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DVR21 数据表(PDF) 291 Page - Microchip Technology |
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DVR21 数据表(HTML) 291 Page - Microchip Technology |
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291 / 416 page ![]() 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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