a really basic analog query

I am a beginner in analog electronics side.
I want to get clear of the doubt that i have

it is general that a transistor(MOSFET) will be set the proper operating point by choosing a point from the voltage transfer characteristics.(Vout Vs Vin) by choosing a certain load resistance value and the same resistance value is used when that transistor is used as an amplifier

what my doubt is, wont the circuit performance change if you try to increase the load resistance by means of active load or some other confirgurations in the load for the amplifier circuit in order to get a higher gain. But most of the time, its not mentioned about that at all in almost all the literatures.

please dont get angry if my question is very trivial. I thought I would ask this here.

thanks people


Dan

The performance will change the stage gain will goes up!! The important quiescent (DC) conditions will not. Suppose you need 6V at a current of 1 mA flowing into the collector or drain of a transistor. If you start of with a Vcc of 12V, then the quiescent load is (12 - 6)/1 K = 6 K ohms. So if the current due to the signal is 2 micro A P-P, then the output voltage is 6 K X 2 micro A = 12 mV P-P. If instead you use some sort of active device (or boot strapping) so the dynamic impedance is 100K ohms, then the output voltage is 2 micro A X 100K = 200 mV P-P. A worthwhile gain. These techniques normally lead to a narrowing of the useable bandwidth, as the active load becomes "less active" as the frequency rises.
Frank

Thanks for the reply
But if you use an active load(dynamic impedance(ac impedance..?) is still an impedance which will make a voltage drop as per its value right?)(100 KOhms) in the place of a 6KOhms resistor, wouldn't the Id be ((12-6)/100K)= 0.06 mA instead of 1 mA..? This should also change the operating point as well.
Why isnt happening like that

Its bugging me as I am unable to figure it out as how the performance of a circuit can be made better though operating region get changed due to the increase in the load resisitance which is basically used in the first palce to compute the operating point..

thanks again chuckey

Dan

Hi Dan,

it's important to distinguish between static and dynamic resistance - and to remember that the transistor acts as a CURRENT source.
That means: The collector(drain) current is NOT determined by the collector(drain) resistance!
What happens if you increase the collector/drain resistance:
1.) The current Ic (Id) remains (nearly) constant and the gain goes up (as desired)
2.) The region for proper linear ac amplification drops because the operating point (assumed to be optimum in the middle of this region) is shifted due to a larger dc voltage across Rc (Rd). Up to a certain degree this could be corrected by a corresponding increase of supply voltage - however, there are practical limitations.

A better method is to use a dynamical load resistance with a fixed (and rather small) dc resistance allowing a relatively small supply voltage (as before) - however with an increased ac gain due to a higher dynamic ac resistance. OK?

Regards
LvW

Thanks LvW for the explanation.
I will try reading the chapter that I was reading before I got cleared about this fact and get back to here

When you say dynamic resistance, do you mean to mention about the channel resistance or the intrinsic rds that you could acheive with an active load.?

Dan

analog_curious said:

When you say dynamic resistance, do you mean to mention about the channel resistance or the intrinsic rds that you could acheive with an active load.?

Dan

Click to expand...

Yes, the dynamic resistance (or differential resistance) is effective for ac signals only. Simple example: A coil has R=1 ohm static resistance for dc and much more (frequency dependent) for ac signals.

HI LvW,
I see that the recent thread you posted was removed from here along with my recent question. May I know as what the reason would be.I just want to make sure that I am following the forum's rules

Thanks

Hi dear Dan*
As i understood, your problem is active loads.
for this you can use current source .
if you use a current source for drain resistance your out impedance will increase .

best wishes
green tree