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how to detect zero volt ??

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andy2000a

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lm393 input impedance

in single power supply CMOS asic , how to detect "zero volt"
zero cross circuit need it ..
 

Feed input signal through a capacitor to a voltage divider such that mid point of the voltage divider is 0.5Vcc. Input signal at zero volts is now 0.5Vcc. Just a dd a comparator to detect signal at +/- some small value around 0.5Vcc.
 

A comparator with a depletion NMOS or enhancement PMOS input pair can have a common-mode input range that includes 0V in a single-supply configuration.

Simply use a comparator of this type. In the specs, it will state whether the common mode range includes GND.

--RussMax
 

sorry .. my problem is zero detect need fast speed and small volt

input < 20mv and delay < 30ns

and only single power 3.3v
 

hi
try with this.
two stage balnced OTA with pmos input pair
DIAGRAM attached.
this might solve ur probs..but how much power u can spend (to get lesser than 30ns)..
 

Come on... the circuit to detect 0 volt can be very simple. Throw away all the circuits presented above. Obviously there is one important thing: where the input signal comes from ? You can use any known comparator (for example LM393) that can operate from a single power suppy. The circuit shown in the diagram is independent of the supply voltage. The voltage applied to the inverting input is always equal to half a supply voltage (VCC/2). Tho detect 0V, the voltage at the non-inverting input must also be polarized to VCC/2. It is expressed by an equation: Vneg = (VCC + Vinput) / 2. One must remember that R1/R2 = R4/R5. R6 resistor ensures some hysteresis that eliminates oscillations at the output. The supply voltage may be in 3,3 to maximum voltage supply for LM393 (I don't remember the accurate value, you must check this in LM393 datasheet). If you need higher input resistance, increase R4 and R5 (they must have the same values !)
 

The last circuit proposed only work in cases where the input source has low output impedance. Why not use a fast rail to rail comparator? Does the signal need to be centered to Vcc/2?
If you need to build your own circuit for this purpose, and you need 30ns or faster for 20mV overdrive, you need a fast input stage of low gain, (avoid the slow current mirror load), followed by a high gain stage, use positive feedback where applicable.
 

Input impedance can be increased by choosing greater values of R4 and R5. There are many fast comparators powered from single power source. Many of them can be found at www.national.com. The circuit doesn't have to be centered to VCC/2 but this is the most comfortable way (the resistors used have the same values). Besides, CMOS output drivers have quite low impedance to drive the circut. But I agree with the fact that if high impedance is needed, the circuit must be improved.
 

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