# Why the resistor in this op-amp circuit?

Here's a circuit diagram for a signal response test circuit from the spec sheet for a HA-5195 op amp and looks like a non-inverting amplifier circuit with a gain of 5, plus the 200Ω resistor between Vout and ground:

It's the canonical (I assume?) non-inverting amplifier circuit with R1 = 400Ω and Rf = 1.6kΩ, plus the resistor I'm asking about.

Can someone explain the purpose of the 200Ω resistor?

• Maybe to give a fixed, known impedance to the opamp? – todbot Nov 15 '09 at 21:54

It is a load resistor probably. Often op-amp circuits have the "load" placed as a resistor.

The current an op-amp drives can greatly increase it's non-ideal property. The finite gain becomes more apparent as you drive more current, along with the output resistance.

When simulating an OP-amp you should always attempt to place a load resistor across the output for the effective load you are connecting to. If you want to look up a method of doing something like this, Thevenin equivalent circuits are a good example.

Community Wiki is on if anyone wants to expand.

• Glad to hear it! – Kortuk Nov 16 '09 at 2:23

Right above that diagram, in the NOTES section, it says RL = 200Ω. "RL" means "load resistor". You'll see that they show it in all the other diagrams, too.

This op-amp is specified for high-frequency video applications, and in these cases you generally have low impedance loads like this so the sources and loads can be matched to avoid reflections back up cables.

In Recommended Test Procedures for Operational Amplifiers they describe using a load resistance when measuring transient response, and have a table of recommended values for each part (0.2 kΩ for this part). I guess the high-speed transient response is affected by the load (I don't work with high-speed stuff), so they're showing it in-circuit to show a real-life application.

The overall gain will also be decreased, since the output impedance of the op-amp is 25-30 Ω (as shown in page 2 of the datasheet), and the maximum output level will be decreased, as shown in Figure 13. In the application notes, it says:

In Figure 19, RIN is usually the terminating resistance for the input cable, and it is usually 50Ω or 75Ω . RM is the matching resistance for the cable being driven, and RT is the terminating resistance for the driven cable. RT is often shown here for gain calculations while it is physically placed at the cable end.

In this case, RT is the same as RL in the datasheet. So it's being shown "here" for its effect on the gain.

So, in general, they're showing the load in the circuit to demonstrate that their measurements were tested in a real-life video situation.

• I would have suggested you just edit my post with your extra information. You have a very good grasp of op-amp theory, something I rarely see in people anymore. – Kortuk Nov 17 '09 at 3:18

Test specifications always go into great detail describing test conditions and the test environment, otherwise you the test wouldn't be repeatable with the same results. For this case the test spec apparently says load must be 200$\Omega$.
But since the 1600$\Omega$ + 400$\Omega$ are parallel to the 200$\Omega$, the actual load is 182$\Omega$, and it's not likely that this is what they wanted. They simply could have used 160$\Omega$ + 40$\Omega$ instead of 1600$\Omega$ + 400$\Omega$, and they would have exactly 200$\Omega$ without needing the third resistor.

In a test environment this is not so important, but in a design for production the third resistor would be an extra cost.

• I agree someone just used a common 5x gain configuration, then added a specific load without thinking about it too hard. However, there is one little wrinkle that makes this different from using 160 and 40 Ohm feedback resistors, although very unlikely that's the reason in this case. Sometimes you try to keep both opamp inputs driven with the same impedance to ballance out the input bias currents. In this circuit, the minus input sees 320 Ohms, wheras in the 2 resistor circuit it would see 32 Ohms. Again, I really don't think that's what's going on in this case. – Olin Lathrop Jul 3 '11 at 13:37
• @Olin - You're right, and for a test circuit you would expect that they thought of that, so that the impedance unbalance doesn't influence the measurement. But I don't see the 320$\Omega$ on the non-inverting input either. – stevenvh Jul 3 '11 at 15:31