0
\$\begingroup\$

I'm extending my question from previously.

How to find SMD resistor and capacitor value and correct size

I'm doing some calculation and find something confusing below is the calculations:

V = 5

R = 1000

I = V/R = 5/1000 = 0.005A

P = IxV = 0.005A x 5V = 0.025W = 1/40W

1/40W is output for me to find the right size of SMD now confusion kicks in when I change the values see below

now let's change this to

V = 5

R = 10000 (10k)

I = V/R = 5/10000 = 0.0005A

P = IxV = 0.0005A x 5V = 0.0025W = 1/400W << that's 400 m'i doing something wrong from the SMD power rating chart i cant any bigger than 1/20 ? how to resolve this?

Now let's come to my real problem see below.

V = 3.3

R = 0.10 ohms

I = V/R = 3.3/0.10 = 33A << ??

P = IxV = 33A x 3.3V = 108.9W = 108 9/10 << ??

See the Circuit and focus on R2 and R3 value. Can someone help to understand me this? or correct me what I'm doing wrong? There is no way its 33 Amps of current for this what will be the correct power rating for this R2 and R3 based on the inputs mentioned on the circuit, V = 3.3 and R2, R3 = 0.10 ohms

enter image description here

if anyone can show me the calculation for the above problems and advise me the correct way of calculation, I have created an excel sheet to make life easy but unable to figure it out whats is wrong when changing the value and if any of you can help me to improve that sheet for the public I'll appreciate

here is the link https://1drv.ms/x/s!AuEcOHVa1BRjgch-V1bGfitSW3QNbA?e=EJMghY

\$\endgroup\$
5
  • 1
    \$\begingroup\$ why you assume that over r2 and r3 would be 3.3v? help me to understand this. \$\endgroup\$ – marcosbc Jan 31 at 1:38
  • \$\begingroup\$ I really don't know why I assume... hmm, so I need a value across the resistor not what I'm supplying as input pfft, poor me! @marcosbc \$\endgroup\$ – Adeel Rizvi Jan 31 at 1:51
  • \$\begingroup\$ no problem friend, we are here for help. lines below Frog gives you the exact answer! \$\endgroup\$ – marcosbc Jan 31 at 1:56
  • 1
    \$\begingroup\$ 1/400 is much smaller than 1/20 which in turn is much smaller than 1/8, so a 1/8 W resistor is more than ample for either your 1 K or 10K examples. Look at the decimal values rather than the fractional values. \$\endgroup\$ – Peter Bennett Jan 31 at 2:31
  • \$\begingroup\$ Righto! That's the first mistake I'm doing not reading decimal!! good point out @PeterBennett \$\endgroup\$ – Adeel Rizvi Jan 31 at 2:37
1
\$\begingroup\$

These are current-sense resistors, they won't ever have the full 3.3V across them, for the reasons you've identified. If you study the datasheet you should find the nominal voltages for ISENA and ISENB, and substitute these in to your calculations. Spoiler alert: you'll very likely need larger than 0603 and you'd be well advised to specify resistors that are designed for current-sense purposes, as these are likely to be more stable than general purpose parts.

\$\endgroup\$
3
  • \$\begingroup\$ Check out page 12 of the datasheet for the device. In any case, the max current that the device can handle is 1.6A, so if you size the resistors for that current you'll be safe. 1.6 * 1.6 * 0.33 = 0.84W so I'd suggest using a 1W resistor. 1.6 * 1.6 * 0.1 = 0.256W. Those figures assume that the resistors will both be carrying the maximum current continuously, which they obviously won't but it seldom does any harm to give yourself a good safety margin. \$\endgroup\$ – Frog Jan 31 at 1:46
  • \$\begingroup\$ From where did you get that 1.6 Max current rating? I've opened the datasheet on another screen and I'm on page 12, I can't see any except that Ichop equation. Maybe I'm blind how did u get that 0.33? buh! I'm totally lost in the world @Frog \$\endgroup\$ – Adeel Rizvi Jan 31 at 2:05
  • \$\begingroup\$ Section 7.1 absolute maximum ratings. The datasheet may or may not spell out that the A and B motor windings’ currents pass through the A and B current sense resistors, but that’s the way it is. \$\endgroup\$ – Frog Jan 31 at 9:33
0
\$\begingroup\$

In your first examples, if you increase the resistance by a factor of 10 (from 1kOhm to 10kOhm), the power that is dissipated over the resistor decreases by a factor of 10 (from 0.025W to 0.0025W). If you pick a resistor form the SMD power rating chart that can dissipate 0.025W then dissipating 0.0025W will obviously not be a problem.

Regarding your question about your circuit with R2,R3 and high current you calculated: R2 and R3 are current sense resistors (hence their low value). And yes, the power dissipated will be large. As per the Texas Instruments datasheet for the DRV8824, the layout proposal for those resistors requires a fairly large space.
Layout Example according to TI datasheet for DRV8824

There are SMD versions of current sense resistors (shunts). However, you might want to consider another form factor as well.

\$\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.