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I'm getting 6 volts open circuit voltage. Whenever I connect a load (resistors), there is a huge drop in voltage values. Does it signify power input is less or is there any other reason for the same?

I'm using a thermoelectric module as a power source.

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closed as unclear what you're asking by Kevin Reid, DoxyLover, Neil_UK, Edgar Brown, Blup1980 Feb 19 at 12:50

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    \$\begingroup\$ google ohm's law to learn the relationship between voltage, resistance and current ...... hint: the power supply has some internal resistance \$\endgroup\$ – jsotola Feb 18 at 5:13
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    \$\begingroup\$ What are you connecting this load to? There is not enough information for an answer that might be relevant to your case. \$\endgroup\$ – Edgar Brown Feb 18 at 5:30
  • \$\begingroup\$ Shivam - Although we can guess at the likely cause, there are several possible variations. Therefore please edit the question and add more details and context e.g. What circuit are you building and why? Add some clear, in-focus photos of your hardware so that we can see the length and type of wiring. Add photos of the battery and resistors. What is the battery type / model (at a guess, for 6V it's SLA)? What value are the resistors? Why did you choose that value? Do you have a multimeter and experience using it? If so, did you take any measurements already? See here: How to Ask Thanks. \$\endgroup\$ – SamGibson Feb 18 at 5:56
  • \$\begingroup\$ When voltage drop is 50% you know that source resistance is equal to load. Load regulation error defines this relationship from resistance ratio or simply Ohm’s Law sharing the same current. If there is active current limiting occurring, then it may be non-linear. \$\endgroup\$ – Sunnyskyguy EE75 Feb 18 at 6:32
  • \$\begingroup\$ Because your thermoelectric module is not enough of a power source. Edit your question stating resistance, current of load. \$\endgroup\$ – StainlessSteelRat Feb 18 at 16:26
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Like some of the comments say your problem seems related to the internal resistance of your power supply. All voltage sources have some internal resistance that effectively limits the amount of voltage they can deliver to a load. For example, take a look at the following circuit.

schematic

simulate this circuit – Schematic created using CircuitLab

In an ideal world, V1 would have no internal resistance. We do not live in an ideal world, so this voltage supply has an internal resistance, which I've made to be 1 ohm. In this circuit, where we have a 1k ohm load, that internal resistance has little effect on circuit performance. To calculate the voltage across the load, we can use the voltage divider equation, so Vload = V1*(Rload/(Rload+Rint)), so Vload = 1*(1000/1001) = .99 volts. So the large Rload means almost all of the voltage from V1 is appears across Vload. Internal resistance becomes a problem when Rload becomes too small.

schematic

simulate this circuit

For this circuit, Vload = 1*(1/1+1) = .5. Bad! In short, if you have a crummy voltage source (high internal resistance), use large valued resistors (1k+). You could also read about ohm's law, voltage dividers, and thevenin equivalent circuits to have a better idea of whats happening.

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Every power supply has an internal resistance even if it is very small.

if there is not an internal resistance you don't get a drop of voltage in the load

Let consider this circuit with : R int=0 ohm, R load=0.1 ohm enter image description here if we calculate the total current we get

I=V/R I=6/0.1= 60A

the voltage across the load equal V=I*R load

V load= 60*0.1= 6v ie: no drop of voltage

but if we change this circuit and we use a resistor of 10 ohm in load and Rintarnal considered as 2 ohm enter image description here

we get:

I=V/(R load+R int) I=6/(2+10) I=0.5 A

the voltage across the load equal V=I*R load

V load= 0.5*10= 5v ie: there is a drop of voltage

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