Rectifier: Why is there a leakge in current?

[CDTOE]

Hi, Everybody!

I was tasked by my lab instructor to design an AC to DC converter using a 4 diodes model. I started doing the simulation in Proteus ISIS7 part by part to get the needed values for components to get a final DC output of 4.3V.

Currently, I'm doing a simulation of the first block of the converter, which is the rectifier. I have started simulating with a half-wave rectifier to get familiar of how the diode I'm using (1N 4007) may work, and from there I would use the same type of diode to build a 4 diode rectifier.

I connected a 12V peak-to-peak sinusoidal AC voltage source in series with a 1N 4007 diode, and a 1Kohm resistor. I used a built-in Oscilloscope to display the output waveform across the resistor. It has a peak value of 11.25 V, which means there's a 0.75 voltage drop across the diode, but I also found that there's -500mV (-0.5V) in the negative cycle, whereas in reality it should be 0V, because the peak inverse voltage of 1N 4007 is set to be 1KV according to ISIS7 software, so how come there's a leakage in current from the reverse side? is it something I have to worry about in the real life design? or is it just something related to the simulation software?

Thanks.

Edit1: I used Multisim to run the simulation, and I got a negative value for the output voltage when the diode is in reverse mode, but this time, it was in micro-volts. Does this really reflect the application in reality? also, I did a full-wave rectifier, and it was more precise in calculating the output voltage, with no negative value at all! still curious why there's a leakage in current for the half-wave rectifier.

 

[DragonPetter]

Hi, Everybody!

I was tasked by my lab instructor to design an AC to DC converter using a 4 diodes model. I started doing the simulation in Proteus ISIS7 part by part to get the needed values for components to get a final DC output of 4.3V.

Currently, I'm doing a simulation of the first block of the converter, which is the rectifier. I have started simulating with a half-wave rectifier to get familiar of how the diode I'm using (1N 4007) may work, and from there I would use the same type of diode to build a 4 diode rectifier.

I connected a 12V peak-to-peak sinusoidal AC voltage source in series with a 1N 4007 diode, and a 1Kohm resistor. I used a built-in Oscilloscope to display the output waveform across the resistor. It has a peak value of 11.25 V, which means there's a 0.75 voltage drop across the diode, but I also found that there's -500mV (-0.5V) in the negative cycle, whereas in reality it should be 0V, because the peak inverse voltage of 1N 4007 is set to be 1KV according to ISIS7 software, so how come there's a leakage in current from the reverse side? is it something I have to worry about in the real life design? or is it just something related to the simulation software?

Thanks.

Edit1: I used Multisim to run the simulation, and I got a negative value for the output voltage when the diode is in reverse mode, but this time, it was in micro-volts. Does this really reflect the application in reality? also, I did a full-wave rectifier, and it was more precise in calculating the output voltage, with no negative value at all! still curious why there's a leakage in current for the half-wave rectifier.

Your leakage can come from a number of sources. Also remember that you're using AC voltage here.

A semiconductor junction has leakage from random movements of electrons and holes in the depletion region, which is a function of temperature. There is also leakage through dielectric layers (in mosfets) used where electrons can tunnel through. Also two different materials in contact can have an electron jump between them if the electron is at a high enough energy state, which can happen in thermal vibrations.

The other factor in your design is that you are using AC voltage across the diodes, and every component has parasitic capacitances that take charge even when the rest of the component is working as you'd expect. So when you reverse bias the diode, you are still charging its parasitic capacitance and will see a transient drop in voltage. Also, when you are picking larger diodes with higher voltage ratings, they tend to have worse capacitive side effects.

 

[DragonPetter]

And your multisim might not be taking into account all of the parasitic capacitances that exist in your real circuit, such as wiring and interconnects.

And yes, leakage and parasitics are real things you have to worry about in a lot of electronics designs. In RF design they actually treat parasitics as components to work to their advantage.

 

[CDTOE]

The phrase 'Parasitic capacitance' is new to me. I haven't dealt with its technicalities before. The question is, how to overcome this small leakage when the diode is in reverse bias mode? if I applied a half-wave rectifier in reality, and I found that there's a small current leakage resulting in a negative output voltage across the load, but only within micro-volts, Does this have an effect on the design? and again, how to overcome this issue?

 

[AlephZero]

If you attach images your actual circuts (including showing where you connected the measuring instruments in your simulation) you will probably get getter answers. There are plenty of people here who know how "standard" rectifier circuits work, but we don't know what circut you have come up with, unless you show us.

Assuming your circuit is operating at AC mains frequency (50 or 60 Hz), parasitic capacitances etc are very unlikely to be the cause, unless there is something else wrong with your design.

 

[Averagesupernova]

I suspect you have some AC coupling (through a capacitor) somewhere in this circuit. As previously stated, a schematic would help us a lot.

 

[CDTOE]

Here's an image of the half-wave rectifier in Multisim:

When I connect the Oscilloscope across the load resistance, the waveform has a peak negative component of around -710 μV. What's the cause then?

The following image is for my final AC--->DC converter design:

When I measure the output voltage at the load using a built-in multimeter, it displays an exact peak value of 4.3 V, which is the required output I was tasked to get by my lab instructor, but when I use the Oscilloscope to see the output wave form, it looks a purely straight line, only with a small variation in the displayed value extending from 4.268V to 4.317V. Is this normal?

 

[AlephZero]

When I connect the Oscilloscope across the load resistance, the waveform has a peak negative component of around -710 μV. What's the cause then?

In a "real" diode there is a small leakage current in the reverse direction. This data sheet http://www.fairchildsemi.com/ds/1N%2F1N4001.pdf gives the maximum reverse current for a 4007 at room temperature as 5μA. If you have -710 μV across a 1K resistor, that is s leakage current of 0.710μA which is quite reasonable at your low voltage, compared with the maximum value in the data sheet.

When I measure the output voltage at the load using a built-in multimeter, it displays an exact peak value of 4.3 V, which is the required output I was tasked to get by my lab instructor, but when I use the Oscilloscope to see the output wave form, it looks a purely straight line, only with a small variation in the displayed value extending from 4.268V to 4.317V. Is this normal?

The smoothing capacitor C1 is charging and discharging on each half cycle of the AC voltage. So the voltage across C1 will not be constant. A "real" zener diode is not a perfect voltage regulator, so a small variation in the voltage would be expected.

Put the oscilloscope across C1 to see how much the voltage changes there, compare with the change across the zener. You could also measure the changes in current through the zener and check the voltage and current changes against a data sheet for the BZX85.

 

[CDTOE]

Thanks for your clarifications.

On the same topic, I've decided to use my basic knowledge on AC-->DC converters to design a battery charger for my cell phone. The problem is, on the adapter that came with my cell phone, it says that the input range is AC 110-240V/180mA, and the DC output is 5.37V/800 mA. What's there in the circuit that increased the DC output rating to 800mA? I had an initial guess that it's a couple of transistors, or maybe an operational amplifier? also, is it correct to divide the voltage rating (5.37V) over the current rating (800mA) to get the load resistance of the battery?