Measurement of Raspberry Pi voltage and current at the 5V pin under load
If you connect a 5V pin to Raspberry Pi, then you expect a voltage of 5 volts. Unfortunately, the voltage and current behavior changes under load. How exactly can be determined with a series of measurements.
|Volt (V)||Amps (A)||ohm||ohm||Amps (A)||Volt (V)||Amps (A)|
|0||5.0||Comparative measurement without resistance||5,027||deleted|
The table consists of a total of 3 parts. The first part consists of the 5 volt (column 2) voltage applied to the 5V pin and a current from 0.001A to 0.016A (column 3). In order for the given current to be set below the specified voltage, a corresponding resistor must be used for this purpose, which was calculated for this purpose (column 4).
The second part of the table is that the calculated resistance values are not real resistances, but real resistances with other values have to be used. In this case from the E12 series (column 5). In some cases, there are overlaps with the calculated resistance values (between columns 4 and 5). Since a given current is set at a specified voltage at a given voltage, it was calculated (column 6). This current is what would be expected at the fixed voltage of 5 volts.
The third part of the table deals with the actual measurement of voltage and current (columns 7 and 8). Here, the voltage at the 5V pin was first measured (column 7) to see if there really were 5 volts across the resistor. The current measurement was then performed (column 8) to see if the calculated current actually flows through the resistor.
Measurement 0 was made to check if there were really 5 volts. The current measurement is omitted here, because there is no closed circuit here. Subsequently, the measurement was carried out 1 to 22. In each case once for the voltage and the current.
Observation and evaluation of the measurements
- As the resistance decreases, the voltage decreases.
- As the resistance decreases, the current increases.
- As the current increases, the voltage drops.
Findings and conclusion
The measurements show that the 5V pin has a lower voltage under load than expected. When a current flows, the voltage drops. The actual flowing current differs only slightly from the calculated current.
However, you can only conditionally load the 5V pin because the voltage drops across the current. That is, the 5V pin is only partially suitable for the power supply of the external wiring.
Subsequent circuit components must be dimensioned for less than 5 volts or require their own power supply.
The extent to which these findings are disadvantageous for a particular circuit depends on the exact requirements and the dimensioning of the circuit.