video card

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I would like my computer to be able to display text and images on a monitor. I will design and build a video card, which will connect between the computer and a monitor.

The video card needs a number of different stages. The first stage to consider is an interface, what conenction and standard am I going to aim for? This output interface will connect the card to the monitor.

Over the years there have been many standards of output interfaces. The main ones are CGA, EGA, VGA and SVGA. Of these VGA (Video Graphics Array), which was introduced with the IBM PS/2, in 1987, became ubiquitous. VGA is what I have chosen to implement. I chose this because many devices currently in use (monitors, TVs, projectors) have VGA connections and it is era-appropriate. Also, it would be possible to interface with other output interfaces, including BNC and DVI. BNC will work because it is just a breakout of the connections that I will be using in the VGA standard. DVI is a digital standard but is workable because it is compatible with analogue signals, via a mode called DVI-A.

Figure 1 shows a VGA connection. The VGA connector is a 15 pin D-subminiature, with 3 rows – their functions are listed in table 1.

The signal is analogue but the video card will be digital. Therefore, another stage that I need to design is a circuit that converts the digital signals that I generate to analogue signals that a monitor requires. This type of circuit is known as a digital to analogue converter (DAC).

Although there are 15 connections, I will only need to use 10 of these, as shown in figure 2 and table 2. These are red, green, blue, horizontal sync and vertical sync, plus five ground connections.

1	RED
2	GREEN
3	BLUE
4	ID2/RES
5	GND
6	RED_RTN
7	GREEN_RTN
8	BLUE_RTN
9	KEY/PWR
10	GND
11	ID0/RES
12	ID1/SDA
13	HSync
14	VSync
15	ID3/SCl

1	RED
2	GREEN
3	BLUE
13	HSync
14	VSync

The implementation that I would like to aim for is a video card that satisfies the requirements of the original VGA specification, which are in table 3.

screen refresh rate	60 Hz
vertical refresh	31.46875 kHz
pixel frequency	25.175 MHz
HORIZONTAL
visible area	640	25.422045680238
front porch	16	0.63555114200596
synch pulse	96	3.8133068520357
back porch	48	1.9066534260179
whole line	800	31.777557100298
VERTICAL
visible area	480	15.253227408143
front porch	10	0.31777557100298
sync pulse	2	0.063555114200596
back porch	33	1.0486593843098
whole frame	525	16.683217477656

There are some problems that will need to be overcome to be able to implement VGA. The first is the VGA signal timing. To keep each frame in time, there is a signal timing that needs to be adhered to. What signal timing is required is dependent on the resolution and refresh rate and in term is a function of the pixel frequency. The required pixel frequency is 25.175MHz. There are a number of ways of generating this, I will use a crystal oscillator.

I need to design a circuit that takes the 25.175 MHz pixel frequency and uses it to generate a horizontal sync and a horizontal blanking pulse.

I need to design a circuit that takes the 25.175 MHz pixel frequency and uses it to generate a vertical sync and a vertical blanking pulse.

Figure 3 shows the anatomy of a frame. Each frame is created by starting in the top-left corner and moving to the right. When one line is complete, the next line begins, progressing from left to right. When all of the lines have been drawn, one frame is complete. The process then repeats. As well as the visual information, there is information that corresponds to horizontal and vertical timing. These are necessary to ensure that the image is drawn at the correct point! These timing signals can be broken down into a back porch, sync pulse and a front porch.

The front porch occurs after the line of video information has ended and blanks the signal, ready for the sync pulse that follows. The back porch occurs after the sync pulse and gives time for the beam (of a CRT) to move from right to left, ready for the next line. There is no beam with an LCD monitor but the timing still needs to be followed, to keep the timings correct. These pulses also apply to vertical timing. However, the back porch for the vertical signal is also needed for the beam to return not just from right to left but also to the top!

Looking at table 3 we can see that each line needs to 800 pixels and lasting 31.777557100298 µs. I need to design a circuit to meet this criteria, it must know when 31.777557100298 µs has elapsed and the reset to 0. There are a lot of ways of achieving this but in any scenario, I need to create some form of counting circuit. 800 in binary is 1100100000, which is a 10-bit number. What I will do is use three 4-bit counters and cascade them so that they form a 10-bit counter.

The other criteria is that the circuit is synchronous, We can split counters into two categories. asynchronous and synchronous. With an asynchronous counter a clock is provided at the first stage and ripples through to the following stage(s). With a synchronous counter a clock is provided to each stage, making everything clock in time. I need to use a synchronous counter, to keep everything in time.

To build my counter circuit, I will use three 74161. These can be chained together and they are synchronous counters. I will tie all of their clock pins together, making each count together. Since the 74161 is a 4-bit counter, it can count to 16. With the three cascaded together, the circuit will create a 12-bit counter, allowing me to count up to 4096! Therefore I will modify it to stop at 256 and reset back to 0. I will also tie all of the clear pins together, so that all of the counters are reset to 0 at the same time. To detect when I have reached the required numbers, I will connect the relevant outputs of the counters to gates, to get the required combination.

HORIZONTAL

264

I have a counter that counts from 0 to 4096. Now I need to modify it to count to 264 and then reset it to 0. I will use logic gates to detect when 264 has been reached. In binary, 264 is 100001000. If I connected a 9-input AND gate to the outputs of the counter and an inverter at the input of the 4th and 9th bit, the AND gate will have a 1 at its output when it has counted to 100001000. This will work but can be improved on with some optimisations. I could reduce this to just a 2-input AND gate! This is possible because I only really need to know when the 4th AND 9th bit is 1. Any 1s in the 5th, 6th, 7th and 8th bit are irrelevant because the counter will never count that high! I could then connect the output to the clear pins of the 74161s. However, the clear pin is active low, so I would need to invert the output. However, there is another solution, to use a NAND gate to replace the separate AND and NOT gate. I will use a 7400, which contains four 2-input NAND gates. Now when 264 is reached both inputs of the NAND gate will be 1 and the output will be 0, causing the counter to reset to 0.

It would be good to check if the timing is correct. One way to do this is with a logic analyser. If you look at figure 4 you will see the results. This shows that the scan line that I have generated is 26.42µs. I need 26.4µs, which will be close enough.

Now that I have a counter that can count to 264 and then resets to 0, I need circuitry to detect when the front porch, sync pulse and back porch have been reached. I will use the same method of using NAND gates to achieve this.

242

Now I need to add logic to detect when the counter has reached 242. 242 in binary is 11110010. Using the same method as I did to detect when 264 had been reached, I need to detect when the 2nd, 5th, 6th, 7th AND 8th bit has changed to a 1. There are no ICs in the 74 series that have NAND gates with 5-inputs. There is an 8-input (7430) NAND gate. This will work fine because I will tie the unused inputs high, behaving the same as a 5-input NAND gate!

210

Now I need to add logic to detect when the counter has reached 210. 210 in binary is 11010010. I need to detect when the 2nd, 5th, 7th AND 8th bit has changed to a 1.

200

Now I need to add logic to detect when the counter has reached 200. 200 in binary is 11001000. Using the same method as I did to detect when 264 had been reached, I need to detect when the 4th, 7th AND 8th bit has changed to a 1.

VERTICAL

I will build two counters, one that counts the pixels horizontally and one that counts the pixels vertically. To count the pixels I will use a 74161. I need it to count up to 264, which will require 9 bits. The 74161 is a 4-bit counter, so I will need 3 of them.

The clear pin will reset the counters to 0. I will need them all to clear at the same time, so I will tie them together. All of the counters will use the same clock, so I will tie all of the clock pins together.

I won’t be using the data inputs because I won’t be loading any data into this. The 75161s will be used purely as counters. I will connect all of the enable pins high, as I want the chips to always be enabled. Because I will not be using the data inputs, I will not need to use the load pins. The load pins are active low, so I will tie all of them high. The enable T is what is used for the ripple carry. So I will use the first one, so that the 4 bits in the first counter rolls over to the second counter and then to the third. The ripple carry output of the 1st is connected to the enable of the second. Then the ripple carry of the second chip to the enable of the third chip. This creates a cascade, so I can count 12 bits, although I only care about the first 9 bits. I should now have a 12-bit ripple counter. To check that it is working as intended, I will connect it to 12 LEDs. I can’t connect the clock signal from the computer because that would be too fast to see anything happening. I will use my 555 timer circuit to provide the clock signal. It’s counting in binary, which is what I want.

We want to start at 0 and know when we have reached 200, 210, 242 and 264. Then I want to reset back to 0.

IMAGE

The pixel data needs to be stored somewhere in memory. I will store images in an eeprom. eeprom can give the byte stored at address according to the x and y position. The byte can represent the pixel data. VGA expects a voltage between 0 and 0.7 V. This range controls the intensity of each colour. I will connect each output from the counters to the address bus of the eeprom. The data bus will be connected to the inputs to the VGA connector. I will use 6 lines of the data bus, 2 for each colour. The horizontal sync signals generated from the counter circuitry will be connected to their corresponding VGA connections.

INTERFACE

Now that I have a working graphics card, I want to connect it to the computer, so I can run programs and load pictures that output to it. I plan to remove the eeprom that I used for testing and use the eeprom in the computer to store programs and pictures. I will then use the ram in the computer to load these into. For this to happen, I need a way of isolating the graphics card from the computer when the CPU is accessing the RAM. The reason for this is that if I did not do this, then whatever is being outputted to the RAM would cause ‘garbage’ to be displayed!

There are a number of solutions to this problem, the one that I will use is DMA – direct memory access. DMA is a way of allowing a connection to be made to the RAM, without the involvement of the CPU. This means that only visual data will be outputted and also allows the CPU to do work at the same time.

To implement DMA I will use a bus transceiver. A transceiver allows two-way communication between 2 buses. Transceiver is a portmanteau of transmitter and receiver. I will use a 74245, which is an octal bus transceiver. The 74245 contains a direction pin, which is used to control the direction of data. There is also an enable pin, which can be used to isolate circuits. Using these 2 pins I can control the connections between the graphics card and the RAM. This is called bus mastering, controlling which bus has control. The plan is to give the graphics card control during the period that a frame is drawn and to give the CPU control during blanking periods.

As well as controlling when the graphics card has access to the RAM, I need to be able to stop the CPU from running when it should not be accessing the RAM. The 6502 has 2 pins that I can use to implement this: RDY and BE. RDY is the ready pin. BE is the bus enable pin.

The plan is to allow the graphics card to use RAM, stop the CPU from carrying out instructions and to stop the CPU from using the data and address buses during display time. During the blanking period the CPU will continue as normal.

The BE pin only exists on the CMOS 6502, not the NMOS version! This is another good reason to use the CMOS version. The BE (bus enable) pin allows external control of the buses.

The RDY (ready) pin can be used for a number of operations, here I wish to use it to connect to my DMA controller. When this pin is high the processor runs, when it is low it halts the processor (in its current state).

It is the use of the combination of the BE and RDY pins that will allow me to implement DMA as it will force the 6502 to give up control of the buses. I will design circuitry that will be able to make the RDY pin go low, forcing the 6502 to halt at its current state. The circuitry will also be able to make the BE pin go low, putting all of the 6502’s output lines into a high-impedance state – meaning that it cannot have any effect on the address and data lines (and some other pins).

I can use a slower pixel clock by compensating – by sending fewer pixels. I can’t change the timings but they are based on the pixel frequency and the number of pixels. What matters Is how long is the visible area, front porch, synch pulse and back porch.

DAC

What needs to be done is to take the output from the EEPROM, which is digital and transform it into a voltage that is between 0 and 0.7V. What is needed is a Digital to Analogue Converter (DAC). The type of DAC that I will use is an R2R DAC. The name is derived from the fact that it is composed of resistors that are either one value R or 2R. The EEPROM will be outputting an 8-bit digital signal – this needs to be converted to a signal that a VGA monitor can understand! The VGA specification (table 1) expects a 75ohm between 0 and 0.7V. The value between this range determines how much of each colour there is. To get the correct voltage I will use a voltage divider. Point A is 5V and point C is 0V. Point B will be in between these values, determined by the size of R2 compared to the total resistance (R1 + R2). If R2 is half of the total resistance (which means that R1 = R2). then point C will be half of the voltage (2.5V). R2 / R1 + R2. Imagine inside a monitor the colours are connected to ground through a 75ohm load. If we have 5V and want to get it down to 0.7V for the monitor, then I can add a resistor, creating a voltage divider between the resistor I add and the resistance inside the monitor. Therefore, at point C I want it to be 0.7V. The internal 75ohm is fixed by the resistor that I add can be whatever I want it to be. To work out what value of resistance would make 0.7V I can use the formula. 0.7V = 75 / R + 75. Need to solve for R. 5 x 75 = 375. 375 – 52.5 = 322.5. 322.5 / 0.7 = 460.7.

Bus contention.

conclusion

64 colour graphics card. 6-bit RGB.

HSync and VSync are digital.

testing

If you look at figure 5, you will see that the video card horizontal and vertical sync is working, as the monitor has correctly said that it is setup for 640 x 480, 60 Hz.