This blog is terminated as the new blog is created.

London's Subway Inferno

http://www.youtube.com/watch?v=Way6Adl1B0Y&feature=related

LC passive circuit

notch filter

AC sweep

bode plot

bandwidth is very narrow. 

 

looks like low pass filter

except it resonate near (LC)^-0.5

at resonant frequency, output goes unstable.

 

log amplifier, exponential amplifier, precision amplifier

log amplifier with diode

dc sweep log scale linear

Vout = -Vt*ln(Vin/(Is*R)), Is saturation current of diode.

log amplifier with transistor

dc sweep log scale linear



exponential amplifier

linear scale exponential

Vout ~ -R*Is*exp(Vin/Vt).

full wave rectifier

~1.4v drop across diodes

pricision rectifier

rectifier with no voltage drop on diode











bistable, astable, monostable circuit

bistable

dc sweep -10 to 10v, transition happens at (L+)*[R1/(R1+R2)]. (L+ amplifier supply V+, in this case 15v)

dc sweep 10 to -10v, transition happens at (L-)*[R1/(R1+R2)].  (L- amplifier supply V-, in this case -15v)

astable

excited with impulse, disturbing the banlance point at 0v, we got a viberator.Viberation period ~ R3*C1 = 0.1s

output is also 0.1s periodic. when output is positive, capacitor is charged, U1- increases till it hits the threshold (L+)*[R1/(R1+R2)]. Then output plummets to negetive, capacitor is discharged, U1- decreases untill it his the threshold (L-)*[R1/(R1+R2)]. Then output bounces to positive, capacitor is charged...

monostable

D5 is used to isolate V3 after the pulse goes through.

excited with positive pulse, U1+ is positive, (Uout - U1+)/R2 = U1+/R1, Uout = [(R2+R1)/R1]*U1+, Uout is also positive. D4 is reversely biased, and is like an open. Capacitor is charged. U1- increases untill it hits (L+)*[R1/(R1+R2)]. Then Uout plummets to negative. Capacitor is discharged. U1- decreases to -0.7v. Diode is foward biased. U1- can decrease no more. Circuit becomes stable.

Even with high energy pulse, after a jerk, Vout settles down.

cross-over distortion

output stage of an op-amp. when -0.6<v0<0.6v, vout = 0. cross-over distortion because  Vbe of Q1 and Q2 are too low to turn the transistors on.

By adding two diodes, we can increase Veb of Q1 and Q2 and  force them to turn on.

Increase current at current source will increase voltage across diodes. At I = 0.1mA, we see the distortion is reduced.

At I = 2mA, voltage across diodes are high enough to turn Q1, Q2 fully on. Distortion eliminated.

Alternertively we can replace diodes with transistors, same principle.

When I = 0, distortion.

when I = 10uA, distortion reduced.

When I = 40uA, distortion eliminated.

Input 1mV sin wave.

When I = 0, working within distortion range of input, output is unresponsive. (few uV).

When I = 40uA, get rid of distortion, output is significant. (400mV).

want2bprettythin

http://want2bprettythin.xanga.com/?uni61409786-direction=p&uni61409786-nextdate=8%2F5%2F2011+8%3A29%3A11.970

BJT output current protection

There is a current limit near the output.

 Increasing current across the resistors above 0.7v will turn Q1, Q4 on. Excess current drained from Q1 and Q4.

circuit resilient to current surge, but not permanent current change. As output should remain at 0 vdc. Ideas base current for Q2 and Q3 is 175mA.

BJT active load

R12, R13, Q8, Q9 form wilder current source. Ic(Q8)*R12 = Vt*Ln( I(R13)/Ic(Q8)).

DC input at base of Q1, Q2 raise from 0 to 100V, current at wilder current source varies little.

DC voltage at output almost remain constant. (High voltage protection by active load).
Vout = Vc(Q4) = Vc(Q3) ~ V(R10) ~ V(R9 or R11) = I(R9)*R9 ~ (Ic(Q8)/2)*R9.
Current output from wilder current source varies little => Vout varies little.

When we delete the feed back loop between active load and the wilder current source, we lose over voltage protection.

Ib(Q3) + Ib(Q4) + Ic(Q3) + Ic(Q4) ~ IC(Q8). If Ic increase, Ib will decrease, which in turn result in Ic decrease. The feedback loop of the active load keep system stable.



also the current source becomes unstable.