{"id":396591,"date":"2024-06-29T12:55:50","date_gmt":"2024-06-29T12:55:50","guid":{"rendered":"http:\/\/savepearlharbor.com\/?p=396591"},"modified":"-0001-11-30T00:00:00","modified_gmt":"2024-06-29T12:55:50","slug":"","status":"publish","type":"post","link":"https:\/\/savepearlharbor.com\/?p=396591","title":{"rendered":"Active Termination Drivers<\/span>"},"content":{"rendered":"
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The easiest way to build a driver with specified output impedance is to use an amplifier with high load compatibility and add a resistor to its output. The penalty is a voltage drop across this resistor, so there is power loss and we need a higher supply voltage. If our driver is able to deliver the same voltage and current to the same load, but the extra resistor will have a lower value, our device will be able to deliver the same output power at a lower supply voltage. Less power losses, less heat, and longer working time when a battery is used.
There is an idea how to solve this problem: active termination. We can synthesize the output impedance!<\/p>\n

Now when we know what we want, go to design our drivers!
<\/a> <\/p>\n

Single-Ended Driver<\/h2>\n

Maxim Integrated\u2019 engineers shared equations to design active termination single-ended drivers in [1]. The idea is to add positive feedback to reduce the output resistor value, use it to measure current, and synthesize the desired output impedance. As the result we have lower power losses. It is interesting that this circuit looks exactly as Improved Howland Current Pump! We just use another set of equations to get the desired behavior.<\/p>\n

Let\u2019s design single-ended drivers with parameters: gain G is 1 when unloaded, output impedance Zout is 50\u00a0Ohm. It means that if we load our driver with 50\u00a0Ohm resistor, the output voltage drops by half and the gain is 0.5.<\/p>\n

Inverting Active Termination Driver<\/h3>\n

<\/p>\n

\n\n\n\n
\"'Circuit<\/td>\n<\/tr>\n
Inverting Active Termination Driver<\/td>\n<\/tr>\n<\/table>\n<\/div>\n

Select Ro = 22 Ohm<\/nobr>, which is much lower than 50\u00a0Ohm. Now we can choose either R1 or R2. Let\u2019s choose R2 = 3 KOhm<\/nobr>. <\/p>\n

      Ro    22 Ohm K = \u2014\u2014\u2014\u2014 = \u2014\u2014\u2014\u2014\u2014\u2014 = 0.44     Zout   50 Ohm       R2     3 kOhm R1 = \u2014\u2014\u2014 = \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014 = 6818 \u2248 6.8 K (E24)      K G   0.44 \u00d7 1 <\/pre>\n
  Now we need to choose R3, let\u2019s set it to 4.3\u00a0kOhm.  <\/p>\n
 R4 = R3 [(1 + K G)\/(1 \u2212 K) \u2212 1] = 4.3 KOhm \u00d7 [(1 + 0.44 \u00d7 1)\/(1 \u2212 0.44) \u2212 1] = 6757 \u2248 6.8 K (E24) <\/pre>\n
  <\/p>\n
\n\n\n\n
\"'Circuit<\/td>\n<\/tr>\n
Inverting Active Termination Driver, Unloaded Gain = 1, Loaded Gain = 0.5, Zout = 50 Ohm<\/td>\n<\/tr>\n<\/table>\n<\/div>\n
  Enter the values into a simulator. The output voltage drops by half when 50\u00a0Ohm load is connected, so the circuit works exactly as expected.  <\/p>\n
\n\n\n\n
\"Inverting<\/td>\n<\/tr>\n
Inverting Active Termination Driver, simulation<\/td>\n<\/tr>\n<\/table>\n<\/div>\n
Why are we wasting our time and money?<\/h3>\n
  We had to do calculations and add 2 resistors. What are the benefits?
  Let\u2019s compare an inverting driver and inverting active termination driver with the same parameters.  <\/p>\n
\n\n\n\n
\"Common<\/td>\n<\/tr>\n
Inverting Driver Vs Inverting Active Termination Driver<\/td>\n<\/tr>\n<\/table>\n<\/div>\n
  The output voltages, green and yellow lines, are the same 0.5\u00a0Vpp. But the common driver output voltage is 1\u00a0Vpp, the cyan line, and the active termination driver output voltage is 0.72\u00a0Vpp, the purple line. So the power losses of the common driver are:  <\/p>\n
 P1 = Urms^2 \/ R = [\u221a2 \u00d7 (1 Vpp \u2212 0.5 Vpp)]^2 \/ 50 Ohm = 10 mW <\/pre>\n
  and for our active termination driver:  <\/p>\n
 P2 = Urms^2 \/ R = [\u221a2 \u00d7 (0.72 Vpp \u2212 0.5 Vpp)]^2 \/ 22 Ohm = 4.4 mW <\/pre>\n
  More than 2 times lower! And a lower supply voltage can be used.
  Obviously, the lower the Ro value, the lower the power losses.
  The Zout\/Ro ratio defines the R1\/R2 ratio and there is some output impedance of a used Op Amp. When the output is shorted to the ground, Ro acts as a protection. And the lower Ro value, the more the positive feedback must be, which can cause instability and distortions. So Ro should not be too low.
  Thus, the most used Zout\/Ro ratio in real circuits is about 2 and practical limit is about 10\u2026<\/p>\n
Choosing R3 and R4<\/h3>\n
  We see that R3 and R4 are arbitrary values linked with the equation:  <\/p>\n
 R4 = R3 [(1 + K G) \/ (1 \u2212 K) \u2212 1] = R3 N <\/pre>\n
  So what values to choose?
  Maxim Integrates\u2019 engineers write in [1]:  <\/p>\n
Ensure that the time constant of the negative feedback loop dominates the overall loop control.
  \u2026
  TC(\u2212) = Ccom (R1 || R2)
  TC(+) = Ccom (R3 || R4)
  Set TC(\u2212) < TC(+)<\/p><\/blockquote>\n
  If we design two drivers with unloaded gain of 1 and 50 Ohm output impedance, take parameters from the OPA2134 datasheet, for example, we will see 2.2 dB peak at 2.8 MHz when (R1||R2) > (R3||R4)<\/nobr>, the top circuit and green line on the chart.  <\/p>\n
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\"Inverting<\/td>\n<\/tr>\n
Inverting Active Termination Driver, simulation with an Op Amp model<\/td>\n<\/tr>\n<\/table>\n<\/div>\n
  Stability check shows that the phase margin is about 40\u00b0 for the top circuit and 84\u00b0 for the bottom circuit.  <\/p>\n
\n\n\n\n
\"Inverting<\/td>\n<\/tr>\n
Inverting Active Termination Driver, stability check<\/td>\n<\/tr>\n<\/table>\n<\/div>\n
  Of course, a capacitor can be added in parallel with R2 to increase the phase margin.<\/p>\n
  R3 and R4 are connected in parallel with a load, so it is the reason to have them not too low. On the other hand, there is resistor thermal noise, which can be important if we design a driver for some audio device, for example, so their values should be as low as possible.
  To solve this problem, a good start value for R3 is to be equal to R1.<\/p>\n
Input Impedance of the Inverting Active Termination Driver<\/h3>\n
  The input impedance of the inverting active termination driver without a load is:  <\/p>\n
                      R2 R3 Rin.unloaded = R1 \u2212 \u2014\u2014\u2014\u2014\u2014\u2014\u2014                     Ro + R4 <\/pre>\n
  and with a load is:  <\/p>\n
                             R2 R3 Rload Rin.loaded = R1 \u2212 \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014                   (Ro + R4) Rload + (R4 + R3) Ro <\/pre>\n
  Thus, unlike the inverting amplifier, the input impedance of the drivver is less than R1 and depends on Rload. In our case Rin.unloaded \u2248 4.9\u00a0kOhm and Rin.loaded \u2248 5.7\u00a0kOhm with 50\u00a0Ohm load.<\/p>\n
  Pay attention that the output impedance of the previous stage can affect the driver parameters, though usually it is not a problem, because the output impedance is low.<\/p>\n
Non-Inverting Active Termination Driver<\/h3>\n
  AN-3390 [1] says:  <\/p>\n
Inverting only practical operation.<\/p><\/blockquote>\n
  Still we will try to design the non-inverting version of the driver.<\/p>\n
  It is the non-inverting amplifier, so we can design the non-inverting active termination driver only with a gain greater than 1. The equations are the same but we must subtract 1 from specified unloaded gain value or 0.5 from a loaded gain value. If we enter the same values as in the previous example, the gain will be 2 when the driver is unloaded.  <\/p>\n
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\"Non-Inverting<\/td>\n<\/tr>\n
Non-Inverting Active Termination Driver, simulation<\/td>\n<\/tr>\n<\/table>\n<\/div>\n
  Our simulation confirms that at the same component values Gain = 2<\/nobr> when the driver is unloaded and 1 when loaded.<\/p>\n
Input Impedance of the Non-Inverting Active Termination Driver<\/h3>\n
  The input impedance of the non-inverting active termination driver without a load is:  <\/p>\n
                     R1 (Ro + R4) Rin.unloaded = R3 \u2212 \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014                          R2 <\/pre>\n
  and with a load is:  <\/p>\n
                   R1 [(Ro + R4) Rload + R4 Ro] Rin.loaded = R3 \u2212 \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014                         R2 Rload \u2212 R1 Ro <\/pre>\n
  In our case Rin.unloaded \u2248 \u221211.16\u00a0kOhm and Rin.loaded \u2248 \u22128.3\u00a0MOhm with 50\u00a0Ohm load.
  Yes, it is negative! It means that the current flows from the driver input, so the previous stage must be able to sink this current.<\/p>\n
Fully Differential Active Termination Driver<\/h2>\n
  A fully differential version of the driver is described in [2], Chapter\u00a016.  <\/p>\n
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\"'Circuit<\/td>\n<\/tr>\n
Fully Differential Active Termination Driver<\/td>\n<\/tr>\n<\/table>\n<\/div>\n
  Let\u2019s design a driver with the same parameters: gain is 1 when unloaded, output impedance Zout is 50\u00a0Ohm.
  Select Ro = 16 Ohm<\/nobr>, R2 = 3 KOhm<\/nobr>.  <\/p>\n
         R2          3 kOhm R3 = \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014 = \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014 = 8.333 \u2248 8.2 K (E24)          2 Ro       2 \u00d7 16 Ohm      1 \u2212 \u2014\u2014\u2014\u2014   1 \u2212 \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014          Zout         50 Ohm <\/pre>\n
  <\/p>\n
                 2 Zout R2 R3                           2 \u00d7 50 \u00d7 3000 \u00d7 8200 R1 = \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014 = \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014 = 4701 \u2248 4.7 K (E24)      G [2 Ro R3 + Zout (R3 + Ro \u2212 R2)]   1 \u00d7 [2 \u00d7 16 \u00d7 8200 + 50 \u00d7 (8200 + 16 \u2212 3000)] <\/pre>\n
  <\/p>\n
\n\n\n\n
\"'Circuit<\/td>\n<\/tr>\n
Fully Differential Active Termination Driver, Unloaded Gain = 1, Loaded Gain = 0.5, Zout = 50 Ohm<\/td>\n<\/tr>\n<\/table>\n<\/div>\n
  <\/p>\n
Differential Active Termination Driver<\/h2>\n
  Another variant is used in ADSL modems and described in [3], for example. Since it is a modified differential amplifier, there is limitation for minimum gain.  <\/p>\n
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\"'Circuit<\/td>\n<\/tr>\n
Differential Active Termination Driver<\/td>\n<\/tr>\n<\/table>\n<\/div>\n
  Let\u2019s design a driver with parameters: gain is 2 when unloaded, output impedance Zout is 50\u00a0Ohm.
  Select Ro = 20 Ohm<\/nobr>, R2 = 910 Ohm<\/nobr>.  <\/p>\n
         R2          910 Ohm R3 = \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014 = \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014 = 4550 \u2248 4.7 K (E24)          2 Ro       2 \u00d7 20 Ohm      1 \u2212 \u2014\u2014\u2014\u2014   1 \u2212 \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014          Zout         50 Ohm <\/pre>\n
  <\/p>\n
                            2 R2 R3                                                    2 \u00d7 910 \u00d7 4700 R1 = \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014 = \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014 = 4320 \u2248 4.3 K (E24)      Ro \u2212 R2 \u2212 R3 + G (Ro + R3 \u2212 R2 + 2 R3 Ro \/ Zout) \/ 2   20 \u2212 910 \u2212 4700 + 2 \u00d7 (20 + 4700 \u2212 910 + 2 \u00d7 4700 \u00d7 20 \/ 50) \/ 2 <\/pre>\n
  <\/p>\n
\n\n\n\n
\"'Circuit<\/td>\n<\/tr>\n
Differential Active Termination Driver, Unloaded Gain = 2, Loaded Gain = 1, Zout = 50 Ohm<\/td>\n<\/tr>\n<\/table>\n<\/div>\n
  The differential input impedance is constant, 2\u00d7R1.<\/p>\n
Balanced Driver or Floating Differential Output<\/h2>\n
  There is the problem with differential drivers described above: if one of their outputs is loaded to the ground, the other output voltage will not change, so the differential output voltage will be lower. But what if we want to have a constant differential voltage, as if it were a transformer?<\/p>\n
  We would like:  <\/p>\n