Precision active half-wave rectifier

[MRW]

Circuit link:
http://sound.westhost.com/appnotes/an001.htm

In Figure 1 of the above link (Basic Precision Halfwave Rectifier), how
would I analyse this circuit? At first, I thought R1 and R2 didn't
really do anything because it seemed like they weren't part of any
feedback networks, but when I ran a simulation without them, the
circuit did not rectify.

Any tips?

I initially analyzed the circuit by replacing the diode with a voltage
drop. I assumed 0.7V, so I wrote my output as Vout = Immediate voltage
output of opamp - 0.7 V. Then, set Vout equal to my Vin. That's why I
had the impression that R1 and R2 did not do anything.

Thanks!

 

[john jardine]

Circuit link:
http://sound.westhost.com/appnotes/an001.htm

In Figure 1 of the above link (Basic Precision Halfwave Rectifier), how
would I analyse this circuit? At first, I thought R1 and R2 didn't
really do anything because it seemed like they weren't part of any
feedback networks, but when I ran a simulation without them, the
circuit did not rectify.

Any tips?

I initially analyzed the circuit by replacing the diode with a voltage
drop. I assumed 0.7V, so I wrote my output as Vout = Immediate voltage
output of opamp - 0.7 V. Then, set Vout equal to my Vin. That's why I
had the impression that R1 and R2 did not do anything.

Thanks!

R1 shouldn't really be needed. It's just defining the input voltage as being
wrt 0V.
R2 in reality wouldn't normally be explicity shown as part of the circuit.
Some kind of lowish value is needed though in order to hold the output
voltage near 0V when the opamp and it's reverse biased diode has flopped
down to near the negative rail (any negative input voltage).

Without the resistor your sim will be measuring the output voltage with some
kind of infinite resistance voltmeter and the multi megohms due to a
reversed biased diode, would look essentially like a short circuit.
john

 

[[email protected]]

Circuit link:
http://sound.westhost.com/appnotes/an001.htm

In Figure 1 of the above link (Basic Precision Halfwave Rectifier), how
would I analyse this circuit? At first, I thought R1 and R2 didn't
really do anything because it seemed like they weren't part of any
feedback networks, but when I ran a simulation without them, the
circuit did not rectify.

Any tips?

I initially analyzed the circuit by replacing the diode with a voltage
drop. I assumed 0.7V, so I wrote my output as Vout = Immediate voltage
output of opamp - 0.7 V. Then, set Vout equal to my Vin. That's why I
had the impression that R1 and R2 did not do anything.

Thanks!

Download LTSpice from linear.com. Then you can see the waveforms
everywhere in the circuit.

- Tom Gootee

 

[Tam/WB2TT]

Circuit link:
http://sound.westhost.com/appnotes/an001.htm

In Figure 1 of the above link (Basic Precision Halfwave Rectifier), how
would I analyse this circuit? At first, I thought R1 and R2 didn't
really do anything because it seemed like they weren't part of any
feedback networks, but when I ran a simulation without them, the
circuit did not rectify.

Any tips?

I initially analyzed the circuit by replacing the diode with a voltage
drop. I assumed 0.7V, so I wrote my output as Vout = Immediate voltage
output of opamp - 0.7 V. Then, set Vout equal to my Vin. That's why I
had the impression that R1 and R2 did not do anything.

Thanks!

Do you actually want the half wave rectified sine wave, or a DC voltage
equal to the peak?

Tam

 

[Jamie]

Circuit link:
http://sound.westhost.com/appnotes/an001.htm

In Figure 1 of the above link (Basic Precision Halfwave Rectifier), how
would I analyse this circuit? At first, I thought R1 and R2 didn't
really do anything because it seemed like they weren't part of any
feedback networks, but when I ran a simulation without them, the
circuit did not rectify.

Any tips?

I initially analyzed the circuit by replacing the diode with a voltage
drop. I assumed 0.7V, so I wrote my output as Vout = Immediate voltage
output of opamp - 0.7 V. Then, set Vout equal to my Vin. That's why I
had the impression that R1 and R2 did not do anything.

Thanks!

R1 is there for capacitive/inductive coupling., the input of the OP-amp
could be bipolar, thus you need to rid that diode effect and give some
in range impedance for the AC source.
the R2 is needed to create load on the OP-amp since, at the zero state,
you will not have any load coupling for the next stage. A Oscope with
AC input set for example, would not work very well because of the
diode in line.
there are other more in death reasons for the state of the op-amp,
but i won't get into that.

That is about as simple as i can explain it.