DC Spannungswandler?
Ich würde mir gerne einen Spannungswandler holen und habe dazu 2 fragen:
- Sagen wir, ich löte da einen Anschluss für DC Netzteile an. Dann schließe ich ein Netzteil an, das 12V liefert und 1A. Jetzt stelle ich das Modul auf genau 5V ein und ohne das Poti zu bewegen schließe ich ein neues Netzteil an, das 9V und 800mA liefert. Ändert sich der Ausgangswert dann? Sprich ist der Output gleich oder ändert er sich wenn ich ein anderes Netzteil nutze?
- Falls er gleich ist, gibt es eine gute Möglichkeit zu berechnen, wie groß der Widerstand vom poti sein muss, damit ich genau 5V Ausgang habe, oder müsste ich messen? Ich will immer nur 5V haben und nichts aus Versehen verstellen oder so
Falls jemand eine bessere Idee für Stromversorgung hat, gerne her damit. Es muss nur wenig Platz wegnehmen und ich will mich und meine Bauteile nicht rösten. Es sollen 5V und 1,5A ca sein. Da soll ein Arduino Mega und 10 Relais dran. Die Relais sollten laut Datenblatt 40mA ziehen, falls ich richtig gerechnet habe
(1 votes)
Loading...
I don’t understand your problem.
The pot was deliberately installed so that you can adjust the output voltage.
It should therefore always be used in conjunction with a voltmeter and the output voltage should be set to the desired value.
I don’t want anything to change. It should not be a laboratory power supply but just a 5V power supply for my system and this will not change
How about you get that part and just experiment with it? If it’s not what you’re presenting, you’ll send it back. You will certainly be able to find out in two weeks whether it meets your requirements or not.
Yes, however, I can also save myself 6€ shipping, waiting time and work for riches if I just ask someone who knows
My suggestion, if you have the right pot and have found your setting and the source with 9V and 12V does not make any difference, can take two fixed resistors in total the same value as the pot and make sure the output 5V remains stable.
But would the voltage remain the same if I changed the input voltage?
To keep the output voltage stable is the task of such a voltage converter. However, you searched out a step-up converter that increases the output voltage compared to the input voltage. What you need would be a step down converter.
Otherwise, this is not a regulated voltage converter.
You have to test it. I’m guessing, with little deviations.
When applied to the 5V relay 5V, a current of about 28mA flows, due to the coil resistance of 178Ω. 5V times 28mA then makes 140mW.
If you were to apply the 5V relay 6V, a current of about 34mA would flow through the 178Ω coil resistance. 6V times 34mA then makes a power of about 202mW.
The 390mW relate to the 3V relay when it is operated with 5V because the coil resistance of this relay is only 64Ω. At 5V a current of approx. 78mA. 5V times 78mA then makes a power of about 390mW.
https://de.wikipedia.org/wiki/Ohmsches_Gesetz
I=U:R or A=V:Ω or mA=V:kΩ
With an ohmic resistance, the flowing current depends on the applied voltage and in a direct current circuit, a coil is also only an ohmic resistance if we let out the on and off torque.
For each of these relays, a voltage range is specified in which the respective relays can be operated with a specific nominal voltage. However, the power consumption is not the same over the entire voltage range, but depends on the actual applied operating voltage. 140mW require the relays only when they are operated with their corresponding nominal voltage. If the applied operating voltage is lower than the rated voltage, then a smaller current flows and consequently the power also drops, or if the applied operating voltage is higher than the rated voltage, then a larger current also flows and because the power is the product of current times voltage, then the power also rises over the specified 140mW.
It confused me that the resistance is on the left. I thought the voltage then depends on the coil resistance and there are several versions or so. This then makes much more sense. But where do the 390mW come from? For the 5V version there is 140mW. By the way, we didn’t do anything with data sheets, but I still regret that I didn’t watch in physics at the time
What is not to understand? First, the table must be taken to the hand, which fits the execution of the relay(s). The first table relates to the monostable embodiment of the relay and the second and third table relate to the two bistable versions. I suppose you use the monostable version. The first table is then used.
In the first column, the coil resistance and the last column are the rated operating voltage, corresponding to the respective coil resistance. The 5V version of the relay has a coil resistance of 178Ω. At 5V a current of approx. 28mA through the coil of the relay, which corresponds to a recorded power of about 140mW. The current through the coil depends on the applied voltage. With increasing operating voltage, the current flowing through the coil (coil resistance) also increases. This is caused by the Ohmic Law.
If you use the 3V version of the relays, the coil resistance is 64Ω. The operating voltage of this relay is permitted in a voltage range of 2,3-7,5V, so that it can also be operated with 5V. At 5V, however, a current of about 78mA flows, which corresponds to a recorded power of about 390mW.
I’d think I’d get a power supply with 9/12V and 1/1.5A. Then I definitely have enough buffers with tension and strength. As I said, there should be an arduino Mega and 9 relays with 0.04A each. However, I did not fully understand the data sheet. Since for the individual embodiments there are around the 7 values for the different resistances of the coils, however nowhere is there anything about which resistance the coil has. If you want to watch
https://www.reichlt.de/index.html?ACTION=7&LA=3&OPEN=0&INDEX=0&FILENAME=C300%2Ffrt5.pdf
Yeah.
Yes, but the input voltage must not fall below a certain value. In this voltage converter, the input voltage should be at least 1.7-2.0V above the set output voltage so that the set output voltage can be kept stable.
So… der quasi. That means if I connect a power supply with other voltage, the output voltage remains 5V?
Voltage converters normally have a large input voltage range. It will have to be larger than the desired output voltage but also no “high voltage”. For maintaining a stable output voltage, a reference voltage must be generated internally. The output voltage is thus compared and regulated with fluctuations in the load and the input voltage.
There are always the smallest deviations. Then it’s not 5,00V anymore, but maybe 4,99 V or…
You don’t have to turn. A bad power supply might supply 10V if there is 9V on it. Or if the mains voltage changes.
I knew there was a voltage with which the converter compares, but I didn’t know how to generate it and whether it changes when you suddenly have another power supply
such a reference voltage is generated with Z diodes. At a certain voltage flows, because a very large current flows at the Z diode when there are small voltage changes. If the current has to flow through a resistor beforehand, the voltage at the Z diode becomes smaller again. Thus the value again has the same as before a cause (voltage change at the input/new power supply …). This is in no time. Fluctuations then occur practically only by heating/cooling of the Z diode. For an Arduino something is meaningless, it could also carry 4.5 to 5.5 V (must be read).
Of course, you can place a 5.1V Z diode parallel to the procedure, which I consider superfluous.
I would actually run the arduino directly over the 5V Pin and not the VIn PIN. Should I, then maybe add an overvoltage protection? No jamming or something? Or isn’t it likely what happens? The device will be active for a few hours
that made my smartphone. That’s right.
From a specific voltage flows, then a very large current at low voltage changes at the Z diode through the Z diode.
1. You don’t need a step-down converter.
Two. Yes the output voltage remains the same
3. 5V is eh standard there are such voltage converters with 5V fixed output there you no longer have to adjust.
ZB
https://amzn.eu/d/avf2RrS
There are others.
Of course, you can also stick the pot on your hot glue or replace it with corresponding resistors.
However, I think those types without USB port have already found
It looks pretty good. Especially small. I would then just record the USB port
As long as the input voltage is 2-3 volts above the output voltage, there are normally no problems.