Page: Module 4 : Unit 2
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Module 4

SRAM DRAM
Retained stored info as long as power supply On. Content is lost as soon as supply is removed loses its stored info in a few milli sec even though its power supply is on
Uses conventional flip Flop to store bit 0 or 1 store info in the form of charge on a cap
6 transistor to form A mem cell Low packing density 1 transistor per mem cell high packing density
Faster slow
Expensive cheap
1. 6T STRAM & stability criteria for read & write. write operation Write ‘0’

1. assume ‘1’ is stored at V1

2. M1 &M2 = off &M2 & M5 = on

3. V1 =Vdd& V2=0 before M3 & M4

4. Activate WL M3 &M4 =on

5. Since V2<Vt ,1 ,V2 cannot be used to turn on Ml</p>

We need to turn ON ML, so that the path is created from V1 to ground and voltage at V1 will drop to 0 since path pulled down to gnd

1. Think turning off M2

i.e V1<Vt, 2 to off the M2</p>

To achieved V1 is forced below threshold voltt

when V1 =Vtd , M3= lin& M5=sat

$\frac{KP5}{2}(0 - V_{DD} - V_{TP})^2 = \frac{K_{n,3}}{2}(V_{DD - V_{Tn}})$

Due to this, M2 will be forced to cutoff& M1 turns on

And hence ‘0’ is written at Vl

1. SRAM using resistive load =>write ‘1’

• Make WL=1 (activate)

• M3 &M4 =on

• force data =1& data=0

• voltage at Q , Vq=1, Vq=1

• hence logic 1 is written

=> write ‘0’

• activate WL M3 & M4 = on

• force data = 0 & data =1

• Vq =0 ,M2 = off &Vq= 1 , M1 = on

• Vq =0 , “0” is written

• Data and data are prechargedVdd

• Wl is activated

• M3 and M4= on

• Logic 1 is started at q

• Vq=1 &Vq=0

Hence M1-off & M2- on

• $V_Q \gt V_Q$, It is reads as logic ‘1’ by sense amplifier

• Data and data =precharged

• Activate wl =M3 and M4 on

• Logic ‘o’ is started at Q

=> VQ=0 &VQ=1

M1-on & M2-off

$V_Q \lt V_Q$

It reads as logic ‘0’ by sense amplifier

1. 3T DRAM =>write ‘1’

• Data= 0 because data to be written is logic 1

• Word select is pulled high

=> M1 turn on

• with ml conductance charge of C1 is shared which side since cap C1 is very large compared to C1, C attains approximately same logic Hai as column cab C1 at end of charge sharing process and M2 turns on

[ 7 shares the charge with C till both caps are not same logiclevepu

• C stores the logic 1 in form of charge

• A frteroperation WS is made 0

=>write ‘0’

• Data=1 because data to be stored is 0

• MD is pulled to gndi.e logic 0 (MD is on)

• WS is pulled high M1 =on

• Voltage level AT C1 and C is pulled to logic 0 through M1 and MD

• M2 terms off ,capChas no charge left ,0 is written

• Make read select high

• M3- On

• M2 and M3 create a conduction path between the cap C2 and gnd. The cap C2 discharges through M2 and M3.

• This decreases in voltage and C2 is interpreted by data read circurity as stored logic 1 • Main RS- high

• M3 on and M2 off

[ M2 is off because when 0 was being written in write 0 operation it turns off M2]

-M2=off there is no conducting pass between C2 and ground C2 do not discharge and thus high-level C2 is interpret as logic 0 by data read circuitry

1. 1T DRAM • Before read operation data line is charged to vdd/2

• Activate wl =M1 on

• If logic one is stored at C1 then charge is shared with C2

• This result in change in the voltage of data line

• The sense amplifier senses this change and generates valid output

• If voltage at DL increases then stored bit is 1 and DL decreases stored bit 0 in DRAM cell

• Direction of voltage change determines what is stored in cell

=> write

• To write 0 make data line 0 to write one make DL=1

• Activate the word line

• Based on data line C1 is either charge or discharge

• If C1 is charging logic 1 is being written

• If C1= discharging =logic 0

=> refresh

• Perform a dummy read operation after every read or write operation

=> Basic working of EPROM, EERROM & flash

• consists of 1 torwith floating Gate whose vth can be changed recently by applying and electric field at it Gate

• Thus when electron accumulated at floating Gate its voltage increases and it regarded to be in 1 state

• when electron are removed it is 0 state

• this injecting or removing of electron is done by programming technique

[ programming technique remains same for EPROM, EEROM and flaws]

1. Programming techniques

=> Hot electron injection • When high voltage Eg 12 volt is applied at control gate and across [Vds (Eg 6V)] drain to source electron are heated by high lateral electric field

• Avalanche breakdown occurs near train and electron hole pair are generated by impact ionization

• the control that attracts the electron and inject into the floating Gate and holes flow to the substrate as substrate current

• this is known as hot electron injection

=> Fowler- Nordheintunneling • The floating rate is programmed or arranged by the tunneling current of the oxide by high field of >10 Mv/cm

• 10 volt and high voltage (Eg 2) volt are applied that control gate and source respectively electrons at the floating Gate are ejected to the source by tunneling effect

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 modified 3 days ago  • written 4 days ago by Mayank Aggarwal ♦♦ 0