0
\$\begingroup\$

I am working on a circuit and I need to divert the current to higher resistance components in parallel without changing their voltage input. There are 4 thermoelectric devices (T1, T2, T3, T4), a heatsink, and a pump. They all ideally need 12 volts but the heatsink needs 0.09 A, the fan needs 1.1 A, and the thermos can take any current. The battery is 12V and 2.9 Ah.

Since the thermoelectric devices have such a relatively low resistance, nearly all the current goes through them and none will go through the pump or heatsink as shown in the schematic below. Is there some component that will help divert the current to the pump and heatsink? We are avoiding using extra batteries due to the weight and ease-of-use.

enter image description here

We have run out of ideas since we have limited experience with circuits. Any suggestions would be greatly appreciated. Thank you!

\$\endgroup\$
  • \$\begingroup\$ What current do the thermo devices draw when supplied with 12 votls? What is the voltage across the loads when all are connected to the battery? \$\endgroup\$ – Peter Bennett Mar 14 '18 at 18:31
  • \$\begingroup\$ We physically tested the initial resistance when supplied with 12 V but not the current. I will test that today. We just made the assumption it drew the full 2.9 A. In the theoretical schematic above they draw 0.7 A each. \$\endgroup\$ – s007 Mar 14 '18 at 18:38
  • \$\begingroup\$ The problem is: will the thermoelectric devices (what are those? Heaters? Peltier cells?) work with less current? Because, if you can estimate the minimum current required, you could put a rheostat in series with them to limit the current to that value. A rheostat like this electronics.stackexchange.com/questions/358173/… could work. \$\endgroup\$ – Sredni Vashtar Mar 14 '18 at 18:54
  • \$\begingroup\$ If the battery rating is 2.9 Ah (Ampere-hours), that is the capacity (stored energy) of the battery. Theoretically, a 2.9Ah battery will be fully discharged in 1 hour if you draw 2.9 Amp. The Ah rating has no bearing on the current the battery can deliver. The current delivered by the battery will be whatever the combined load demands. \$\endgroup\$ – Peter Bennett Mar 14 '18 at 18:56
  • \$\begingroup\$ Also, where did you get those resistance values? \$\endgroup\$ – Sredni Vashtar Mar 14 '18 at 18:58
2
\$\begingroup\$

If the battery can maintain 12 volts while delivering the total required current, each device will draw the current it requires. There will be no need to divert current.

If the battery cannot maintain 12 volts while delivering the required total current, you need a bigger battery.

\$\endgroup\$
  • \$\begingroup\$ So in practice, assuming the battery maintains 12 V, the higher resistance components will still draw the current they need even if the resistances of the thermoelectric devices are 1000x less? \$\endgroup\$ – s007 Mar 14 '18 at 18:49
  • \$\begingroup\$ With the resistances you used in your schematics, the heatsink and the pump will draw that same small current even if they were individually connected across the battery. 12V / 16.2kohm = 736 uA. \$\endgroup\$ – Sredni Vashtar Mar 14 '18 at 19:04
0
\$\begingroup\$

If your battery is powerful enough to supply all the current those loads need, then it will maintain a nominal 12v across the rails. If your battery is not powerful enough, then your rail will drop, and you're in trouble.

At 12v, each parallel load will take whatever current it wants to. The fan will still take its 1.1A, the heatsink still take its 90mA, and the thermos will take whatever they want as well.

There is no need to 'divert' current into the fan and heatsink, just because the thermos are taking a large current. There is no way to 'divert' current without changing the voltage on various components.

You might want to reduce the amount of current your thermos are taking, by regulating their voltage down. At the high currents involved, a buck converter is usually the best way to do this.

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

I might not understood your question, but as long as you have resistors - or circuits very similar to resistors - their resistance will determine their voltage to current ratio. If you have parallel resistors, and you want to change the current through them, you have to change the voltage on them. For example by inserting a buffer with the desired voltage gain.

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

For most components/circuits the current is dependant on the voltage.

For resistors there is a simple relationship \$ I = \dfrac{V}{R} \$ this is also almost true for a lot of other devices but lots of things are not resistive.

A zener diode for example will take almost no current below its zener voltage and current will rapidly increase if you try to increase the voltge across it until either your power supply can't provide more current or the zener gets too hot and fails.

My PC power supply will take less current as you increase the supply voltage as it needs to supply the same power to my laptop whether I have 100V mains in Japan or 230V mains in the UK.

You can not divert current between parallel devices. You can add series resistance to divert current but then the devices have different voltages across them so are not in parallel any more.

\$\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.