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.

Register Log in

Class AB amp, how does thing even work with a negative base voltage?

Status
Not open for further replies.
M

maark6000

Member level 5
Joined
Jan 15, 2013
Messages
92
Helped
1
Reputation
2
Reaction score
1
Trophy points
1,288
Activity points
2,270
Not sure I understand how this circuit operates. The voltage divider is creating a negative voltage at the base of Q3 (-6.75 v), which from what I've been taught would not forward bias the transistor (+0.7 v ?), and thus there would be no current in the middle leg.

The circuit does work, I was able to use the pot to balance the node between Q1 and Q2, and the circuit put out a 8.5 v (pp) signal before clipping.

what am I missing?

thanks as always!

 

maark6000 said:
Not sure I understand how this circuit operates. The voltage divider is creating a negative voltage at the base of Q3 (-6.75 v), which from what I've been taught would not forward bias the transistor (+0.7 v ?),
Click to expand...

Transistor operation requires a voltage BETWEEN base and emitter of approx. 0.7 Volts. Look at the emitter potential!
 

ah! there it is. thank you. with the small 2.7k emitter resistor, the emitter voltage would be around -7.5 v, enough to forward bias the transistor. So in calculating IE (VE/RE), I would get -2.78 mA. But what does the negative sign mean... current is still flowing from Q3 down to -Vcc, right? and if IE roughly equals IC, is IC also -2.78 mA?
 

FYI the centre of the two diodes is always very close to 0V, else one output transistor is turned on more then the other, with the output terminal then not being at 0V. So the collector of Q3 is .7(diode volt drop) more negative then 0V i.e.- .7V.
Frank
 

Status
Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top