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Topic: Voltage source in parallel (Read 330 times) previous topic - next topic

hanyc93

Hi,

Anyone knows how to get 2.18 V at node1?

Thank you.

tack

#1
Jan 10, 2013, 12:06 pm Last Edit: Jan 10, 2013, 12:11 pm by tack Reason: 1
Still not woke up properly, but does that arrangement give you 7.5V at Node 3?

What is the purpose of the circuit, because you're going to be trying to charge that 1.5v cell from the 9v source. 9v across a 1.5v cell probably isn't a good idea. Maybe a diode is required in there? A diode will have a forward volt drop so you might have to have two cells and two diodes in series (2 x 0.7v drop) to get back to an effective +1.5v from that source.

If your 300R is a fixed value then put another resistor between Node 1 and Node 2 to drop the required voltage (9-2.18). You will be dissipating power as waste heat in this though. More than in your 300R.

Basically, you're just making a voltage divider. Just rough and ready, it's about 4.5:1 so I'd guesstimate A little bit less than 1.35k.

OR insert an adj VR into the circuit, set to output 2.18v across 300R.

dhenry

That's an expertly designed circuit.

:)

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