# Homework Help: Bridge rectifier (conceptual)

1. Oct 20, 2011

### sandy.bridge

1. The problem statement, all variables and given/known data
I am in the midst of preparing for my midterm which is tomorrow, and I was going over a past midterm that has the solutions. However, there is one conceptual problem that I am not understanding, in fact, I think it is wrong...

The diodes in a bridge rectifier have a maximum dc current rating of 2A. This means the dc load current can have a maximum value of?

The answer is 4.... But that doesn't make sense to me. The current in a bridge rectifier goes through the diode before it goes through the load, and the diode is not going to increase the current by 2 A, so my logic is the max load current through the resistor is 2.

What am I missing here?

2. Oct 20, 2011

### rude man

Hard to answer without the schematic. Is it a half wave or full wave rectifier? How many diodes? Is the load paralleled by an arbitrarily high value capacitor or unfiltered?

3. Oct 20, 2011

### sandy.bridge

Well it's a multiple choice question with no diagram. I am assuming it is the simplest form I have been exposed to, which with a positive input has a diode then a resistor then another diode. With negative voltage it is the same, just going the opposite way.

4. Oct 20, 2011

### dimpledur

If it is as you described, then I would agree with you in that the maximum current through the load is 2A. If the maximum current through the resistor was 4 A, then that 4 A would first pass through the diode which is supposed to have a maximum current of 2 A through it.

5. Oct 20, 2011

### Staff: Mentor

Unless there is some other information, it sounds wrong to me as well. The DC load current would be maximum for a DC input to the bridge, and in that case the load current is going through two of the bridge diodes in series. If the bridge is rated at 2A, that's all you can pass on to the load.

6. Oct 20, 2011

### Staff: Mentor

The problem as stated is rather vague. IF they are presuming an AC input to the bridge they could be implying that the load is seeing twice the average current that the diodes do, or perhaps they are referring to the RMS currents (equivalent DC power delivered). Without knowing what the load is (it could be a filtered load with capacitive smoothing) it's hard to provide a definitive answer.

7. Oct 20, 2011

### nsaspook

8. Oct 20, 2011

### rude man

I agree, except possibly the idea is that since each diode only conducts the current half the time, they only have to be rated at half the load current, which runs all the time.

9. Oct 20, 2011

### sandy.bridge

I'll quote the question, which has an answer, considering there is no "other" option.

The diodes in a bridge rectifier each have a maximum dc current rating of 2 A. This means that the dc load current can have a maximum value of
a)1
b)2
c)4
d)8"

That's all.

The question doesn't say anything about a filter, so I assume there isn't one... Here's an example of the rectifier given in the text:

10. Oct 20, 2011

### nsaspook

Correct, each diode pair would have 4 amps at a .5 duty cycle. As long as the diode instantaneous forward current limit was not exceeded it would be be copacetic.

11. Oct 20, 2011

### rude man

That's exactly right. The problem states "dc current", I think they meant "dc component" or "average current". Then everything is OK.

Thing is, which is what nsaspook is saying - if a diode has a dc current rating of x amps then it does not mean you can exceed x amps for a short time while leaving the current zero the rest of the time, so long as the average current does not exceed x amps. Diodes usually have a surge rating as well as a dc rating.

Last edited: Oct 20, 2011
12. Oct 21, 2011

### sandy.bridge

Let me get this straight: due to the alternating input current the maximum resistance going through the diode will be four, since 2 A will be coming from either side of the cycle?

Last edited: Oct 21, 2011
13. Oct 21, 2011

### rude man

No. We're talking duty cycle, which is the percentage of the time the diode conducts.

Take 1 of the diodes. Assume 60 Hz sine voltage. The current thru it per 60 Hz cycle will be iL = iL0*sin(wt) for 8.33 ms, and zero for the other 8.33 ms (w = 2pi*60). So the average current is half of what it would be if it conducted iL for the entire 16.67 ms. Heating of the diode is half also.

Forget "diode resistance" altogether. Diode resistance is a function of its instantaneous current, moreover there is static resistance and dynamic resistance. For your purposes the diode looks like a constant-voltage (Vd) drop device for the entire range of load current iL. Power dissipation is thus Vd*iL.