Zero Crossing Signal

[rayhall]

I have made hardware based on the circuit below. When I test the signal with my oscilloscope the signal does not go below zero volts. I can confirm I have +10v on pin 4 and -5v on pin 11.

Ray.

**broken link removed**

 

[crutschow]

Seems normal to me. You have a non-inverting linear amp with a gain of 11. Why would you expect the output signal to go below 0V if the input signal doesn't?

 

[rayhall]

Seems normal to me. You have a non-inverting linear amp with a gain of 11. Why would you expect the output signal to go below 0V if the input signal doesn't?

I expect it to cross below zero as the amp has positive 10v and negative 5 volts on pins 4 and 11.

What needs to be changed to make it do what I expect ?

Ray.

 

[simonbramble]

They are the supply pins. If you have a unipolar input signal, you will get a unipolar output signal (for a non inverting configuration). BTW - you are not getting the gain you want (11) because your op amp output is hitting the supply rail. You can change R90 to 20k (or greater) and see no difference in the signal

 

[crutschow]

I expect it to cross below zero as the amp has positive 10v and negative 5 volts on pins 4 and 11.

What needs to be changed to make it do what I expect ?

Ray.

My point was that, in that non-inverting circuit, the output signal won't go below zero if the input does not. Why do you think it should? What output voltage do you expect (or want) and why?

 

[alec_t]

If you want an output that crosses zero you could bias the input at, say, +2.5V and couple the input signal through a capacitor (the appropriate value will depend on your input frequency).

 

[wkrug]

You also can tie the Resistor R90 not to ground , but to a little positive Voltage.
That let the Output be negative at 0V input signal.
That voltage can be generated by a voltage divider between the positive supply voltage and GND.

Another Question - Why using a linear Amp for detecting Zero crossings?
I think a comparator configuration will give better results.

 

[rayhall]

My point was that, in that non-inverting circuit, the output signal won't go below zero if the input does not. Why do you think it should? What output voltage do you expect (or want) and why?

Why I think it should is...someone a lot smarter then me designed the circuit. I asked for a circuit that would convert a 0 to 5 volt square wave into a square wave signal that would cross zero. Requirements were -5 volts to +24v volts peak. The peak to be adjustable with a pot. As far as adjusting the peak this works as planed. It is just does not go below 0v as required.

The signal needs to cross zero so the engine management ECU can read the signal as a variable reluctor type signal.

 

[crutschow]

Why I think it should is...someone a lot smarter then me designed the circuit. I asked for a circuit that would convert a 0 to 5 volt square wave into a square wave signal that would cross zero. Requirements were -5 volts to +24v volts peak. The peak to be adjustable with a pot. As far as adjusting the peak this works as planed. It is just does not go below 0v as required.

The signal needs to cross zero so the engine management ECU can read the signal as a variable reluctor type signal.

Your "someone" may be smarter than you but he didn't know how to design the circuit to do what you want.

You can get a negative signal by adding a small DC offset voltage such as wkrug suggested.

Adding a series capacitor will also give offset but the amount of offset will be determined by the duty-cycle of the input pulse.

 

[Mosaic]

Adding a series capacitor will also give offset but the amount of offset will be determined by the duty-cycle of the input pulse.

Can u elaborate on this please?

 

[rayhall]

You can get a negative signal by adding a small DC offset voltage such as wkrug suggested.

Is this correct ?

**broken link removed**

 

[crutschow]

Can u elaborate on this please?

Certainly. The zero DC level of an AC signal after is is passed through a capacitor is equal to the average value of the AC waveform. Thus the amount of the signal above zero and below zero for a pulse signal will be determined by the duty cycle of the waveform. Thus, for example, if the pulse at the capacitor input were 5V for 25% of the time and 0V for 75% of the time, then the capacitor output (if connected by a resistor to common with a long time constant) average value will be 0.25 x 5V = 1.25V. The pulse high level will then be 5V - 1.25V = 3.75V and the pulse low level will be 3.75V - 5V = -1.25V.

If you change the duty-cycle then you will change the average value of the voltage at the capacitor output and thus the positive and negative value of the pulse.

 

[wkrug]

It looks right.

When I've calculated right, the voltage at the output give out 0V when the input signal is +1.37V. At higher Values it goes positive, at lower Values it goes negative until saturation of the OP is arrived ( Supply Voltage ).

 

[rayhall]

It looks right.

I will make a test circuit to see what I get.

Thank you

Ray.