[PATCH] iio: adc: ad4080: allow debugfs register access to both channels

Antoniu Miclaus posted 1 patch 1 month ago
drivers/iio/adc/ad4080.c | 13 +++++++++++--
1 file changed, 11 insertions(+), 2 deletions(-)
[PATCH] iio: adc: ad4080: allow debugfs register access to both channels
Posted by Antoniu Miclaus 1 month ago
The dual-channel parts (AD4880/AD4883/AD4884) are built from two
independent ADC dies, each with its own SPI chip select and register
map. They are presented as a single IIO device because the FPGA
interleaves both dies into one DMA stream, so only one direct_reg_access
debugfs file is created and it can only reach channel A's regmap. Channel
B's registers are therefore inaccessible, which blocks bring-up and
debugging of the second die.

Select the target die through the debugfs register address: the low 8
bits are the register offset and the upper bits are the channel index
(0x0nn selects channel A, 0x1nn channel B). A read returns the real
register value of the selected die and each die is addressed
independently; single-channel parts only accept channel 0.

Signed-off-by: Antoniu Miclaus <antoniu.miclaus@analog.com>
---
 drivers/iio/adc/ad4080.c | 13 +++++++++++--
 1 file changed, 11 insertions(+), 2 deletions(-)

diff --git a/drivers/iio/adc/ad4080.c b/drivers/iio/adc/ad4080.c
index 04cd6628ebff..994625a7cf8b 100644
--- a/drivers/iio/adc/ad4080.c
+++ b/drivers/iio/adc/ad4080.c
@@ -145,6 +145,10 @@
 #define AD4080_MAX_SAMP_FREQ					40000000
 #define AD4080_MIN_SAMP_FREQ					1250000
 
+/* debugfs direct_reg_access channel windowing: 0x0RR = ch0, 0x1RR = ch1 */
+#define AD4080_DEBUGFS_REG_CH_MSK				GENMASK(15, 8)
+#define AD4080_DEBUGFS_REG_OFFSET_MSK				GENMASK(7, 0)
+
 enum ad4080_filter_type {
 	FILTER_NONE,
 	SINC_1,
@@ -210,11 +214,16 @@ static int ad4080_reg_access(struct iio_dev *indio_dev, unsigned int reg,
 			     unsigned int writeval, unsigned int *readval)
 {
 	struct ad4080_state *st = iio_priv(indio_dev);
+	unsigned int ch = FIELD_GET(AD4080_DEBUGFS_REG_CH_MSK, reg);
+	unsigned int offset = FIELD_GET(AD4080_DEBUGFS_REG_OFFSET_MSK, reg);
+
+	if (ch >= st->info->num_channels)
+		return -EINVAL;
 
 	if (readval)
-		return regmap_read(st->regmap[0], reg, readval);
+		return regmap_read(st->regmap[ch], offset, readval);
 
-	return regmap_write(st->regmap[0], reg, writeval);
+	return regmap_write(st->regmap[ch], offset, writeval);
 }
 
 static int ad4080_get_scale(struct ad4080_state *st, int *val, int *val2)
-- 
2.43.0
Re: [PATCH] iio: adc: ad4080: allow debugfs register access to both channels
Posted by Jonathan Cameron 4 weeks ago
On Wed, 26 Aug 2026 14:16:13 +0300
Antoniu Miclaus <antoniu.miclaus@analog.com> wrote:

> The dual-channel parts (AD4880/AD4883/AD4884) are built from two
> independent ADC dies, each with its own SPI chip select and register
> map. They are presented as a single IIO device because the FPGA
> interleaves both dies into one DMA stream, so only one direct_reg_access
> debugfs file is created and it can only reach channel A's regmap. Channel
> B's registers are therefore inaccessible, which blocks bring-up and
> debugging of the second die.
> 
> Select the target die through the debugfs register address: the low 8
> bits are the register offset and the upper bits are the channel index
> (0x0nn selects channel A, 0x1nn channel B). A read returns the real
> register value of the selected die and each die is addressed
> independently; single-channel parts only accept channel 0.
> 
> Signed-off-by: Antoniu Miclaus <antoniu.miclaus@analog.com>
Applied to the testing branch of iio.git

thanks,

Jonathan