Precise Half-wave rectifier problems

[cvicente]

I'm actually trying to build a precise full wave rectifier. But after a first failed try, I though I would better be off break down things to analyse every part of the circuit.

So, I started building a half-wave rectifier...

and even at this point I'm seeing something I was not expected:

So, has you might guess, the yellow wave-form is my Vin and the blue is my Vout (both channels are AC coupled).

Here's the difference when I change my scope settings to have both channels to DC coupled:

I was sincerely expecting to get something more like this:

Currently I'm using 20k (1%) resistors, ahottky diodes (1N5818) and the TLC272ACP op-amp.

Any thoughts on what is going on? Why am I still getting a negative portion of the rectified wave form? I must have messed something up...

I'm also quite a noob in what regards to scopes, so... is this something I should account when reading in AC and DC coupling? How do I really know what to expect from Vout?

Any light on the subject would be greatly appreciated!

 

[(*steve*)]

My first bit of insight is to suggest that you DC couple the inputs to your scope

 

[(*steve*)]

My second bit of insight is that your input is to the inverting input of the op-amp. So it's going to flip the negative part of the waveform.

Set you scope up so that

a) both inputs are DC coupled

b) both traces (with no signal) are coincident

c) (if possible) invert one input.

 

[duke37]

Should one diode be connected the opposite way?

 

[Harald Kapp]

@duke: The diodes are correct.

It seems the OpAmp is operated from a single supply. While the TLC272 is rated for single supply operation, this circuit is not suitable for single supply operation:
Assume a positive input voltage. This will force a current into the 20k input resistor. The OpAmp in turn will try to keep the voltage difference between the "+" and the "-" input close to 0V. It does so by sinking a current at the output such that the input current flows through the feedback diode directly into the output of the OpAmp. Across the diode a voltage drop of ~0.6V develops. Add up the voltages:
- Input should be at 0V (if the feedback loop is closed and operates as expected).
- Across the diode there is ~0.6V.
- This requires the output to go down to -0.6V. It can't do that with a single supply (lowest available voltage is 0V)!

Try the circuit with a dual supply voltage.

 

[cvicente]

Thank you all for answering!

@Harald So, let me see if I got this right. The only reason I actually need a dual supply voltage, it's because the voltages I'm working with are quite low... low enough that that ~0.6V (actually it's a little bit less since I'm using shottky diodes) make quite a difference. Correct?

Is there a op-amp you would recommend for dual supply voltage? I would required a dual op-amp, since my main objective is to create a full wave rectifier.

@Steve

a) both inputs are DC coupled

b) both traces (with no signal) are coincident

c) (if possible) invert one input.

a) Check... done that. Actually you have the screen-shot on my question.

b) Could you clarify please? To which traces do you refer to?

c) I'm afraid I don't understand (I'm a bit limited in electronics). This is an AC signnal, What do you mean by inverting one input?

 

[Harald Kapp]

It's not +0.6V that matters, it's -0.6V that can't be output by the amplifier because its lower supply voltage is 0V.

You can use (almost) any opamp. You can even use the TLC272 as long as the total supply voltage does not exceed the limit (18V) of the chip. This is shown in the datasheet on page 16.
E.g. +-5V would be fine (corresponds to 10V single supply).

You could also use an asymmetric +12V -3V supply since you need only a little bit of negative voltage to push current through the feedback diode.

 

[cvicente]

@Harald

Thank you. To tell the truth I got hanged by the first schematic regarding the op-amp pins layout, that specified "ground" and not Vdd-. My fault!... Thank you for pointing this out.

In this case, I should be fine using this op-amp, since I'm only +5VDC on Vdd+, and the input is not supposed to go higher than 1V (unless I decide to amplify it).

Sincerely, now... I got a little confused. As far as I can see there's not a real difference between a single and dual power supply op-amp...

 

[cvicente]

So, I've followed Harald suggestions and acquired a an inversion regulator to be able to provide de the op-amp with the necessary negative voltage instead of ground. Nevertheless, and before applying it I noticed something wrong on my circuit.

As I'm using this circuit to power up an A/C appliance, I wanted to use the same power supply to power up the circuit. So basically I've connected a AC/AC transformer to drop-down voltage from 230 to 6V (or 12V), and then got the AC converted to DC through a bridge rectifier and later regulated to 5V.

The thing is, I noticed I was using "different grounds" in different parts of my circuit. I was giving ground passed through the voltage regulator to the op-amp and giving ground directly out of the bridge rectifier to the bias the CT sensor.

For some reason that is not clear to me, between the "regulated ground" and the ground output directly from the bridge rectifier there's a difference of 2.3mV.

Well, know after having this corrected I get this on the scope:

Once again, I was expecting a closer match in terms of Max voltage...