Resistor Values

[Riley Dyen]

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Standard resistor values are calculated using the simple formula given below. Round the results to the proper number of significant figures (three for 1% and 2%, two for 5% and 10%). As the chart at the right shows (created in Excel), plotting the values on a logarithmic scale results in a straight line due to the exponential in the equation.

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I am just getting started with the student edition of Multisim. Firstly, when I place a component such as a jack, there is a large title with the compontent that gets it the way of the schematic. How do I remove it? Second, it seems that there is only 1 resistor value, 82K, in this edition. Do I have to make a new component for each value? If so, how is it done? Thanks for the help!

1. When you place a component on the work area, you can double click on it and select the "Display" tab, uncheck the "Use schematic global setting" and you have the option to turn on/off various component label.

2. Multisim have a lot of resistors, go to the "basic" group and select the "resistor" family, you should see many resistor values. Directly below "Component" there is a filter, if you type "8" Multisim will filter out other values and only 8 appears in the list, place your mouse pointer in the filter and press the backspace button on your keyboard to clear the 82K and Multisim will show you the complete list. I prefer to drop any resistor on the work area and use the copy/paste function, when I double click on the resistor, I can change the value.

Strange one. I created my own resistor library, importing a symbol from a working library. I went ahead and created my package, then device, etc. I have a device with a symbol, package, connections and prefix as I understand from the video series. But when I place the symbol on the schematic, it i the value (R1206), not a blank value. I go to the properties, and there is no value field to modify. Just device and package - both of which are named R1206. Name shows correctly - R1, position in the schematic, Gate G$1, what library it is from. It is smashed. I bring it a part from another library - resistor-power supplied by Eagle, and I get the same thing except there is a value field and some attributes. What is going wrong here? I really need to enter resistor values.

I have been having the same issue. I have a schematic with lots of custom parts. On some of them the value is not showing up. The radio button is definitely set to on. As far as I can tell, all of the other parameters are the same between the parts with values showing up, and the parts with values NOT showing up. any suggestions?

The internal pull up isn't a true resistor. It can be thought of as an 'active circuit' or a current source, with the rating being held in the electrical characteristics table show below. this information was take from the datasheet.

Is there a place that I can get a hold of a table with common voltages by using standard resistors? (Sorry that this is poorly worded but I cant figure out how to say it better, so I will give an example)

This page lets you enter the desired voltage and then calculates optimal values for the resistors. You can choose the resistors' series. For instance, if you select the E12 series, it will find the best match with resistors from the E12 series. Obviously E96 series will give better matches.

I know that you could have infinite number of combinations for these resistors but why would some one use specific value. i.e Rf = 100Mohm, Rin = 10Mohm gives a gain of 10V/V but also Rf = 10 ohm and Rin = 1 ohm gives the gain of 10V/V. What difference would it make to the design?

My thoughts say that higher value resistors are not precise so it wouldn't give you precise gain and using lower value resistors sink higher current from the source (Vin). Are there any other reasons? Also, let me know If I am right or wrong as well.

There are downfalls with choosing very large resistors and very small resistors. These usually deal with the non-ideal behavior of components (namely Op-Amps), or other design requirements such as power and heat.

Small resistors means that you need a much higher current to provide the appropriate voltage drops for the Op-amp to work. Most op amps are able to provide 10's of mA's (see Op-amp datasheet for exact details). Even if the op-amp can provide many amps, there will be a lot of heat generated in the resistors, which may be problematic.

On the other hand large resistors run into two problems dealing with non-ideal behavior of the Op-Amp input terminals. Namely, the assumption is made that an ideal op-amp has infinite input impedance. Physics doesn't like infinities, and in reality there is some finite current flowing into the input terminals. It could be kind of large (few micro amps), or small (few picoamps), but it's not 0. This is called the Op-amps input bias current.

The problem is compounded because there are two input terminals, and there's nothing forcing these to have exactly the same input bias current. The difference is known as input offset current, and this is typically quite small compared to the input bias current. However, it will become problematic with very large resistance in a more annoying way than input bias currents (explained below).

Here's a circuit re-drawn to include these two effects. The op-amp here is assumed to be "ideal" (there are other non-ideal behaviors I'm ignoring here), and these non-ideal behaviors have been modeled with ideal sources.

However, notice that if R1 and R3 are very large, the current flowing into the inverting input is very small, on the same order as (or worse, smaller than) I1. This will throw off the gain your circuit will provide (I'll leave the mathematical derivation as an exercise to the reader :D)

All's not lost just because there's a large bias current though! Look what happens if you make R2 equal to R1R3 (parallel combination): if I1 and I2 are very close to each other (low input offset current), you can negate out the effect of input bias current! However, this doesn't solve the issue with input offset current, and there are even more issues with how to handle drift.

There's not really a good way to counteract input offset current. You could measure individual parts, but parts drift with time. You're probably better off using a better part to begin with, and/or smaller resistors.

In summary: pick values in the middle-ish range. What this means is somewhat vague, you'll need to actually start picking parts, looking at datasheets, and deciding what is "good enough" for you. 10's of kohms might be a good starting place, but this is by no means universal. And there probably won't be 1 ideal value to pick usually. More than likely there will be a range of values which will all provide acceptable results. Then you'll have to decide which values to use based off of other parameters (for example, if you're using another value already, that might be a good choice so you can order in bulk and make it cheaper).

In your specific op-amp circuit, the voltage on junction of Rf and Rin is the same as the voltage on the non-inverting input. This has to be so - it's called a virtual earth. Given that fact, this means that your signal (Vin) sees an input impedance of exactly Rin. It also means that your output (without connecting to anything else) has to drive an output load that is Rf.

There are some common resistors that make good ratios for gain and better yet, common precision resistors with low temperature coefficient and nice resistance ratios. I like to use the precision parts if at all possible. (Same is true for caps in op-amps like for integrators - polystyrene precision and temperature stable). Like 10K/1K or 33K/3.3K. Beyond 100K/10K the Resistance gets high enough that the small capacitance in the circuit starts to turn your circuit into an integrator or differentiator (or low pass filter).

Very low Rin values load the input and high Rf values increase output impedance. These problems are easily overcome. Most op-amp packages have more than one OA. Use one as a voltage follower and as the input to your OA that has gain. Your total circuit presents a very high input impedance and your OA with gain sees a very low impedance on its input and you can use low values for Rin. You can also use an OA follower on the output to have high drive current and low impedance output. You can even easily configure the output to match impedance of the next circuit or a coaxial cable, etc. I like to use high precision low tempco resistors or low tempco pots (or digital pots) for Rf and trim for gain.

I have used 1M/1K for a gain of 1000 (2 op-amps in a row gives gain of 1 million) with low-pass for seismology, but this is a few Hz bandwidth and works even with the lowly uA741. LM308 requires much less trim. Good modern OA's are great by comparison. If you get into the 10M to 100M area for Rf, your bandwidth will drop off and noise will go up.

Note, firstly, that the tolerance on the ratio is higher than the tolerance on the individual resistors. This is good to keep in mind if you want a precise gain. However, the gain tolerance does not increase with the nominal resistance values, as long as the ratio is constant.

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