1
\$\begingroup\$

I'd like to use an MCP73831 to charge my lipoly battery. In section 6.2 I see they recommend short wires to the battery.

I bought my battery with wires, 20 gauge and approximately 30cm long. The maximum charge current will be 500mA. I know that there will be a voltage drop over the wire and I therefore will lose power. This power loss will be very small: \$0.60\cdot0.03331\cdot0.5^2=0.0049965 W\$ (length, resistance / meter, current). So I don't see how this is a good reason to keep the traces as short as possible.

So besides reducing power loss, is there any reason to keep the traces as short as possible?

\$\endgroup\$
  • \$\begingroup\$ do you remember answering this question? Calculation of power lost in the wires is the same in this case. \$\endgroup\$ – Phil Frost Jan 21 '13 at 15:02
  • \$\begingroup\$ Uh, yes. Sorry for that. I'll change my question to: is there any other reason to keep the traces as short as possible besides power loss? \$\endgroup\$ – Keelan Jan 21 '13 at 15:12
  • \$\begingroup\$ @PhilFrost I also added a calculation for the power loss. It's ~5mW, so that isn't a very good reason to keep the wires that short. Is there another reason? \$\endgroup\$ – Keelan Jan 21 '13 at 15:21
  • \$\begingroup\$ Maybe the application note you are referencing is trying to make sure that you do not try to use wires 100m long. \$\endgroup\$ – Michael Karas Jan 21 '13 at 17:00
2
\$\begingroup\$

Think of the short wire recommendation as a 'best practice'. This application is fairly low current, so the risks in disobeying are somewhat minimized. That being said, here are a few 'best practice' reasons to keep the wires short:

This particular IC doesn't have separate load voltage sensing inputs, so any loss in the cabling will manifest itself as a load setpoint error which varies as a function of the current (higher current means higher \$I^2R\$ losses and lower voltage at the load).

Another effect: very long cables could have enough inductance to resonate with the \$4.7 \mu F\$ capacitor at pin 3 of the IC, especially if a charged battery is plugged in and the IC has no supply voltage. The IC may get damaged if the voltage on the charging pin gets excessively high.

\$\endgroup\$
0
\$\begingroup\$

The problem with longer traces isn't only the power lost, but the voltage dropped over the wires. Part of the charging strategy involves measuring the voltage of the battery. However, any voltage drop in the wires will manifest as an error in the battery voltage measurement.

A more sophisticated circuit could use one circuit to supply current, and a separate, low-current circuit to measure voltage at the battery. However, this device is "for use in space-limited, cost-sensitive applications," so the extra complexity was probably deemed unnecessary.

\$\endgroup\$
  • \$\begingroup\$ Okay, but the voltage drop would be \$0.60\cdot0.03331\cdot0.5=0.009993 V\$. But thanks for your reply! \$\endgroup\$ – Keelan Jan 21 '13 at 19:50
  • \$\begingroup\$ @CamilStaps actually twice that, about 20mV, since you drop that voltage in each wire. If 20mV is "close enough", then your wires aren't "too long". \$\endgroup\$ – Phil Frost Jan 21 '13 at 20:40
  • \$\begingroup\$ No, I used 0.6m because I have twice 0.3m. But 10mV is close enough, isn't it? \$\endgroup\$ – Keelan Jan 21 '13 at 20:47
  • 1
    \$\begingroup\$ @CamilStaps ah, I see. I don't know enough about these kinds of batteries to offer any expert advice on how critical the voltage is, but I wouldn't assume it's irrelevant without more research. \$\endgroup\$ – Phil Frost Jan 21 '13 at 20:56

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.