Single supply 5V LM324 Half-Wave Rectifier

[ezea.audio]

Hi there

I have this problem in my half-wave rectifier. I have a 20-200mV sine as an input signal, centered on 1.5V. So, I'm using one op-amp of the quadruple LM324 (supplied by 0-5V), a fast-rectifier diode 1N4148 configured as a super diode (between the op-amp's output and the V- input/feedback loop), and a 10Kohm pull-down resistor (this resistor is at the rectifier's output working as the load of the circuit, and it goes down to the same 1.5V virtual reference). The problem is that this design I've made doesn't rectify for some reason which I don't actually know, and the output (rectified) amplitude at the oscilloscope doesn't match with the input amplitude.

If anyone could help me out with this, I'd be grateful.
Thanks in advance,

Elias

 

[duke37]

Show us a circuit
The LM324 can provide an output to within 1.5V of the supply i.e.3.5V, The diode will need 0.7V to turn on. Have you run out of voltage?

 

[ezea.audio]

Sorry I didn't post the circuit. I thought I did. Here it goes:

http://images.elektroda.net/63_1294689800.png

With the only difference that the ground terminal of the RL resistor is attached to a 1.5V virtual reference , which is the same voltage where the input signal is centered.

And I don't understand your last question. I'm supplying the circuit with a voltage source (not with batteries yet, if that's your concern).

 

[(*steve*)]

Does the ground shown on the diagram reference the 0V rail of a split supply, or is it your most negative supply rail?

Is the input signal relative to your 0V rail or your most negative supply rail?

edit: if you can show us the input and the output as it appears on the scope, that would be useful.

 

[duke37]

The amp can only give an output of 3.5V (supply -1.5V)
The diode will take 0.7V, giving a maximum output voltage of 2.8V.
Your resistor is referenced to 1.5V not ground so the maximum across the resistor is 1.3V.
Seems it should work. Is this right Steve?

 

[(*steve*)]

Seems it should work. Is this right Steve?

No, I don't think so.

Answers to my questions by the original poster would be handy, but assuming what is told to us and drawn in the circuit diagram, the input signal never goes negative with respect to the reference voltage, so the output will not be rectified.