Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

question on diode load amplifier

Status
Not open for further replies.

analogic

Junior Member level 1
Joined
Nov 24, 2004
Messages
17
Helped
1
Reputation
2
Reaction score
1
Trophy points
1,283
Activity points
164
common source with diode load

one kind of simple common-source amplifier, is diode load in parallel to an active load. the active load reduces the current flowing in the diode load, therefore, reduce the overdrive voltage. the diode load can make the gain linear.

so it seems this kind of structure has the advantage of both good linearity and good gain (high gain will not result in large degradation of output swing as it does for purely diode load common-source amplifier). so this structure is suitable for moderate gain amplifer, am i right?

welcome comments.
 

diode load sharing

The logic behind it seems fine but you have to bias your active load such that it takes more current. Or else, the diode load will always dominate(they come in parallel). Hence this can be used as a moderate gain amplifier.
 

if so, why not many people use it?
 

Well, in some cases, it is useful. You can supply the input device with more current than the loads.I mean to say that each load is sharing the current. But in most cases, people want all the current into a single load transistor to meet the slewing conditions.
 

For the diode load amplifier,

1.good linearity;
2.small gain;
3.small output swing;

Since for a diode load differential amplifier
The gain is

Av=(Gm(in)/Gm(load))

Gm(in): input pair tansistor transconductance;
Gm(load):input pair tansistor transconductance

Gm=√2k(W/L)Id
or
Gm=Id/(Vsat)

Therefore , the linearity is good. But the maxim output level is larger a threshold than the general active(current mirror) load amplifier.
 

jiang is right. You also have to look into the fact of a threshold voltage loss. Normally you would want all your amplifiers to have very good common mode ranges.
 

Linearity seems good, but gain and input common mode range is not good. You lost a threshold voltage. But it has higher bandwidth.
 

One of the main problem using diode load is the gain of the amplfiier will be sensitive to process variations, since Av =gm1/gm2. If a high precision gain is required, the diode load cannot be used.
 

Status
Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top