23 January 2018, 20:34 | #21 |
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This circuit use operational amplifiers so resistors can be pretty high value - in passive matrix you need use different values.
Assuming that impedance load on RGB input will be 75ohm then you must calculate resistors as reciprocal i.e. for Green (1/0.59)*75 = 127 Ohm, Red (1/0.3)*75 = 250 Ohm and Blue (1/0.11)*75 = 682 Ohm. Combined signal from those three resistors will give you grayscale signal that can be feed to Green input of your display. hope that math is correct - calculated this ad hoc. Of course if you wish you can use different resistor values and you may need also active circuit if necessary - nowadays life is so easy - there is plenty of nice circuits to do such things. Forget to mention that those resistors value are not highly agnostic - small variations are acceptable (so 130, 240 and 680 from E24 5% will work OK). Above resistor values seem to be correct however signal loss need to be compensated by using amplifier with gain = 2. Seem pure passive should be quite OK with resistors values divided by 8 (they will be pretty low values but this is good for bandwidth) R=33 Ohm, G=16 Ohm, B=86 ohm. Last edited by pandy71; 26 January 2018 at 19:06. |
29 January 2018, 00:10 | #22 | |
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Quote:
The Apple IIe PAL/NTSC output has the colour carrier signal, controlled by UB8 in the schematics. If you feed this into a video chip expecting a non composite video signal, you will have a speckly display, caused by the high frequency components of the colour information in the signal. If you wanted greyscale video, it's possible to filter out the colourburst but you will lose some video information, with a crude RC filter. |
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29 January 2018, 14:04 | #23 | |
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29 January 2018, 17:42 | #24 |
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29 January 2018, 18:35 | #25 |
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Then you will loose information - for example red and blue... Old CRT's don't loose information even if there is particular phosphor color involved - it is like grayscale seen trough translucent color.
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