Continue to Site

Welcome to our site!

Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

  • Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

Design Help --- Bubba Oscillator

Status
Not open for further replies.
The max base-emitter voltage of the driver transistor is 0.65V without a load on the project and 0.8V with a 100W load.

the max base-emitter voltage of the output transistor is 0.7V without a load and 1V with a 100w load.

The two voltages are in series so the outputs of the LM358 are limited to 1.35V to 1.8V.
If the transistors are good and are connected properly then 13V at the output of the opamps is impossible.
You must have the wrong transistors or have their pins connected wrong.
 
for the transistors i am using the exact same transistors an in the schematic

2n3055
2sc1061 i also have tip31
lm358n
tell me what you thiink about these chips and transistors.
 
In the year 2009 are you still using darlington connected BJTs for switching high power? The Vcesat of up to 2 volts is a terrible waste of precious 12V energy. The 2N3055 runs out of useable β at a collector current of only 10A.

You can replace each darlington (e.g. Q1, Q11, R4, D1) with a high current MOSFET with no other design changes. If you cannot afford a low Rdson MOSFET then put as many as necessary in parallel to give a value of 0.01 ohms or less. You can use 6 per side BUZ0105 for us$0.55 each at N-CHANNEL MOSFET, 55V 22A .06OHM | AllElectronics.com.) Or a single POWER MOSFET MTP75N03HDL | AllElectronics.com for each side for us$1.75.

At 30A, 50% duty cycle, the power wasted in 0.01 ohms is 4.5 watts each so you only need a small heat sink.
 
The circuit using old transistors connected as darlingtons was designed in a 3rd-world country for use in the 3rd-world country where power Mosfets are not available.

The poor people use inverters because they do not have electricity. A guy on a motorcycle picks-up discharged batteries and replaces them with charged batteries.
 
So all you guys agreee that if i change the design from darlington to sue of a single mosfet i will get the expected result. without having to change the controller circuit and the LM358 opamp???? and would the the output of the opamp just go directly to the mosfet with its output going directly to the transformer. i kam not really concerned about cost at this point. i was just concerned with a working model; modifications could be done afterward as it pertains to power consumption and things of that sort.
 
You may still have connection problems if you get no output; the 2N3055 should work well enough to show the concept. You can finish making sure the connections are right using the 2N3055.

Even if MOSFETs are cheap it still would be a shame to blow them up. You can make the switch at any time.
 
A single Mosfet will not make a working high power inverter.
You need two Mosfets to drive the center-tapped transformer.

You are modifying an extremely simple inverter circuit that does not turn off if it is overloaded (instead it blow up) and does not have voltage regulation so its output voltage is too high if the load is light or if the battery is recently charged and the output voltage is too low if the load current is high or if the battery is running down.

The inverter has an extremely simple square-wave output instead of a modified one or a sine-wave one. It can drive an incandescent light bulb or a heater.
 
Hi guys,

sorry iwas actually referring to adding one mosfet on either leg. that is what i was trying to get. it makes it really simple if that is the case.
 
guys i was able to make a little progress. i couldnt get a battery with a rating of 10 amp so i pulled out a power supply from a computer with a current rating of 12V and 8.5 amps which is a little less than required. the output on the transformer now is a pulse at the 100 volt mark there about. is it because the currrent is not sufficient. or what do you guys think the problem is?
 

Attachments

  • 1109091209a.jpg
    1109091209a.jpg
    368.2 KB · Views: 380
guys i was able to make a little progress. i couldnt get a battery with a rating of 10 amp so i pulled out a power supply from a computer with a current rating of 12V and 8.5 amps which is a little less than required. the output on the transformer now is a pulse at the 100 volt mark there about. is it because the currrent is not sufficient. or what do you guys think the problem is?
1) The circuit is cheap, simple and old-fashioned.
2) A 12V lead-acid battery is 12.0V when it is dead. It is 12.6V to 13.8V when fully charged.
3) Maybe the transformer is cheap with poor voltage regulation so they made it 14V to 120V without a load.
4) The 2N3055 output transistors have a max saturation voltage loss of 3.0V.

Use a 10V-0-10V transformer instead.
 
1)
4) The 2N3055 output transistors have a max saturation voltage loss of 3.0V.
If you are getting the proper waveform, then converting to properly rated MOSFETS could give you about 20% increased voltage (assuming your source current is adequate and your transformer can handle it).
 
well my guess is the transformer i am using is right. i got a power supply that will deliver the expected current. actually it deliver 13amps. i will try it tmor and see what it gives. hopefully it will work if that doesn't i might just have to switch the transformer like audio guru suggested.
 
gotta a question about the inverter circuit. what is the purpose of the 4.7k and the 1k resistors in the circuit. the 4.7 is coming out of the astable vibrator outputs and the 1k is in parallel with the diodes respectively.
 
gotta a question about the inverter circuit. What is the purpose of the 4.7k and the 1k resistors in the circuit. The 4.7k is coming out of the astable vibrator outputs and the 1k is in parallel with the diodes respectively.
The original circuit used the useless 4.7k resistors and they were left in.
The 1k resistors turn off the 2N3055 transistors.
The diodes conduct if a negative voltage spike occurs on the transformer which forward-biases the collector-base diode of a 2N3055 transistor in series with a diode which becomes forward-biased.
 
I have a few questions about the darlington setup in the inverter circuit as well as the op amp added. what exactly does it contribute to the overall inverter circuit. in essence what i am asking is, why couldnt i have sent the output of the vibrator to the transformer immediately. what exactly is it going to achieve....
 
Look on the data sheets and examine how much current each device can provide. If you want 30A in the transformer it cannot be supplied by a device which only gives 1mA.
 
well i guess i stated my question incorrectly. but i was just trying to find out how exactly the current is boosted from the op-amp to the transformer. like what exactly is going on at the op amp stage and also at the darlington stages.


also, i have received the mosfets. how do i connect the two mosfets into the circuits do i connect the gate where the base was, source where emmiter was and drain where collecter was or is there a different configuration. also how do i introduce regulation in such a circuit and also use pre cautionary protection measures...
 
Status
Not open for further replies.

Latest threads

New Articles From Microcontroller Tips

Back
Top