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DVR21 数据表(PDF) 291 Page - Microchip Technology

部件名 DVR21
功能描述  16-Bit General Purpose Microcontrollers with 256-Kbyte Flash and 16-Kbyte RAM in Low Pin Count Packages
PDF  416 Pages
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制造商  MICROCHIP [Microchip Technology]
网页  http://www.microchip.com
标志 MICROCHIP - Microchip Technology

DVR21 数据表(HTML) 291 Page - Microchip Technology

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 2016-2018 Microchip Technology Inc.
DS30010118D-page 291
PIC24FJ256GA705 FAMILY
24.4
Achieving Maximum A/D
Converter Performance
In order to get the shortest overall conversion time
(called the ‘throughput’) while maintaining accuracy,
several factors must be considered. These are
described in detail below.
• Dependence of AVDD – If the AVDD supply is < 2.7V,
the Charge Pump Enable bit (PUMPEN,
AD1CON3<13>) should be set to ‘1’. The input
channel multiplexer has a varying resistance with
AVDD (the lower AVDD, the higher the internal
switch resistance). The charge pump provides a
higher internal AVDD to keep the switch resistance
as low as possible.
• Dependence on TAD – The ADC timing is driven
by TAD, not TCYC. Selecting the TAD time correctly
is critical to getting the best ADC throughput. It is
important to note that the overall ADC throughput
is not simply the ‘Conversion Time’ of the SAR. It
is the combination of the Conversion Time, the
Sample Time and additional TAD delays for
internal synchronization logic.
• Relationship between TCYC and TAD – There is
not a fixed 1:1 timing relationship between TCYC
and TAD. The fastest possible throughput is funda-
mentally set by TAD (min), not by TCYC. The TAD
time is set as a programmable integer multiple of
TCYC by the ADCS<7:0> bits. Referring to
Table 32-25, the TAD (min) time is greater than the
4 MHz period of the dedicated ADC RC clock
generator. Therefore, TAD must be two TCYC in
order to use the RC clock for fastest throughput.
The TAD (min) is a multiple of 3.597 MHz as
opposed to 4 MHz. To run as fast as possible,
TCYC must be a multiple of TAD (min) because
values of ADCSx are integers. For example, if a
standard ‘color burst’ crystal of 14.31818 MHz is
used, TCYC is 279.4 ns, which is very close to TAD
(min) and the ADC throughput is optimal. Running
at 16 MHz will actually reduce the throughput,
because TAD will have to be 500 ns as the TCYC of
250 ns violates TAD (min).
• Dependence on driving Source Resistance (RS) –
Certain transducers have high output impedance
(> 2.5 k
). Having a high RS will require
longer sampling time to charge the S/H cap
through the resistance path (see Figure 25-3).
The worst case is a full-range voltage step of
AVSS to AVDD with the sampling cap at AVSS. The
capacitor time constant is (RS + RIC + RSS)
(CHOLD) and the sample time needs to be six time
constants minimum (eight are preferred). Since
the ADC logic timing is TAD-based, the sample
time (in TAD) must be long enough, over all condi-
tions, to charge/discharge CHOLD. Do not assume
one TAD is sufficient sample time; longer times
may be required to achieve the accuracy needed
by the application. The value of CHOLD is 40 pF.
A small amount of charge is present at the ADC
input pin when the sample switch is closed. If RS is
high, this will generate a DC error exceeding
one LSB. Keeping RS < 50
 is recommenced for
best results. The error can also be reduced by
increasing sample time (a 2 k
 value of RS
requires a 3 µS sample time to eliminate the error).
• Calculating Throughput – The throughput of the
ADC is based on TAD. The throughput is given by:
where:
Sample Time is the calculated TAD periods for the
application. SAR Conversion Time is 12 TAD for
10-bit and 14 TAD for 12-bit conversions. Clock
Sync Time is 2.5 TAD (worst case).
Example: For a 12-bit ADC throughput, if using
FRC = 8 MHz and the Sample Time is one TAD, the
use of an 8 MHz FRC means the TCYC = 250 ns and
this requires: TAD = 2 TCYC = 500 ns. Therefore, the
throughput is:
Note that the clock sync delay could be as little as
1.5 TAD, which could produce 121 KS/sec, but that
cannot be ensured as the timing relationship is asyn-
chronous and not specified. The worst case timing of
2.5 TAD should be used to calculate throughput.
Example: A certain transducer has a 20 k
 output
impedance. If AVDD is 3.0, the maximum sample
time needed would be determined by the following:
If TAD = 500 ns, this requires a Sample Time of 4.95 µs/
500 ns = 10 TAD (for a full-step voltage on the
transducer output). RSS is 350
 because AVDD is
above 2.7V.
Throughput = 1/(Sample Time + SAR Conversion Time +
Clock Sync Time)
Throughput = 1/(500 ns) + (14 * 500 ns) + (2.5 * 500 ns) =
114.28KS/sec
Sample Time = 6 * (RS +RIC + RSS) * CHOLD
= 6 * (20K + 250 + 350) * 40 pF
= 4.95 µS



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