Microprocessors Lecture 17

SRAM and DRAM

We have looked at some of the ROM devices which can be used in a small microcomputer system and at some of the different technologies which are used to make them. The same technologies are also available for read/write memory, commonly called RAM.

In addition when we look at the most common type, NMOS, there are two categories of memory device; static RAM and dynamic RAM.

Static RAM

In a static memory device the data bits are stored in a simple flip/flop circuit consisting of two cross-coupled transistors. The data is thus stored permanently as long as power is supplied. The advantage of this arrangement is simplicity - the disadvantages are that each memory cell consumes power all the time so the overall power consumption is high and also the memory cell requires two transistors plus load resistors (usually also transistors) in addition to select transistors making the circuit relatively large to construct.

One example of a CMOS static RAM cell is shown below.

Dynamic RAM

The alternative is dynamic RAM which uses a very simple single transistor memory cell in which the data is stored on the gate of an MOS transistor. This charge turns the transistor on if it is present thus making it possible to read the data. The main problem is that the charge leaks away slowly and data will be lost if it is not refreshed periodically (typically every 2ms). The process of reading the data also refreshes it, so effectively every bit of memory must be read every 2ms.

To reduce the amount of time spent in refreshing each memory chip is organised as a matrix of cells and when a bit in any row is read the entire row (of perhaps 256) cells is refreshed. Thus to refresh the entire chip one read operation has to be performed on every row (again typically 256). Also all the chips in the system can have their row addresses connected together so that the same row in every chip is refreshed at the same time. Thus only 256 read cycles are needed every 2ms to keep the memory intact. A dynamic memory controller ensures that at regular intervals a memory refresh cycle takes place on the next row - often this can be done when the processor is not accessing memory but sometimes it is necessary to make the processor wait while the refresh takes place.

The big advantage of dynamic memory which makes the extra complication of refreshing worthwhile is that because of the simplicity of the memory cell very many more bits can be fitted on the same size of chip (typically a factor of 8). Thus while 1MBit memory chips are commonplace (and 4 MBit available) static memory devices 64 kByte is a more typical size. Another advantage is that unless the cell is being read from or written to it does not consume any power. Thus the overall power consumption is much lower.

Another problem in dynamic memories is that the capacitor on which the charge is stored is so small that a single ionising particle passing through it can sufficiently disrupt the charge pattern so as to cause a mis-read of the data. Thus in large memory systems, and ones where data integrity is crucial memory checking is necessary - either a simple parity check by adding one more bit to each word or a more complex check allowing correction which will add several bits to each word.

The address inputs to a DRAM are multiplexed into ROW and COLUMN addresses by a DRAM controller (or other multiplexing hardware).

For example, a small 64K x 1 DRAM chip (4164) has the pinout below:

This is a 64K bit device and 64K bits of memory needs 16 address lines. First, half of the 16 address lines are applied to the DRAM's A0-A7 with /RAS (and latched internally). Then the other half of the 16 address lines are applied to the DRAM's A0-A7 with /CAS as shown below.

Bit and Word Organised Memories

There are two ways in which any memory device can be organised. Either each chip can store one bit of a word, thus requiring 8 such chips to make a memory system capable of storing 8-bit words, or each chip can store a number of complete 8-bit words. In a memory system storing say 64 Kbytes the total number of chips would be the same, either 8 chips each storing 64 Kbits or 8 chips each storing 8 kBytes. There are three reasons for choosing the former. The first is that the total number of connecting pins on the one bit chip is fewer than the byte chip. 

                             Address lines       Data lines      Total

              1 bit chip          16                   1          17

              8 bit chip          13                   8          21
This means that the bit organised chip is physically smaller than the byte organised one.

The second reason relates to the connections to the chips on the printed-circuit board. In the first case each data bus line only connects to one chip while in the second each data bus line connects to 8 chips. This means the loading on the data bus is considerably higher (therefore requiring extra drivers or a degradation in speed) and also that considerably more space is required for the connections on the PCB

The third reason is that extra address decoding is required since only 13 of the address lines go to each chip therefore the other 3 have to be decoded (1 of 8) to drive chip select inputs.

Why have byte or word organised chips at all then? The answer is that in some systems, particularly small dedicated microprocessor ones, only a small amount of memory is needed. For example if only 16 kBytes is needed then this can be satisfied with two 8 kByte chips whereas using bit-organised would require 8. Thus both types of device have a role in system design.

Address Decoding

When an address is presented to a memory chip it must obviously select one bit or one word depending upon which type of chip it is. This process of selecting which memory cell is selected is known as address decoding. The process is usually carried out in stages. As we have already seen, one stage is selecting one of the many memory or interface devices in the system. For a given address only one chip will be selected if we have byte organised memory or one group of 8 chips will be selected if we have byte- organised.

The on-chip decoding takes place in two stages. The address connections are divided into two approximately equal groups one group selects the row in the memory matrix in which the required cell is located and the other group selects the column. The row address decoder wil consist of buffers and inverters which make each address line and its complement available. AND gates then pick out every combination of addresses - thus the first row (row0) will be:- 


               row0         /A7./A6./A5./A4./A3./A2./A1./A0

               rowl         /A7./A6./A5./A4./A3./A2./A1. A0

               row2         /A7./A6./A5./A4./A3./A2. Al./A0

		..

               row254        A7. A6. A5. A4. A3. A2. A1./A0

               row255        A7. A6. A5. A4. A3. A2. A1. A0
When a particular row is selected then the outputs of every cell in that row will be made available on the column data lines. The column address is decoded in the same way using addresses A8 to A16 and this decoded signal is used to select one out of these 256 outputs to be fed to the single output pin (assuming bit organisation). If the chip is byte organised then only five column address lines will be used and decoded to give 1 out of 32. This signal then selects a group of 8 out of the 256 available outputs to be fed to the output pins.

Chip Select Inputs

All memory chips have at least one chip select input. If this input is FALSE then the chip does not respond to any address on the address bus. Thus this is used both for higher level address decoding and for timing.

In a small system it is desirable to reduce the number of chips as far as possible and this can be done by partial address decoding as we have already seen. This can be taken a step further by providing several chip select inputs on memory and interface devices, preferably some active high and some active low. For example suppose we have small 128 byte word organised memory devices with six chip select inputs CS0 to CS5 where CS0 and CS3 are active high and the rest are active low. In other words the chip will only respond if the pattern 100100 is present on the CS inputs. Suppose also that we want to divide memory space such that below $8000 we have RAM, from $8000 to $BFFF is for interfaces and above $C000 is ROM. In the RAM area we could easily have a number of such memory chips without requiring any additional address decoding. 


A15 A14 A13 A12 All A10  A9  A8  A7  A6  A5  A4  A3  A2  Al  A0

CS1 CS2                      CS4 CS5  |   To all chips	      |

CS1 CS2                      CS4 CS0  |			      |

CS1 CS2                      CS3 CS5  |			      |

CS1 CS2                      CS3 CS0  |			      |



                          A15 A14    A8   A7

Chipl  $0000 to $007F      0   0      0   0

Chip2  $0080 to $OOFF      0   0      0   1

Chip3  $0100 to $017F      0   0      1   0

Chip4  $0180 to $01FF      0   0      1   1

Chipl  $0200 to $027F      0   0      0   0

   etc.
Notice that we have in fact used all six of the chip select inputs in different combinations. Notice also that because we have not completely decoded the addresses the memory chips will appear repeatedly for every different combination of A9 to A13 i.e. each location will have 32 different addresses.

Take a different example, where a system requires just one 128 byte RAM, a 2Kbyte ROM and 3 PIAS. Each of them has a different number of chip select inputs 


               2708       2 Kbyte ROM        /CS

               6810       128 byte RAM       CS0 /CS1 /CS2 CS3 /CS4 /CS5

               6821       PIA (4 locations)  CS0  CS1 /CS2
As we know the ROM must be at the top of memory and since we have only one chip select we have to allocate half of memory from $8000 to $FFFF to this one device. We use A15 through a NAND gate with E (Enable) to do this.

It is normal to put the RAM at location 0 upwards - we can do this by connecting A15 and A14 to /CS1 and /CS2. Enable is connected to CS0 to enable RAM only when Enable is 1.

The PIAs can be allocated memory from $4000 upwards by connecting A15 to /CS2 and A14 to CS1. The remaining CS0 must separate the three so we can connect A2 to CS0 in one case, A3 in the second and A4 in the third. Thus we have:- 


                     PIA1 $4004 to $4007

                     PIA2 $4008 to $400B

                     PIA3 $4010 to $4013
A danger in doing this is that if the processor outputs an address with A2, A3 and A4 all '1' then all three devices will respond. Such an address would be $401F. Clearly the programmer would have to take great care to avoid such a situation. This situation only occurs when partial address coding is used, <
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