problem with PWM signal

[maicael]

hello,
apart from isolation from AC mains,is it completely safe to step down a 120VAC or 220VAC by connecting it directly to a diode bridge without first stepping it down to an appropriate voltage level using a step down trafo.i dont want to use a step down trafo so can i connect the voltage directly to the bridge as long as the diodes can handle the high input AC voltage?

 

[MrAl]

Hello again,

Bridge circuits are used with higher voltages like 120vac, 220vac, 240vac, and higher. The key factor is the voltage rating of the diode, as it has to be high enough to handle the peak and should be even higher. For example for 120vac diodes rated for 200v have been used but diodes rated for 400v are better because of the way diode capacitance affects the circuit too. So for 220vac use 600v or 800v diodes and you should be just fine.

There is always a serious safety concern when using an off line power supply (one that is not isolated electrically). The concern is that if the human operator is grounded and they touch a node in the circuit they can get a serious electrical shock resulting in death. For this reason, extra attention to safety must be exercised when using an off line power supply in a product.

In particular, a circuit that uses an off line power supply should have no point in the circuit that can ever contact a human body even if a small problem occurs with the product like a volume control knob breaking. If the knob breaks and there is a metal shaft and the metal shaft contacts any part of the circuit the human may start turning it without replacing the knob and that could lead to dangerous shock. Thus it makes sense to use a pot with a plastic shaft so even if the pot is damaged it is unlikely to cause a shock hazard.
Likewise switches should have plastic handles, and anything else that can contact the human should be non metallic also.

Of course we can not protect against every kind of problem. If the user slams the unit into a brick wall and the fuse doesnt blow it could cause a safety issue, but that's an extreme. We just want to protect against normal usage with some attention to what happens if the unit is dropped. A warning on the outside of the unit that says not to plug the device into the wall if the case is damaged is also a good idea. If however something goes wrong with the circuit inside it should also not cause a fire either.

 

[maicael]

hello,
pls i want to understand something a bit clearer as it has to do with duty cycle and transformer.my questions goes like this
in normal square wave inverters i noticed that its flip flop output give out max 50% duty cycle as seen from a scope as the ON and OFF time are both equal so when you view the output AC square wave form you see that the spacing of between the positive and negative cycle of the output AC square wave form is equal amounting to 50% D.C
1)what is the maximum duty cycle a transformer can take into its windings especially in inverters like this?
is it possible to pass in 60% or even 70% duty cycle into the upper and lower windings of the trafo assuming that the flip flop oscillator can output this duty cycle?
2)if question 1 is possible then how does the trafo alternate this wave form to create postive and negative AC cycles at the trafo output since a 60% or 70% duty cycle will not have equal ON and OFF time so when the trafo alternates this how will its output be if it is possible at all as i seem to have this thought that the positive will overlap even while the negative has started. with 50% duty it is easy to imagine the output wave form.
Am i making any sense at all.

 

[MrAl]

Hello again,

The best bet is to use 50 percent and 50 percent because the input requirement is to maintain equal and opposite volt seconds, and that is easiest to get at least close to by using a 50 percent duty cycle where the positive half cycle is +Vcc for 50 percent of the time and the negative half cycle is -Vcc for the other 50 percent of the time. But for reduced output the duty cycle can drop lower, like say 40 percent, and then it would be +Vcc for 40 percent of the time followed by 10 percent dead time followed by 40 percent -Vcc followed by another 10 percent dead time. That's the usual method. For a DC output the two half cycles are simply full wave rectified and filtered and that gives close to 100 percent duty cycle overall. For an AC output the PWM can be made equal and opposite over the period of one AC cycle.

 

[audioguru]

A Transformer passes AC, not DC. A squarewave is AC. The positive pulse widths are the same as the negative pulse widths in PWM so it symmetrical and is also AC.

Here are my sketches of PWM. The top trace is DC PWM for a DC motor speed or for a light bulb brightness.The bottom trace can be passed through a transformer because it is symmetrical and also controls the speed of an AC motor or the brightness of a light bulb.

 

[maicael]

Hello again,

The best bet is to use 50 percent and 50 percent because the input requirement is to maintain equal and opposite volt seconds, and that is easiest to get at least close to by using a 50 percent duty cycle where the positive half cycle is +Vcc for 50 percent of the time and the negative half cycle is -Vcc for the other 50 percent of the time. But for reduced output the duty cycle can drop lower, like say 40 percent, and then it would be +Vcc for 40 percent of the time followed by 10 percent dead time followed by 40 percent -Vcc followed by another 10 percent dead time. That's the usual method. For a DC output the two half cycles are simply full wave rectified and filtered and that gives close to 100 percent duty cycle overall. For an AC output the PWM can be made equal and opposite over the period of one AC cycle.

thanks guys for the explanation.i guess it is safe to use a max of 50% but permit me to ask what could happen if the volts seconds are not equal and opposite like using say 60% instead of the best bet of 50%?
i am guessing the transformer will suffer for it.

 

[MrAl]

Hi,

The primary ends up with a net DC current which makes it more likely that the transformer core saturates and ruins everything.
Even with 50 percent positive and 50 percent negative it is not guaranteed that it will work perfectly, as some net DC may still be present due to a slight difference in the positive and negative half cycle times and/or voltage levels. It then depends on the primary series resistance.

For a quick example, say we have exact 50 percent positive and 50 percent negative time periods, but the voltage during the positive half cycle is 1 volt higher than the negative half cycle. The average voltage over the entire cycle is then 1/2 volt, which doesnt seem like much, but if the primary resistance is only 1/2 ohm, that means we have a net DC current of 0.5 volts divided by 0.5 Ohms which equals 1 amp of current in the primary. Depending on the construction this may be ok or it may cause saturation.

 

[maicael]

hello,
something bugs me which is the fact that while working i somehow mistakenly applied positive voltage to both the anode and cathode of a diode 1N4148 at thesame time but i was quick to stop when i realized my mistake but somehow the diode survived and i was wondering why since normally we apply positive voltage to the anode while the cathode is left alone so current flows as usual.
in my thinking the diode should be destroyed or something like that.
what could happen when this situation occurs?

 

[audioguru]

A diode normally conducts current when the anode voltage positive and is higher than about 0.7V more than its cathode. Then the anode is +12V and the cathode is +11.3V maybe like you had or the anode is +0.7V and the cathode is 0V or the anode is +0.4V and the cathode is -0.3V or the anode is -10V and the cathode is -10.7V.

The datasheet for the 1N4148 diode says its maximum allowed continuous DC current is 200mA but 4A is allowed for a short-duration (one micro-second) before it melts.

 

[MrAl]

hello,
something bugs me which is the fact that while working i somehow mistakenly applied positive voltage to both the anode and cathode of a diode 1N4148 at thesame time but i was quick to stop when i realized my mistake but somehow the diode survived and i was wondering why since normally we apply positive voltage to the anode while the cathode is left alone so current flows as usual.
in my thinking the diode should be destroyed or something like that.
what could happen when this situation occurs?

Hi,

audioguru explained the diode action pretty well, i'll just add a little more of my own personal thoughts.

After reading your post i get the feeling that you might believe that the diode should only be biased one way and only one way. But actually the diode can be biased two different ways and that's when we see the main action of the diode.

If we make the diode anode more positive than the cathode (forward bias), we see the diode conduct a significant current. If we make the diode cathode more positive than the anode (reverse bias) then we see the diode conduct only a very very small current as long as we dont exceed the reverse voltage rating of the diode.

Because we have different reactions of the diode depending on the polarity at the two terminals, we see the basic diode action which means it conducts more in one direction than the other, and that is what gives us rectification. This is the most important quality of the diode, that it can rectify a bipolar signal into a unipolar signal and that's how we get raw DC from an AC source.

So the diode can be biased either way, but how much current that flows is dependent on which direction we bias it. With an AC signal it gets biased both ways in turn and only when it is forward biased do we see significant current flow.

Diodes like the 1N4148 have fairly low resistance when correctly forward biased but in reverse the resistance can be as high as 500 Megohms. Note you can measure the reverse resistance at low reverse voltage with a multimeter on the Ohms scale (if it goes up that high), but you may not be able to measure the forward resistance because that requires a certain minimum voltage level and varies quite a bit with the actual voltage. If you happen to have a 'diode check' function on your meter that may help measure the voltage drop at a particular test current that is put out by the meter.

If you've never worked with diodes before this might sound strange, but after doing just a few simple experiments it will be come very clear how amazing the simple diode is.

A simple experiment is to connect a 9v battery in series with a 1k resistor in series with the diode. Measure the current with the diode forward biased, then measure the current with the diode reverse biased. Note the huge difference in current levels.
[Note the current can be measured with a high input impedance meter by measuring the voltage across the 1k resistor. The conversion is 1v equals 1ma of current.]

 

[maicael]

hello,
hope this question does not sound dumb but i was wondering if when using a driver circuit like a totem pole config. to drive a mosfet if there is any need for a gate to source resistor since a totem pole can both source and sink current?
if at all there will be a gate to source resistor it should not be too high i think?

 

[maicael]

one more thing, if i decided to use a PNP transistor as a diode by shorting the base and collector and using the two terminals as a PN junction,would there be any advantage over a real diode like the 1N4148 such as lesser voltage drop.
the 1N4148 has a 0.7V drop if i am correct but would this be lower by trying the above method?

 

[MrAl]

Hi,

If you use a 'totem pole' driver then you dont need a gate to source resistor, but sometimes you need a small transistor output to gate series resistor. This might help if there is an oscillation especially when paralleling MOSFETs.

To get a lower voltage drop for the diode use a Schottky diode instead of a regular diode.

 

[audioguru]

when using a driver circuit like a totem pole config. to drive a mosfet if there is any need for a gate to source resistor since a totem pole can both source and sink current?
if at all there will be a gate to source resistor it should not be too high i think?

The resistor value can be high. It prevents the Mosfet from turning on when there is no input signal.

There is also usually a low value resistor in series with the gate mounted close to the gate pin to prevent high frequency ocillation or ringing.

if i decided to use a PNP transistor as a diode by shorting the base and collector and using the two terminals as a PN junction, would there be any advantage over a real diode like the 1N4148 such as lesser voltage drop?
The 1N4148 has a 0.7V drop if i am correct but would this be lower by trying the above method?

Please read the datasheet of a transistor. Its maximum allowed reverse emitter-base voltage is only 5V to 7V so it cannot be used as an ordinary diode. Also its current rating is very low.
At low currents their forward voltage drops are about the same.

The reverse-biased emitter-base of a 2N3906 PNP transistor breaks down at 5V, the 1N4148 diode at 75V.
The maximum allowed base-emitter current for the 2N3906 is maybe only 25mA but is 300mA for the 1N4148 diode.

 

[maicael]

hello,
i also forgot to ask about the concept of speed up capacitors in parallel to the input resistor of a totem driver or even at the gate input of a mosfet.
any advantage to this as to just using the input resistor only?
if used at all any maximum value of capacitor?
Also i know that mosfets have inbuilt diodes but i was thinking of bypassing this by putting one from source to drain i think and i was wondering what type would do like the regular rectifier diodes?

 

[audioguru]

You can use a speedup capacitor parallel to the series input resistor of the totem pole transistors.
You do not want one parallel to the low value resistor in series with the gate of a Mosfet because then it will probably oscillate

The diode in a Mosfet is part of its structure and is useless. Its load needs a diode parallel to it. A mains rectifier is much too slow to switch when PWM is used.

 

[maicael]

The diode in a Mosfet is part of its structure and is useless. Its load needs a diode parallel to it. A mains rectifier is much too slow to switch when PWM is used

so i have to find ultrafast rectifier diodes then.

 

[maicael]

hello,
i was with a friend of mine who built a sine wave inverter as a test project after reading the concept using a micro-controller programming and push pull config. the switching frequency is 25khz to generate SPWM signals driving the mosfets in a push pull manner we used a totem pole before the mosfet power stage.
anyway everything works fine but there seems a little issue which is when we put a capacitor at the output we can see the waveform but the mosfets are getting unsually warm and when we remove the cap the fets then stop getting warm.
i am wondering is it because the FETs are not getting enough protection as we have not put anything like snubbers or protection diodes from source to drain.
Basically if what i say is correct then in what ways can we protect the mosfets power stage?
even if i am wrong how does one go about full mosfet protection in a push pull configuartion using snubbers and other forms of protection?

 

[audioguru]

When a capacitor is connected across the output then it draws a lot of power charging and discharging which heats the Mosfets.
Maybe the capacitor across the output causes the Mosfets to oscillate at a high frequency which causes them to heat up.

You NEVER connect a diode across an output transistor or Mosfet. Instead, you connect a diode (if it is needed) across an inductive LOAD.
A snubber prevents an output transistor or Mosfet from oscillating at a high frequency.

 

[MrAl]

Hello,

You can not directly load a MOSFET output with a cap where the MOSFET is switching. That is because you will force the cap to charge and discharge too fast.

It's ok if it is a DC output or a sine wave output (sometimes) but not with a pure rectangular output.