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Linus Torvalds1da177e2005-04-16 15:20:36 -07001/*
2 * linux/arch/arm/common/icst525.c
3 *
4 * Copyright (C) 2003 Deep Blue Solutions, Ltd, All Rights Reserved.
5 *
6 * This program is free software; you can redistribute it and/or modify
7 * it under the terms of the GNU General Public License version 2 as
8 * published by the Free Software Foundation.
9 *
10 * Support functions for calculating clocks/divisors for the ICST525
11 * clock generators. See http://www.icst.com/ for more information
12 * on these devices.
13 */
14#include <linux/module.h>
15#include <linux/kernel.h>
16
17#include <asm/hardware/icst525.h>
18
19/*
20 * Divisors for each OD setting.
21 */
22static unsigned char s2div[8] = { 10, 2, 8, 4, 5, 7, 9, 6 };
23
24unsigned long icst525_khz(const struct icst525_params *p, struct icst525_vco vco)
25{
26 return p->ref * 2 * (vco.v + 8) / ((vco.r + 2) * s2div[vco.s]);
27}
28
29EXPORT_SYMBOL(icst525_khz);
30
31/*
32 * Ascending divisor S values.
33 */
34static unsigned char idx2s[] = { 1, 3, 4, 7, 5, 2, 6, 0 };
35
36struct icst525_vco
37icst525_khz_to_vco(const struct icst525_params *p, unsigned long freq)
38{
39 struct icst525_vco vco = { .s = 1, .v = p->vd_max, .r = p->rd_max };
40 unsigned long f;
41 unsigned int i = 0, rd, best = (unsigned int)-1;
42
43 /*
44 * First, find the PLL output divisor such
45 * that the PLL output is within spec.
46 */
47 do {
48 f = freq * s2div[idx2s[i]];
49
50 /*
51 * f must be between 10MHz and
52 * 320MHz (5V) or 200MHz (3V)
53 */
54 if (f > 10000 && f <= p->vco_max)
55 break;
56 } while (i < ARRAY_SIZE(idx2s));
57
Eric Sesterhennd1d8f7d2006-09-26 14:22:00 +020058 if (i >= ARRAY_SIZE(idx2s))
Linus Torvalds1da177e2005-04-16 15:20:36 -070059 return vco;
60
61 vco.s = idx2s[i];
62
63 /*
64 * Now find the closest divisor combination
65 * which gives a PLL output of 'f'.
66 */
67 for (rd = p->rd_min; rd <= p->rd_max; rd++) {
68 unsigned long fref_div, f_pll;
69 unsigned int vd;
70 int f_diff;
71
72 fref_div = (2 * p->ref) / rd;
73
74 vd = (f + fref_div / 2) / fref_div;
75 if (vd < p->vd_min || vd > p->vd_max)
76 continue;
77
78 f_pll = fref_div * vd;
79 f_diff = f_pll - f;
80 if (f_diff < 0)
81 f_diff = -f_diff;
82
83 if ((unsigned)f_diff < best) {
84 vco.v = vd - 8;
85 vco.r = rd - 2;
86 if (f_diff == 0)
87 break;
88 best = f_diff;
89 }
90 }
91
92 return vco;
93}
94
95EXPORT_SYMBOL(icst525_khz_to_vco);
96
97struct icst525_vco
98icst525_ps_to_vco(const struct icst525_params *p, unsigned long period)
99{
100 struct icst525_vco vco = { .s = 1, .v = p->vd_max, .r = p->rd_max };
101 unsigned long f, ps;
102 unsigned int i = 0, rd, best = (unsigned int)-1;
103
104 ps = 1000000000UL / p->vco_max;
105
106 /*
107 * First, find the PLL output divisor such
108 * that the PLL output is within spec.
109 */
110 do {
111 f = period / s2div[idx2s[i]];
112
113 /*
114 * f must be between 10MHz and
115 * 320MHz (5V) or 200MHz (3V)
116 */
117 if (f >= ps && f < 100000)
118 break;
119 } while (i < ARRAY_SIZE(idx2s));
120
Eric Sesterhennd1d8f7d2006-09-26 14:22:00 +0200121 if (i >= ARRAY_SIZE(idx2s))
Linus Torvalds1da177e2005-04-16 15:20:36 -0700122 return vco;
123
124 vco.s = idx2s[i];
125
126 ps = 500000000UL / p->ref;
127
128 /*
129 * Now find the closest divisor combination
130 * which gives a PLL output of 'f'.
131 */
132 for (rd = p->rd_min; rd <= p->rd_max; rd++) {
133 unsigned long f_in_div, f_pll;
134 unsigned int vd;
135 int f_diff;
136
137 f_in_div = ps * rd;
138
139 vd = (f_in_div + f / 2) / f;
140 if (vd < p->vd_min || vd > p->vd_max)
141 continue;
142
143 f_pll = (f_in_div + vd / 2) / vd;
144 f_diff = f_pll - f;
145 if (f_diff < 0)
146 f_diff = -f_diff;
147
148 if ((unsigned)f_diff < best) {
149 vco.v = vd - 8;
150 vco.r = rd - 2;
151 if (f_diff == 0)
152 break;
153 best = f_diff;
154 }
155 }
156
157 return vco;
158}
159
160EXPORT_SYMBOL(icst525_ps_to_vco);