Gain calculations

Hi

I'm just working through Douglas Selfs book and running through some of the calculations.

At the moment I'm looking at LF and HF gain

He gives LF gain to be gm*Beta*Rc

Could someone just clarify that the following assumptions are correct please?

gm in mA/V?

Beta is that of VAs transistor? and if a darlington stage is used, does that need to be used as beta*beta in the equation?

Rc is the output resistance of the VAS current source? I know it's going to be quite high, but can someone tell me how to calculate it for a common 2 transistor CC? I know I should be able to work this one out, but I think I've been looking at textbooks for too long and I can't see the wood for the trees now!

At the moment I'm getting a horrifyingly large number for a circuit setup with gm=95mA/V beta=14400 for 2 2SD669s in VAS, and 5Mohm for the VAS current source. I've done a lot of searching, but failed to find anything that can help yet.

Thanks for looking

Ian
 
For measuring the effective resistance of the current source, say you have a 15V supply. Put a low resistance in line with the current source so there is a voltage drop of 14V across the current source. No measure the current flowing into the collector of the current source. Lets say it is 1mA. Now put a larger resistor inline with the current source so the voltage across the current source is 1V. Now measure the current flowing into the current source. Lets say it measures to 0.991mA. Now do this formula: (14V-1V)/(1mA-0.991mA) which calculates to 1.44Mohms.

So it is basically change in voltage across current source/change in current flowing through current source which is R=V/I. The voltage change across the current source is created by changing the resistance inline with the current source which inturn changes the voltage across the current source. The current is measured at each different voltage level.

The effective resistance would always be changing but would stay high. In the example above, it should always be at least 1.44Mohms or higher in resistance as long as you are staying with in the limitations of the current source. This is a dynamic measurement. A static measurement would be voltage across current source/current flowing in current source which would be 14kohms for the first measurement, unless I am wrong.
 
I choose the VAS based on Pd, Vce, Cob, Ft, and linearity. Linearity of VAS is important in trying to keep non-linear components out of the transfer function (espeacially at HF). A cascode VAS will help with linearity by maintaining a constant Vce on the VAS transistor, but it requires a few more volts. Some people don't like the extra complexity. The advantage is that Vce is constant and Pd is now small for the amplifying transistor so a small signal device can be used. Small signal transistors generally have a larger Hfe and higher Ft. With a higher gain, a darlington VAS may not be needed reducing the number of stages within the feedback loop.:)
 
thanks for the info!

I am a little confused. Ia a cascode VAS isn't the common base tranzistor who is doing most of the amplification? I as i understand it(i might be wrong so please correct me), the common base presents a small impedace to the common emitter tranzistor, thus its voltage gain is limited.
 
Might be a bit late but check out Fairchild's KSA1381/KSC3503 transistors for VAS usage.

300V Vceo, 100mA continuous current, 7W dissipation, 150MHz fT, and 3.1pF Cob for the KSA1381. They're also quite easy to get as you can buy them direct from Fairchild online.

They also have very good SPICE models of them which might be a help!
 
Gm in small signal devices is usually specified as mA/V, but in this equation must be A/V. V=R*I that's volts=ohms*amps, not volts=ohms*milliamps.

You could say the gm term converts the input voltage into a current, the beta term applies gain to that current, and the Rc term converts that current back into a voltage.

Gm in this equation is the gm of the input stage.