Circuit lets you test sample-and-hold amplifiers
Measure voltage drop with a digital voltmeter.
Marián Štofka, Slovak University of Technology, Bratislava, Slovakia; Edited by Martin Rowe and Fran Granville -- EDN, March 4, 2010
Sample-and-hold amplifiers sample an analog voltage and hold it until an ADC can digitize it. A perfect sampling circuit holds a voltage until digitizing is complete. Thus, the amplifier’s output is identical to its input. Real sample-and-hold amplifiers, however, can gain or lose voltage, producing an error. Offset voltages in amplifiers cause a static additive error. Further, there occurs a specific additive error, the so-called voltage pedestal, which originates within the transition from the sample state to the hold state because of a parasitic charge transfer to the hold capacitor.
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By applying a logic-control signal with a duty cycle of D and 1–D, you can measure a mean output voltage difference, [ΔVOUT]=[VOUT–VIN], which the following equations show. [ΔVOUT]=VSTAT+ (1–D)VINJ+½(1–D)²VDROPPEAK, and [Δ
Δ
are the output-voltage differences for D and 1–D, respectively; VSTAT is the steady output-voltage difference for a selected value of the reference input voltage, D is the duty cycle, VINJ is the voltage pedestal, and VDROPPEAK is the peak voltage drop. Figure 1 shows how the voltages in the equations change over time. If you apply a complementary control waveform with a duty cycle of 25%, you can measure another dc component of the sample-and-hold amplifier’s output voltage. Finally, when the sampling switch is continuously on, you can measure the VSTAT voltage, which is a real dc voltage. VOUT and
contain a waveform superimposed onto a selected value of the reference voltage. Thus, you should measure the mean values of these voltages using a series resistor with a value of, say, 10 k(Ω).
Multiplying the voltage pedestal, a simple rectangular waveform, by the duty cycle yields the average value. In contrast, the voltage-drop waveform appears as a sawtooth. Its mean rises as one-half of the duty cycle squared. The peak-voltage-drop value denotes a hypothetical voltage drop at the end of a whole period, T, of the SAMPLE/
You can use the previous equations to find the values of the voltage pedestal and the peak voltage drop. A 75% duty cycle is a convenient value. The following equations are valid for this duty cycle: VINJ=6[ΔVOUT]–2/3[Δ
If you evaluate the equation for a high-performance analog switch, such as the Analog Devices ADG1213, you get a repetition rate of 33 kHz or less. The difference due to voltage drop prevails at low-value repetition rates. In this case, the repetition rate can be the value of the frequency at which
A tested sample-and-hold amplifier using the circuit in Figure 2 uses a supply voltage of –1V, a drain-to-drain voltage of 5V, and a supply voltage of 3.3V for logic circuits in the pulse generator. Two sets of measurements at 25, 75, and 100% duty-cycle values for the AGD1213’s internal switch control used input voltages of 0 and 2.5V. You will measure the output-voltage difference, approximately –0.0366 mV, and the pedestal voltage, approximately –0.0333 mV, at a repetition rate of 1.762 kHz. The value of the residual effective charge injection, QINJ, into the hold capacitor, CH=2 nF, is QINJ=CH×VINJ. The value is negative and doesn’t exceed –75 fC. The following equation defines the difference of charge injection within the 2.5V range of input voltage: ΔQINJ=QINJ(2.5V)–QINJ(0V) and yields a value of –6.7 fC. The following equation determines the residual effective leakage current from the acquired values of peak voltage drop at a repetition rate of 160 Hz: ILEAK=CH×VDROPPEAK×fREP, where ILEAK is the leakage current. A leakage current at the input voltage of 0V is approximately 17 pA, and a leakage current at the input voltage of 2.5V is approximately –17 pA.| Reference |
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