Basic question on output impedance

[snoop835]

gm rout

Hi,

Can someone tell me why low output impedance can drive resistive load and high output impedance can only drive capacitive load? Is there any maths behind this? Anyone can suggest good reading material

thanks

 

[flushrat]

vout=rload/(rload+rout), if rout very large, vout will be small.

 

[snoop835]

It makes sense. But I still don't see how low output impedance can drive resistive load and high output impedance can only drive capacitive load?

 

[Humungus]

Remember that high impedance output stages can be saw as current sources. So, if you connect a small resistor, the output on that resistor, i.e. the effective output voltage is also small. If you connect a capacitor, the output voltage increases linearly (the capacitor charges at constant current) during the slewing phase and then after a certain time it can reach the power rails. Actually, low impedance output stages can also drive capacitive loads as high impedance output stages can drive resistivce loads, but always keep in mind its limitations.

Just make a small signal schematic of a simple amplifier, and get the equations of the output voltage for a resistive and capacitive loads. You'll see the effect for yourself.

 

[surianova]

hi

I answer by using formula

1. if high output impedence,Rout (let say CS cmplifier )driving resistive load (100 ohm)

orignal gain without driving load is A=gm.Rout
but when driving load A=gm.(Rout||100 ohm)
= gm. 100 ohm

We can see that the gain is drop very badly when driving resistive load with high
output impedence.


2. How abt using low output impedence to drive resistive load
Let say we use source follower with output impedence 1/gm.


original gain, A= gm.(1/gm)
= 1

When driving load, A= gm(1/gm|| 100 ohm) and 1/gm << 100 ohm
= 1 ( the gain did not change)



3. Why we can use high output impedence to drive capacitive load ( 1/sc)

Let say we use CS amplifier

A=gm.Rout

When driving 1/sc capacitive load

A=gm. (Rout||1/sc), and we know 1/sc is >> Rout at low frequency
= gm.Rout ( the gain did not change)


This can be apply as well if you use low output impedence to drive capacitive load,
you will still get the same gain.

Hope this will help you


Surianova