PWM to AC Signal. The signal never across the secondary side. Pls help!

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Hi all,

I have tried many times to create AC signal with PWM and I have used transformer to increase the signal at 120 Vpp but I never didn't.

I can't send the signal across the secondary side to see 120 Vpp. (Turn ratio 1:30)

Could you help me please?
 

1. BDX53 is darlington so you may not be giving it enough voltage to drive it into conduction.
2. BDX53 has a breakdown voltage of 45V which almost certainly isn't enough.
3. What frequency are you using? The PWM frequency must be suitable for the transformer you are using.

Brian.
 

hi Brain,
Thank you for prompt reply,

1. How can I know giving enough voltage to drive it. In its datasheet which parameter is driving voltage (Vceo, Vce ??)
2.Means that I have to give the voltage under the 45V?
3. The frequency is 250Hz. (The PWM output frequency is 250Hz)

Thank you,
Mehmet
 

hi bowman1710,

No, I haven't. I think my darlington transistor could not drive. Or I have not giving enough voltage to the mosfet.
I just want to know how can I use its datasheet for driver, breakdown, e.t

Thank you,
Mehmet
 

Hi Mehmet.

Firstly, the BDX53 is not a MOSFET, it is a bipolar Darlington transistor.
The voltage you need to fully turn it on is Vbe(sat) which the data sheet says is 2.5V at 12mA, assuming 3A collector current.
The transistor has a pair of resistors across the B-E pins which draw some of that current. So if you are producing PWM at 5V amplitude (5V supply to the PIC) the resistor in series with the base can be no bigger than (5V - 2.5V)/0.012 = 208 Ohms. If you are using a 3.3V supply it has to be no bigger than 91 Ohms. Your 10,000 Ohms is way too high! Be aware that dropping the resistor will present a risk to the PIC if the transistor C-B junction breaks down, it would be wise to add a Zener diode across the PIC output to help protect it.

The maximum voltage on the collector depends on a number of things, mostly to do with the transformer. It will certainly be higher than the supply voltage by a factor of 2 or more.

Brian.
 

Thank you so much Mr Brian. It helps me for solving too much unknown.

And I just want last things.
Could you please draw a simple circuit with resistor for this?
If you can do I would be happy.

Thank you so much in advance
Mehmet.
 
Last edited by a moderator:

We think your old BDX53 darlington transistor might be destroyed because its voltage rating of only 45V is too low for your supply of 24V where its collector can easily exceed 48V.
Its datasheet also shows that for it to saturate with a collector current of 3A then its base current must be at least 12mA. If the supply voltage of your PIC is 5V then your 10k series base resistor provides a minimum base current of only half of 5V - 2.5V/10k ohms= 0.25mA/2= 0.125mA. The maximum base-emitter voltage is listed as 2.5V.
 

The circuit is the same, just change R1 to less than 208 Ohms (use 180 Ohms standard value) if you use 5V supply to the PIC or less than 91 Ohms (use 82 Ohm standard value) if you use a 3.3V supply to the PIC.

For safety, add a Zener diode, 4.7V from the output pin of the PIC to ground with the cathode (+ or banded end) toward the PIC pin. You can also protect the transistor by connecting a 39V Zener diode across it's collector and emitter with the cathode end toward the collector but it will reduce the circuit efficiency and possibly get quite hot.

Alternatively, you could change the BDX53 to a power MOSFET but if you do, make sure you use one that accepts logic level drive or has a knee voltage well below the PIC supply voltage.

Brian.
 

Hi Mr Brain,

Thank you so much. I really appreciate it.
I have added zener diode as your reference. I hope it is true. (I have also added circuit diagram)

I really want to learn power electronic like you


Thank you again,
Mehmet
 

What is confusing is that you have shown a 130V SINEwave as the output. If you switch the current through the transformer with a square wave, you will get a square wave output with dramatic overshoots at the output.
Betwixt's method of shorting the overshoot energy ( 1/2 L I ^2) to earth via the zener, which causes it to run hot. I have seen this circuit used commercially and all five transistor driver transistors burn't out. I believe this is due to the transistors punching through quicker then the zener conducts.
A better way would be to simply put a diode across the primary. The overshoot energy would be returned to the Vcc line and its reservoir capacitor.
To get a sine wave at the output, you either have to resonate the transformer or follow it with a filter.
Frank
 

The energy has to go somewhere which is why I mentioned it reduced efficiency and get hot. The BDX53 already has a diode across it's C-E pins so reverse polarity isn't an issue but the 45V limit is dangerously close if not less than the peak primary voltage, hence clipping peaks with the Zener. The best solution is probably a diode in conjunction with an RC network across the primary so the excess energy can be dissipated safely in a resistor but that may not solve the transistor Vce/Vcb voltage being exceeded.

Brian.
 

I see that but its a inductance its driving, like a relay coil, Vcc or 0V, if it was a "proper" transformer circuit, like a class A output stage the transistor would have a quiescent current, which would double/go to zero with the Vce going from 0 to 2 X Vcc.
Frank
 

yes, I have burned lots of BDX53C when I used. I have also used RC filter which was low pass filter, (fc=80Hz, R: 2k, C:1uF) but the problem was that when I applied any load on the circuit, the voltage amplitude was so reduce although our load was just 1.5W

And our circuit must draw just 70mA from the supply to escape damage the piezo. Mean that I don't need to much power.

Thank you Frank, It always helps me a piece of information.

Mehmet
 

Your schematic shows a 45V BDX53 darlington so we wasted our time telling you that your supply of 24V is too high for it. Now you say it is actually a BDX53C that has a 100V rating.
Now you also say you used a low frequency filter but did not say where it is connected.

Maybe the filter is causing the darlington output to ramp instead of switch so of course it gets hot.

Is the load a piezo transducer? It will not do anything at the low frequency of 80Hz.
 

Thank you Audioguru,

I have added my last circuit with RC low pas filter in the following photo below,
I have also confused which way is better to use filter on secondary side or primary side.

Thank you,
Mehmet
 

I think things are progressing backwards fast!

Adding a 2K resistor in series with the primary limits the current to about 12mA so the transistor base and collector current will be essentially the same. In other words, it does nothing. Excluding the effects of mutual inductance, the capacitor makes the primary of the transformer resonate at around 159Hz but the PWM is stated to be at 250Hz. Finally, given the highly resonant nature of piezo elements, I doubt any frequency other than it's resonance will work well anyway. Intuitively a 30H inductor is too large to feed it.

Mehmet, to avoid speculation, can you explain exactly what the circuits purpose is? What is the final application?

Brian.
 

It is a class-D audio amplifier with a class-A output transistor driving a transformer.
I have seen many class-D audio amplifiers but I have never seen one with a class-A output transistor and never seen one with an output transformer.

He wants to feed the piezo transducer loud squeak sounds at 130V RMS.
Instead of using a normal push-pull output he uses a single old darlington transistor.
Instead of using a bridged amplifier, he is using an output transformer.
Instead of filtering the output he is severely filtering the input.
 

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