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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?
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?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?
Your "someone" may be smarter than you but he didn't know how to design the circuit to do what you want.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.
Adding a series capacitor will also give offset but the amount of offset will be determined by the duty-cycle of the input pulse.
You can get a negative signal by adding a small DC offset voltage such as wkrug suggested.
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.Can u elaborate on this please?