Translation - Japanese: QDR(QDR, QDR-II/II+ & QDR-II+ Xtreme) SRAMのVOHおよびVOLの理解 – KBA82773 - Community Translated (JA)
Why do the VOH and VOL of QDR SRAMs (QDR, QDR-II/II+ & QDR-II+ Xtreme) have the same range in the DC specs section of the datasheet?
The VOH and VOL specs have a range of (VDDQ/2 - 0.12) V to (VDDQ/2 + 0.12) V, that is defined by the QDR consortium.
QDR SRAMs offer multiple output impedance varying from 35 ohms to 70 ohms that has a corresponding current IOH and IOL. The VOH and VOL values are different wrt the specific IOH and IOL currents.
The DC test specifications for VOH and VOL are shown in the DC Electrical Characteristics section of the datasheet. These specifications are for evaluating the IOH and IOL level of the QDR output drivers for VOH and VOL of VDDQ/2.
The IOH and IOL values are:
|IOH| = (VDDQ/2)/(RQ/5) +/- 15% for 175 Ω ≤ RQ ≤ 350 Ω.
|IOL| = (VDDQ/2)/(RQ/5) +/- 15% for 175 Ω ≤ RQ ≤ 350 Ω.
For example, for a VDDQ = 1.5 V and RQ = 250 Ω
|IOH| = (0.75/50) = 15 mA
|IOL| = (0.75/50) = 15 mA
As explained in the above example, the tester needs to source/sink 15 mA of current (IOH/IOL) at 50-ohm output impedance and 1.5-V VDDQ to bring VOH/VOL = VDDQ/2.
In the Automatic Test Equipment (ATE) testing the tester must act as a constant-current load (sink IOH) for a VOH test and for a VOL test, the tester must act as a constant-current source (source IOL). Figure 1 shows that VOL level reaches to VDDQ/2 at 15-mA IOL load from the tester. The same explanation is true for VOH.
Let’s look at another example for 60-ohm output impedance. The tester needs to source/sink 12.5 mA of current (IOH/IOL) at 60-ohm output impedance and 1.5-V VDDQ to bring VOH/VOL at VDDQ/2.
For example, for a VDDQ = 1.5 V and RQ = 300 Ω
|IOH| = (0.75/60) = 12.5 mA
|IOL| = (0.75/60) = 12.5 mA
The reason +/-0.12 V is added in the VOH and VOL specs range is to account for +/-15% accuracy of the output impedance circuitry.
For example, for 50-ohms output impedance, if the tolerance is -15%, then the output impedance can be as low as 42.5 ohms.
|IOL| = (0.75/42.5) = 17.4 mA
Hence, the difference between the typical and variation is 2.4 mA (17.4 mA – 15 mA).
VOL variation = 2.4 mA x 50 ohm = 0.12 V.
The same explanation is true for VOH.