Continue to Site

Welcome to our site!

Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

  • Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

differential amplifier input stage of op-amp

Status
Not open for further replies.

PG1995

Active Member
Hi

Please have a look on **broken link removed** attachment and help me with the queries, particularly with Q3. I need to understand this quickly and doesn't need broad understanding of the topic. Thanks a lot for your help.

Regards
PG
 
Hate to say that it took me months to grow in my young brain until I understood how different amplifiers (just "DAm" hereafter) does work. Until then, opamps was just a "magic" device that just worked.

Well, often the value of RE is substavtially lower than RC (I mean both if I omit numbers) so that a little change in input will cause a large change on output.

How familiar are you with regular common emittor amplifier (hereafter "CEA") stages?

What took me to understand DAm's was to think of it as a sort of uncommon CEA. Tell yourself that Input 2 has a fixed voltage. How will the outputs swing as Input 1 varies?
Then think about Input 1 as a fixed voltage. Think in same way when Input 2 is changed.
 
it might help a bit if you knew the archaic name of a diff amp, "emitter-coupled amplifier". the emitters are at the same potential, so what happens on one side has an effect on the other side. the resistor feeding the emitters is usually not a resistor, but a constant current source, and the output from the collectors is usually in the form of a current and not a voltage. with a constant current, the two transistors have equal currents through them "at rest". if you increase the current through Q1 by driving it's base with more current, more current flows through Q1. that additional current raises the voltage on Q1's emitter, which is directly connected to Q2's emitter. when Q2's emitter also goes higher, which reduces the bias on Q2's base by the same amount, which reduces Q2's current. the total current remains the same. think of it as a seesaw with a fulcrum in the middle.
 
Status
Not open for further replies.

Latest threads

New Articles From Microcontroller Tips

Back
Top