What's the differents of the two circuit??

Status
Not open for further replies.
Z

ZengLei

Full Member level 1
Joined
Jan 24, 2006
Messages
99
Helped
0
Reputation
0
Reaction score
0
Trophy points
1,286
Location
WuHan China PR
Activity points
2,143
and what's the output value Vout and Vn???

the diode compose a negative feedback or not why??
thanks~~
 

Since Vin=5V, diode in 1.gif will be on, so you will have negative feedback, and Vn=5V, Vout=Vn-Vd.Vd usualy =0.6, Vout=5-0.6=4.4V This will work for all voltages Vin>0.
Lets say that Vin=0, this means V+=0, and as this is ideal opamp V-=0, as there is no difference on opamp input there is no Vout, so the diode will be off. When you apply Vin>0, the opamp go to saturation and Vout=+Vcc, and diode starts to conduct current. When this hepens we will get negative feedback, so V+=V-=Vin, and Vn=Vin, Vout=Vn-Vd.

2.gif is for negative Vin (Vin<0), because if Vin<0 than Vout=-Vcc (negative saturation)...

Answeres:
1.gif Vn=5V, Vout=5-Vd.

2.gif Vn=0, Vout=+Vcc.
 

1) for the 1st diag... 2.gif,
it's in saturation.. i.e., vout=5v. i think vn could not be determined as it is floating.

2)for 1.gif, it is a voltage follower vout=5v. vn=5-0.7=4.3v.
 

For 1.gif Vn=5 volts, Vout=4.3
For 2.gif Dont know...

Added after 1 minutes:

To amriths04:
How u r saying the o/p is saturation and Vn is floating.. Can u pl explain that???
 

To venkat3271:
It is in saturation, but diode is inverse so it is off (for this diode to be on must be Vn-Vout>=Vd) so Vn is "floating" (it is actualy 0V), as Vout=Vcc (saturated).
 

The two circuits are called as "super diodes".Usually a diode has a cut in voltage as .7v using an opamp , would make that cut in voltage to drop to .7/A(≈ µV) where A is the open loop gain of the op amp.So this would enable us to rectify even signals in mv ranges.
 

ZengLei,
the_edge is correct for circuit 1.
.
For circuit 2, the diode is forward biased. The output will go to whatever voltage is necessary to make the Inv input equal to the NI input. As a result, the diode voltage is effectively reduced by a factor equal to the o0pen loop gain of the op-amp. As a practivcal matter, you can ignore the diode drop. The end result is that the output will equal the input voltage plus the input offset voltage of the op amp.
regards,
Kral
 

fig1 vout=5v-vn, vn=vout-vd
fig2 vout=5v,vn=zero
 

Status
Not open for further replies.

Similar threads

Menu
Log in
Register
Cookies are required to use this site. You must accept them to continue using the site. Learn more…