I have a very simple question regarding the 24VDC power supply for the instrument.
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For example, card readers or safety barriers supply power to the field instruments; is this power a regulated 24VDC? Since the power supply is 24V, and after a load is connected in series, the voltage across the load is 8V; so where does the remaining 24-8=16V go? The cable voltage is almost negligible. For 2-wire instruments, the supply voltage is usually around 20VDC, rather than around 24VDC. So where does that extra voltage go? The supply voltage is normal; if the current is too low, the instrument cannot function properly. Could you explain the reason in detail? Formula?
I’m not very familiar with instrumentation and control systems. You’re referring to switching power supplies, right? I once discussed this with the instrumentation and control specialist in my department, and he said that the output voltage isn’t necessarily 24 volts; it can be adjusted by adding resistors to the power supply.
The 24VDC power supply is provided by a voltage regulator； The voltage at the field instruments is not 24VDC because of the resistance in the transmission wires, which results in a voltage drop across those wires. To ensure the proper operation of the field instruments, it is necessary to take into account the magnitude of this voltage drop.
Take a good look at this post below; if you don’t know something, don’t just make things up:P It’s misleading for others:P http://mp.wei*n.qq.com/s?__biz=MzI1NDAxNDYwMg==&mid=2664668189&idx=5&sn=371af3baea44b3eea1e71a17c68e4dfe&scene=1&srcid=0710pFF8Pxx2dSr5w8Ml9OIA#rd
The power supply is, of course, stable. As for the decrease in voltage when a load is connected, there are various reasons for this; it could be due to high resistance in the instrument cables, or low resistance in the instruments in the circuit – all of these can result in a voltage that is lower than that of the power supply. Each instrument has a specific operating voltage range. It doesn’t stop working just because it’s below 24DC.
They’re all junior high physics problems; I’m too lazy to answer them, I’ll have to review them myself.
A 2-wire instrument is electrically equivalent to a variable resistor; when its output current changes, it is as if the resistance value is changing. At this time, the voltage drop caused by the wire resistance also changes, which is why the voltage at terminal 2 of the meter varies as well.
I understand what you mean by cable voltage division, but when the terminal of the safety barrier at the site is removed, 24V is measured on the safety barrier; when it is connected back, 8V is measured. Why isn’t it 24V? The voltages of both the cable and the instruments at the site should be taken into account.
You’ve set safety limits for voltage and current, but that’s not right either; 8V doesn’t suffice to power the gauge at the site. I remember that at least 17–18V is required to power the gauge there. A couple of days ago, the K-TEK level gauge we have here (that annoying piece of junk that really drives me crazy) suddenly stopped showing any readings; the voltage measured at the gauge’s display was less than 18V, while 24V was measured when the display was disconnected. There was no doubt that something was wrong with the display. I took the display apart, removed the circuit board, and went through some troubleshooting – mainly by checking the wires inside. After fixing everything and putting it back together, I powered it on, and sure enough, the voltage was now a little over 23V. It's just my personal opinion, for the original poster's reference.
For 8V. The table definitely won’t get bright.
The 8v is fine, the locator as well