# How does the Pre-charge and Equalizing circuit shown in the below figure work? (from precharge circuit)

Could you please explain the operation of the above circuit regarding how the transistors Q8 and Q9 are responsible for charging the Bit Lines with VDD/2 while Q7 equalizes the voltage between them ?

• Please put enough details and a schematic if required in the question so that we can answer it without having to follow links. You will get more attention from the busy guys/gals and the question will still make sense when the link dies. – Transistor Nov 21 '17 at 18:35

## 2 Answers

Before every read/write operation, the Bit Lines must be pre-charged and equalized. This is done to assure that small voltages can be easily detected by the Sense Amplifier. The transistors Q8 and Q9 are responsible for charging the Bit Lines with VDD/2 while Q7 equalizes the voltage between them. The Pre-charge circuit is activated through the precharge input (φP).

Q7/8/9 act like voltage controlled switches (with resistance). When φP is active the FETs are turned on. Q8 connects ~B to Vdd/2, Q9 connects B to Vdd/2, and Q7 connects ~B to B. So effectively you have this:- simulate this circuit – Schematic created using CircuitLab

The bus capacitances charge up (or down) to Vdd/2 through the Drain-Source resistance of Q8 and Q9, while current can flow in either direction through Q7 to equalize the voltages.

• how does Q7 equalize the voltage? – user2077648 Nov 25 '17 at 8:18
• If the voltages are different then current will flow through Q7, charging up the lower side and discharging the higher side until their voltages are (almost) equal. If you don't understand how this works then study basic electronics eg. Ohm's Law, Capacitors, Charging a capacitor. – Bruce Abbott Nov 25 '17 at 16:19 simulate this circuit – Schematic created using CircuitLab

Assume C1,C2 are the bit bus lines that may have some unknown voltage on each before the 3 switches are gated ON. The result is a rapid RC time constant towards Vdd/2 on each line.

Q7 is slightly bigger with lower RdsOn so they are equalized fast but not necessarily at Vdd/2, while Q8,Q9 are normalized to be equalize RdsOn for Nch, Pch to arrive at Vdd/2 with assistance from Q7. Values are not exact.