I have a resistor diode circuit that I'm powering with a 9V battery. The circuit diagram below is a simplified version of it. The real one has 4 LEDs, 8 resistors.

I want to deliver 30mA to the diode. I had the idea to place two batteries in parallel to increase running time. Will placing two 9V batteries deliver 60mA to the LED? or does the current supplied remain constant and the batteries just drain less?


simulate this circuit – Schematic created using CircuitLab

  • 1
    \$\begingroup\$ 30 mA is absolute maximum at room temperature. Your circuit there pumps out about 20 mA only. Considering diode drop as 2 V. Paralleling cells do not increase the current supplied to the diode. \$\endgroup\$
    – User323693
    Jun 25 '15 at 3:51

This one:

"the current supplied remain constant and the batteries just drain less"

The LED current will be unaffected by the addition of the second identical parallel battery.

V = I x R

In this circuit you are doubling the battery, but not changing the output voltage (two identical 9V batteries in parallel is still a 9V output).

On the load side, the resistor and LED have not changed (that's the R in Ohm's law). Please note an LED is not accurately modeled as a pure resistance, but a complete explanation of that is not necessary to understand the answer to your question.

No change in V; No change in R; ...therefore NO CHANGE in I (current)

E = V x I x t

What does change is the total potential energy in this circuit. If you double the battery count, the total current sourced to the LED will be unchanged, but the current supplied by each battery will be 1/2 of the total. Because the batteries are supplying half the current as before, they will last twice as long.

Energy is voltage times current times the time the current is supplied at that voltage. A 1000mAh Alkaline battery means that it can supply 1A at ~1.4V for ~1 hour.


No change in E; No change in V; ...therefore battery life (time) is INVERSELY proportional to current

  • 1
    \$\begingroup\$ +1. Thanks for clarifying the current related part using Ohm's law equation. \$\endgroup\$
    – WedaPashi
    Jun 25 '15 at 4:43

In general when Batteries are connected in parallel, the voltage remains the same while the current gets divided between the two batteries and so the runtime will increase.

In your case, referring the circuit you have shared, there is no change in resistance. So, V = IR remains pretty much the constant through the time. So, your intended drop of 30 mA across the diode is achieved such that, 9 = I*(300), so I = 30 mA. So, the voltage shall remain 9V and only the runtime is increased by two times (since there are two batteries).

Threats in wiring the Batteries in parallel are subject the matter that whether the cell chemistry is identical or not. As long as the two batteries are of the same state of charge, chemistry, ah rating and voltage, they should do just fine.

  • \$\begingroup\$ thats what i dont get. isnt the Voltage at that node still 9V? and the resistance is the same. \$\endgroup\$
    – scordova88
    Jun 25 '15 at 3:52
  • \$\begingroup\$ @scordova88: See the edit. The voltage at the node you are referring to should remain 9V. \$\endgroup\$
    – WedaPashi
    Jun 25 '15 at 4:00

If higher currents are needed and larger cells are not available or do not fit the design constraint, one or more cells can be connected in parallel. Most chemistry allows parallel configurations with little side effect. Figure illustrates four cells connected in parallel. The voltage of the illustrated pack remains at 1.2V, but the current handling and runtime are increased fourfold.

Battries are in parallel

you can study more about the batteries connected in parallel & series here. http://batteryuniversity.com/learn/article/serial_and_parallel_battery_configurations

In your design, since two cells are connected in parallel voltage will remain 9V only however the current capacity will be doubled but it doesn't mean it will supply double current(however it can depending upon the load). Since load is unchanged current will be the same. From the datasheet, maximum forward voltage you can apply is 2.6V (2V typical) Since you are applying 9V you need to put resistor in series in order to reduce the voltage drop across LED.

if you see the datasheet of LED, it specifies that maximum continuous forward current it can handle is 30mA. it doesn't mean that it will always consume 30mA it simply means it can handle up to 30mA forward current(again forward current will depend upon the current capacity, supply voltage & load ) & brightness of the LED will depend upon the current flowing through it. parallel connection doubles the current capacity but the current in the circuit remains the same but if you change the load, current will change & if this current exceeds 30mA, LED will get damage it won't operate more than 30mA(30mA+30mA=60mA will not be the case)

  • \$\begingroup\$ I know the current capacity increases. But will i see 30mA + 30mA = 60mA current draw to my diode? \$\endgroup\$
    – scordova88
    Jun 25 '15 at 4:29
  • \$\begingroup\$ @DrFriedParts I did. if anything is still wrong you can do the edits to correct me. thanks! \$\endgroup\$
    – nkg2743
    Jun 25 '15 at 5:07
  • \$\begingroup\$ Total charge is x4. Seems like everyone is dodging this key word. Current capacity means nothing to me. \$\endgroup\$ Jun 25 '15 at 7:03

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