I'm an electrician going back to school online to get my electrical engineering degree. I am having a hard time with some of the homework, as the book is extremely confusing, and does not always go into detail as to how the math is done. I need to calculate the minimum value of the surge limiting resistor required for a full bridge rectifier with a peak output of 50 volts. The rectifier is using 1N4933 diodes. I'm not asking for the answer-just the math formula to calculate. I would appreciate any and all help. Thanks!
Surge limiting resistors for full-wave bridge rectifiers
bridgeresistorsvoltage
Related Solutions
I guess the \$\times\$3 rule of thumb refers to the transformer's secondary voltage, the AC voltage before rectifying. If your output is 18 V DC then the transformer will be a 14 V type. That 14 V is the RMS value, peak value is \$\sqrt{2}\$ times higher, or 20 V. Subtract 2 diode drops and you arrive at 18 V.
So applying the \$\times\$3 rule to the 14 V already gives you 100 % headroom, applying it to the 18 V gives you almost 200 % headroom, which is overly cautious IMO. Since 20 V is the highest voltage to expect all three will do fine.
The current is a different story.
If the smoothing capacitor is charged the input voltage will be too low for the greatest part of the cycle to make the diode conduct. Only for a short period you'll have a current that recharges the capacitor. So you can expect a large current peak there, several times the average of 1 A. A 1A diode like the 1N4001 can handle this current, though it's a good idea to have some headroom here as well. For a 1 A output a 2 A diode is a good idea, so also for current all three rectifiers will do.
There is no reason why a DC polarity reversal should take place, and the warranty can basically blame it on the user.
If the device is battery powered, the use of a standard, convention-adhering battery holder with clear markings should prevent such a thing from ever happening. Even users who don't look at markings are trained to put the flat part of an AA battery against the spring, and slide the nub against the leaf contact. 9V batteries have gendered connectors; no way to screw up short of deliberately making a temporary wrong-way contact while the power switch is on. The 99.999% of the users who are able to engage two brain cells cells together when installing a battery don't want to sacrifice battery life for the sake of the remaining 0.001%.
If the device has an AC adapter, then a polarity reversal can never happen if the original AC adapter is used. If a different AC adapter is used, which has a compatible DC barrel jack, but which puts out opposite polarity, or perhaps AC, that's the user's responsibility. Chances are that by the time users have lost the original AC adapter, the item is out of warranty. Possibly, they are not even the first owners, and so do not have the original receipt. So the chances are low of the company having to replace the item or provide a free warranty repair because of damage caused by a wrong polarity (or voltage, for that matter) aftermarket AC adapter.
Internal DC power supplies that run strictly on AC via an AC power cord obviously have no need for DC polarity reversal handling; the only way it could happen is that someone assembles the circuit wrong.
Related Topic
- Electronic – Middle-Tapped Transformer Full-Wave Rectifier Vs. Full-Wave Bridge Rectifier
- Electronic – Peak output current of a Full-Wave Bridge Peak Rectifier
- Electronic – what determines the correct values for bridge rectifiers
- Electrical – Full-wave controlled bridge rectifier and the RMS voltage derivation
Best Answer
The simplest calculation is to assume the worst: that at the instant of switch-on the cap on the far side of the bridge is not charged, and therefore presents almost a short circuit at the instant when power is first applied, for that first half-cycle of 50/60Hz, that's the worse case peak current you have to limit, the "inrush current" (to charge up the cap). Every cycle of the AC after that point, the cap will become more & more charged, drawing less & less current per cycle, until it settles to whatever power is being drawn by the circuit powered.
In that case, the max surge capability for that diode is 30 Amps for 8.3mS (i.e. 1 half-wave of the 60Hz mains frequency). This is straight off the datasheet.
In that case, R=V/I = 50 / 30 = 1.66 ohms. Let's say 1.8 or even 2.2 ohms just to be safe, and to choose a standard resistor value (E12-series).
The only problem with this approach is that resistor is always there, dissipating heat (diode is rated for 1.0A continuous, so presumably this is the design max current for the power supply, so Power = IIR = 1.0 * 1.0 * 1.8 = 1.8 Watts, so it won't be a small resistor! Other real-world solutions would use an NTC resistor whose resistance drops as it heats up, so it limits that inrush current, but then after a few seconds dissipates somewhat less heat (wasted)compared to the 'dumb' resistor. Or, a relay powered by the power-supply turns on after a short delay that shorts out the resistor.