1
\$\begingroup\$

For the following circuit i am confused to as why Resistor R4 when measured on the breadboard is 35k ohms, when it is meant to be 47k Ohms (as it is a 47 k ohm resistor. Also when it is taken out of the circuit and measured it is fine and gives the 47k ohm as expected) . The other resistors when measured match up to what they are meant to be. I assume the reason it is giving this reading is due to the configuration it is in? Such as something to do with the voltage divider, maybe something else?

  • Note for the circuit schematics i have just put 35k ohms just to see if my simulated results would match up to what the breadboard circuit outputted on an oscilloscope. It is actually meant to be 47k ohms

In short

  1. Not understanding why the 47k ohm resistor (R4) reads 35k ohm on the multimeter
  2. How do i calculate what this resistor value will be due to the difference? As in i find what value i want, in this case i needed 47k ohm for the specified application but due to this resistor value difference, how am i meant to see what 47k will actually be. As in will i need to use a 56k or something to get the 47k due to the difference? Hence what formula am i meant to use to find what value this resistor will actually be when in theory its meant to be 47k ohms?

Circuit Schematics enter image description here Circuit implemented on the breadboard

enter image description here

\$\endgroup\$
3
  • 5
    \$\begingroup\$ If you took 300 g of water and mixed it with 100 g of milk, how much water have you got? \$\endgroup\$ – Andy aka May 19 '17 at 15:14
  • 4
    \$\begingroup\$ Unless you have the slightest idea of what you are doing, don't measure any kind of components in-circuit. \$\endgroup\$ – PlasmaHH May 19 '17 at 15:20
  • 3
    \$\begingroup\$ Pull one side of resistor and then make your measurement. \$\endgroup\$ – StainlessSteelRat May 19 '17 at 17:27
10
\$\begingroup\$

Not understanding why the 47k ohm resistor (R4) reads 35k ohm on the multimeter

The LM386 input impedance is \$\text{50k}\Omega\$. What you're really measuring is the 47k, in parallel with 24k and 50k in series. This is the equivalent circuit.

schematic

simulate this circuit – Schematic created using CircuitLab

How do i calculate what this resistor value will be due to the difference? As in i find what value i want, in this case i needed 47k ohm for the specified application but due to this resistor value difference, how am i meant to see what 47k will actually be. As in will i need to use a 56k or something to get the 47k due to the difference? Hence what formula am i meant to use to find what value this resistor will actually be when in theory its meant to be 47k ohms?

It looks like you're just trying to attenuate and filter a square wave. The datasheet only gives typical values for input impedance, so you can't rely on it being tightly controlled. You're going to have to decide how much variance you can tolerate. The easiest thing to do is keep your circuit as designed, and put a cheap rail-rail op amp in as a buffer.

\$\endgroup\$
15
  • \$\begingroup\$ Thanks, so if I have designed the circuit to have 47k ohms, can I just leave it like it is, or do I need to calculate what value will give me 47k ohms when it's in parallel with the 22k and 50k? For this particular task I cannot change the amp and I can only have a maximum of 1 amp \$\endgroup\$ – Student May 19 '17 at 23:21
  • \$\begingroup\$ As in if I leave it as 35k this will change my cutoff frequency won't it for the filter?. Also if these are in parallel, why does the 24k resistor read the correct value? \$\endgroup\$ – Student May 19 '17 at 23:28
  • \$\begingroup\$ Also when I take out the input connection, the resistor reads 46k. Another thing though is in this configuration will the input impedance still be 50k depending on the 47k resistor? As in is the only thing that changes is the 47k value? \$\endgroup\$ – Student May 19 '17 at 23:42
  • \$\begingroup\$ @user6186979 The 24k value is not reading correctly, 20.6k is 14% low. It's not as extreme because 47k and 50k in series have much less effect on 24k in parallel. If you don't want to add a buffer, you're going to have to trade off corner frequency or attentuation. Welcome to engineering. My advice is to draw the equivalent circuit, with the LM386 replaced by a 50k resistor, and derive equations for attenuation and corner frequency based on that. Then you can put the equation into an Excel sheet, and play with the values until you're happy with the compromise. Good luck. \$\endgroup\$ – Matt Young May 21 '17 at 0:03
  • \$\begingroup\$ Thanks. I was just confused still as to how to find R4 due to im unsure still of how to know if its in parallel to the input resistance or not? As in how do i tell if it is. Isnt R4 in series with the 50k, then in parallel with the 24k?Basically i am unsure what configuration it is in? Also ignoring the 50k would R4 and R3 be in parallel? \$\endgroup\$ – Student May 21 '17 at 0:56
9
\$\begingroup\$

When you measure the resistance with the resistor in the circuit, you are measuring the net resistance between two points in the circuit, not just the value of the resistor that happens to be between your probes.

In your case, R2 and R3, and the internal circuit of the LM386 form another resistance path in parallel with R4, reducing the apparent resistance.

In the case of the 56K resistor, remember that resistor values are not precise. You probably are using 5% resistors, so the 56K resistor may be any value between 53.2K and 58.2K, and still be correct.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.