce2 of another transistor:
G= - (rce1||rce2)/[(1/gm)+RE].

If - in addition - the simplificaiton RE>>1/gm=re´ is allowed, we get
G= - (rce1||rce2)/RE.
 
Last edited:

Bordodynov

Joined May 20, 2015
2,603
And here is the same scheme with a dynamic load at the input signal amplitude of 1 mV. This dynamic load has a low output impedance (as opposed to the current source).
Us2N2222tran.png
 

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Bordodynov

Joined May 20, 2015
2,603
Yes Q2 works as an emitter repeater. Its transfer coefficient is close to one. As a result, the current (the variable part but not the constant part) through the resistor is small because it is affected by a low AC voltage. As a result, for the first transistor, it acts as a large-value resistor, the second pin of which is grounded in terms of alternating current.
The equivalent load of the first transistor is equal to the parallel connection of the base resistors of this equivalent resistor and the load recalculated to the repeater input (RLoad*(1+Beta_ekv)).2019-10-02_07-57-45.png
 
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