Speed up differential-output measurements

When you're testing differential line-driver components, you must measure the differential voltage on each driver's two outputs. For critical linear applications, you must perform three different tests to guarantee that the device will function under different loads. At Texas Instruments, we call those tests the VOD1, VOD2, and VOD3 tests (see "Definitions, symbols, and abbreviations ,").

Engineers have traditionally performed these tests in several stages, each of which can require additional load-board components or the connection of special resistive networks. We shortened the test time with a technique that makes extra load-board components or external resistive networks unnecessary. We applied our technique to test TIA/EIA-485-A devices, but you can adapt it for other differential-output devices.

VOD1 test requires a force current of 0 µA at both possible output levels—that is, with inverting output A higher than noninverting output B, and vice versa. In fact, VOD1 is a standard VOH and VOL test. Since VOD1 requires a differential measurement, you first measure VOH and VOL and then take the difference between them:

VOD1 = VOH_{(0 µA)} – VOL_{(0 µA)} (1)

VOD2 is a differential voltage measurement without regard to ground. It's usually performed with one or two separate load resistances (54 Ù and 100 Ù for our application). As Figure 1 shows, the device sees a resistive path from output A to output B. Given a resistive load and a minimum voltage-drop requirement by the data sheet, you can calculate the current (ISS) needed to simulate the load resistance. ISS, which flows from the high-voltage output through the resistance RL and into the low-voltage output, equals VOD2 (the drop between outputs) divided by RL:

ISS = VOD2_min/RL(2)

Figure 1 When configured for a VOD2 test, a differential line driver sees a resistive output circuit.

VOD2 = VOH_ISS– VOL_ISS (3)

VOD2₁ = VOH_{Y(–27.8 mA)}– VOL_{Z(27.8 mA)} (4)

VOD2₂ = VOH_{Z(–27.8 mA)}– VOL_{Y(27.8 mA)}(5)

ÄVOD2 = VOD2₁– VOD2₂(6)

VOC is defined as one half of VOD2 but with respect to ground; you can calculate VOC as follows:

VOC₁=(VOD2₁/2) (7)

+ VOL_(27.8mA)

VOC₂=(VOD2₂/2) (8)

+ VOL_(27.9Ma)

ÄVOC=VOC1-VOC2 (9)

Figure 2 (a) The VOD3 measurement requires a three-resistor network. You can convert the Thevenin equivalent the DUT sees into (b) a delta and then a wye equivalent network. Boldface numerals indicate circuit node numbers.

R ₁ = R_cR_b/(R_a + R_b + R_c) (10)

R ₂ = R_cR_a/(R_a + R_b + R_c) (11)

R ₃ = R_aR_b/(R_a + R_b + R_c) (12)

Voltage on node 1 of the wye network is either VOH or VOL, voltage on node 2 is either VOL or VOH, and voltage on node 3 is V_TEST as specified in the data sheet for our device under test. Now, V1, the voltage between node 3 and node 1, equals the voltage of node 3 minus the voltage of node 1, and V2, the voltage between node 3 and node 2, equals the voltage of node 3 minus the voltage of node 2. Similarly, the node 3 to node 1 current path is IOL, and the node 3 to node 2 current path is IOH. Matrix algebraic calculations from a basic text (for instance, Ref. 1, p. 64) can determine the expressions IOL and IOH:

IOH = [1/(R₁ + R₃)(R₃ + R₂) – R₃ ²] · [(R₃+ R₁)V₂– R₃V₁] (13)

IOL = [1/(R₁ + R₃)(R₃ + R₂) – R₃ ²] · [(R₃ + R₂)V₁– R₃V₂] (14)

For our application, VOD3 is a differential voltage drop across a 60-Ù resistor of the network in Figure 2a, and IVOD is the current through that resistor. Our DUT's data sheet requires that this voltage drop not exceed 5 V and be no less than 1.5 V.

Table 1 shows how VOD shifts as IOL and IOH change due to process shifts. To obtain the Table 1 values, we applied various VOH and VOL levels and measured the current and VOD changes. Table 1 shows two cases: first, where V_TEST equals +12 V, and second, where V_TEST equals –7 V. The VOD3 test must be performed at both of these two V_TEST levels to fully comply with the specs. Again, these are worst-case scenarios but are used to develop this new alternative method for testing VOD3.

Table 1 looks at the two worst-case scenarios—the first with the 5-V maximum VOD3 requirement and the second with the 1.5-V minimum VOD3 requirement. At the maximum requirement, VOH should be at about 5 V, sourcing an IOH of –64 mA, and VOL should be about 0 V, sinking an IOL of 114.6 mA. At the minimum requirement of 1.5 V, VOH is at 2.8 V, sourcing an IOH of –323 µA, and VOL is 1.3 V, sinking an IOL of 53 mA.

These conditions are with a V_TEST of +12 V, where it can be seen that the worst force current is IOL at 114.6 mA. When the V_TEST switches to –7 V, then the IOH has the worst-case current of –114.6 mA.

The table shows different calculated values for a given VOH and VOL to show how IOH and IOL shift in relation to a process shift. From the table you can see that the most important value is the minimum VOD3 requirement of 1.5 V. At this point, measure the actual VOD3 as follows:

For a V_TEST at +12 V with the input high, measure VOD3₁, and then switch the input low to measure VOD3₂:

VOD3₁ = _{VOHY(–1 mA)}–VOL_{Z(60 mA)} (15)

VOD3₂ = VOH_{Z(–1 mA)}–VOL_{Y(60 mA)} (16)

ÄVOD3 = VOD3₁–VOD3₂ (17)

Repeat this step for a V_TEST at –7 V:

VOD3₁ = VOH_{Y(–60 mA)}–VOL_{Z(0 mA)} (18)

VOD3₂ = VOH_{Z(–60 mA)}–VOL_{Y(0 mA)} (19)

ÄVOD₃ = VOD3₁–VOD3₂ (20)

Figure 3 illustrates the relationship of IOL and IOH currents to VOD3. An online appendix  shows the laboratory and production-test results we obtained for SN75ALS176 differential bus transceivers.

Figure 3 For a VTEST level of 12 V, these resistive networks illustrate IOL and IOH calculations for maximum and minimum values of VOD3.

We have found that this method for calculating differential output voltage parameters eliminates the need for the non-standard test configurations and load boards. It has resulted in improved testability, improved yields, standard and simpler hardware, increased production throughput, and a reduction in overall test setup time.