The documentation for a chip that I'm trying to use says it needs 2.7 to 3.6V. When I use a multimeter and put on probe on Vin and the other on GND I read 1.7V. Is this the correct way to read supply voltage? Does this mean that it's not getting enough voltage?
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asked Mar 15, 2011 at 22:46
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2 Answers
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Yes, this is the correct method for reading voltages. You do need to make sure your multimeter is set to DC mode and not AC mode. If it is on DC mode then you are not supply the chip enough voltage.
What are you powering it from? Either your voltage source is low (like if it is a battery) and should be replaced, or if you are using a current limited supply you could be hitting a current limit that is causing the voltage to drop.
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\$\begingroup\$ The power is coming from a voltage regulator which is powering something else. If I take power straight from the power supply, then I get 5.1V. I'm thinking of maybe doing that and using resistors to get the right voltage. I tried some resistors but the voltage is too low. What kind of resistance do I need? \$\endgroup\$ Commented Mar 15, 2011 at 23:16
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1\$\begingroup\$ @200ok404notfound - resistors are an inefficient way to go, and unless your chip takes the same current under all operating conditions, you could have a difficult time finding a single specific resistor that would work in all cases. \$\endgroup\$– JustJeffCommented Mar 15, 2011 at 23:28
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\$\begingroup\$ You should use a second voltage regulator of 3 or 3.3v. You wont be able to get what you want with just resistors. ICs hardly ever pull a constant current, this means that the voltage will always be changing, which is not what you want. \$\endgroup\$– KellenjbCommented Mar 15, 2011 at 23:35
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\$\begingroup\$ I have a 5V voltage regulator. What if I uses this in combination with resistors? \$\endgroup\$ Commented Mar 15, 2011 at 23:53
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\$\begingroup\$ A resistor divider is a poor choice for a voltage supply if what is being supplied draws current, because the voltage at the divided node will change with the current. A simple solution that may get you where you want to go is to use a 3V Zener diode. \$\endgroup\$ Commented Mar 16, 2011 at 0:09
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You don't say what the chip is, but typical power supply will probably be 3.3V, maybe 3V. You say you already have 5V, and want to derive the 3.3V via resistors.
Like others have pointed out that's a Bad Idea. Say you have a series resistor of 2k\$\Omega\$ to drop 2V, because you know you will need 1mA. That would be OK, you'll get 3V, but if for some reason the current increases to 2mA, you'll only have 1V left (2mA x 2k\$\Omega\$ = 4V voltage drop), which is too low. Likewise, if the current decreases to 0.5mA the remaining voltage will be 4V, which is higher than the allowed value of 3.6V and may damage the chip.
A resistor divider will have the same problem. Maybe you used one with the right ratio to get 3.3V, but forgot that the chip is parallel to the lower resistor, which may explain the low voltage of 1.7V.
The neatest way to go from 5V to 3.3V is using an LDO regulator. Common regulators require a few volts difference between input and output; LDO (Low Drop-Out) often only need a few hundred millivolts. The LP2981 is available with 3V or 3.3V output voltage.
An alternative is to use a zener diode. Personally I try to avoid them, because they often need a lot of current, and they don't regulate as well as the LDO. If you want to use one, be sure to select a low-current diode like the MMSZ4684.
