What is full-swing dc voltage?

[khatus]

1. Here What is meant by full-swing dc voltage?

2. what is the significance of using the word cooperative here?

 

[stenzer]

Hi,

(1) Here the MOSFET's on-resistance is considered to be very small, as mentioned by R_on << R_Load. Consequently, almost the complete volatage, in Fig. 3.3 it is +10 V, is dropping across the load (the lamp). So here the Drain-Source on-resistance is neglected (if the load resistance is significantly larger), and there is no voltage drop across the Drain-Source when the MOSFET is conductive (ON).

If the MOSFET is considered perfectly turned off (non-conductive), you are able to switch between 0 and 10 V and you have your full-swing of the DC voltage, as you can apply the whole +10 V to your load, or none.

(2) I assume here the aim is to point out the advantage of not needing to provide sufficient base current, as it is required for a BJT, and also mentioned in the begining of the section. So for a BJT with an h_fe of ~100, and considering a tenth for dimensioning, you would have to limit/provide a base current of about 10 mA (for the lamp in Fig. 3.3). So as you can see, you even do not need a gate resistor, as the only current flowing towards and from the gate is the charging and discharging current of the gate-capacitance. Note, an actual circuitry would include a small valued resistor to limit the in/out-rush current, to reduce the stress on the MOSFET. So here you have a high impedance input and a low impedance output.

BR

 

[khatus]

OK. I am having difficulty understanding the meaning of these lines

1)Why it is called a saturated transistor switch??
2)"Without squandering excessive power......................."
In this line where do we squander excessive power?? is it on the input side??

3) Why this analog switch can not be done with a bipolar transistor ??
4)I can't understand the line "considering worst-case minimum hFE combined with the lamp's cold resistance "?

hFE = IC/IB ,
if hFe to be minimum IB has to be maximum

 

[BradtheRad]

View attachment 181025

3) Why this analog switch can not be done with a bipolar transistor ??

I suppose it's because 1) bjt operates via bias current, and 2) bjt conductivity is adjustable.
It's not saying It's impossible to make a switching circuit using bjt's.

OTOH analog switches are virtual devices, easy to use in simulation. They draw no bias current. They don't turn on halfway (although parameters are settable and predictable such as resistance On & Off).

You could say an fet is easier to use as a real life stand-in for an analog switch.
It still takes effort and ingenuity to arrange either bjt's or fet's in a switching circuit.