[SOLVED] Tip41c/42c arrangement problem

[Audioguru]

Is the NE555 a fake from ebay?
Did you measure the resistance of the 100 ohms resistor? If it is shorted (ebay?) then the 555 gets hot when the circuit has a load.
If the transistors are fake (ebay?) and are simple base-emitter diodes instead then again the 555 gets hot when the circuit is loaded.

 

[Easy peasy]

yes, either partially dead 555 or the xtors are very low gain & nearly dead - or the wiring is wrong somewhere - again - you need higher gain xtors if this circuit is ever to supply a load ...

 

[Audioguru]

yes, either partially dead 555 or the xtors are very low gain & nearly dead - or the wiring is wrong somewhere - again - you need higher gain xtors if this circuit is ever to supply a load ...

That old circuit is begging for Mosfets instead of ordinary low gain transistors.

 

[Easy peasy]

try NJW0281 and its complement NJW0302 gain approaching 150 for good ones ...

 

[Audioguru]

The gain of output transistors is needed in a linear amplifier. But this circuit is not a linear amplifier, its transistors are saturated on-off switches.
The (Japanese?) transistors recommended have a 250V rating that is not needed in this 12V circuit. Their saturation specs are only slightly better than the American TIP41 and TIP42 transistors used in this wrongly designed project.

 

[Easy peasy]

for some one called Audioguru - I am surprised at your statement - if you look closely you will see the xtors are not in saturation - The Vce is not below the Vbe - for saturation to be possible you would need an npn in the lower position and pnp in the top ( suitably driven )

this is why high gain xtors would improve matters - it matters not what the Vce max is - but the gain at operational currents.

 

[Audioguru]

OK, I agree that this squarewave inverter uses emitter-followers and not saturated switches. The original TIP41 and TIP42 transistors have a minimum current gain of 12 at 5A but the transistors you recommend have a minimum current gain of 75. Then the voltage loss of the 100 ohms resistor feeding them will be less.
With a 12V supply and an output of 5A into the transformer, the output current of the 555 will be 5A/75= 67mA and will produce +0.4V to +10.4V. The 100 ohms resistor voltage loss will be 67mA x 100 ohms= 6.7V. Then the base of the transistors swing from +3.7V to +7.1V. The emitters swing from about +4.7V to +6.1V. Then the transformer must have a 1.4V primary winding which is crazy. That is using your high gain transistors.
If you can find transistors having a gain of 150 then the voltage loss from the 100 ohms resistor is 3.33V. The 100 ohms resistor will not be needed, resulting in only a 2V voltage loss of the 555 so then a 10V transformer can be used.

 

[Easy peasy]

Overall the circuit is a poor one, even a bipolar 555 would have trouble providing the base currents needed ( compared to a cmos one - 10mA if you're lucky )

another 555 is needed as an inverter, then both with buffers to drive both sides of the Tx ( with a DC blocking cap ) then the full 12V could be applied to the LV primary rather than the 6V the present ckt allows, via the xtor's I mentioned, 5 x 12V = 60W - so yes 100mA of drive from the buffers would be required.

A push pull is starting to look good here, as either transistors ( in saturation) or fets could be used ( straight from a cmos 555 ) to power the Tx more effectively with the ground referenced drive ...

 

[betwixt]

Even with FETs a bipolar NE555 would struggle to produce enough gate current, I doubt a CMOS one would manage at all.
The push-pull concept is why I queried whether a center tapped transformer was available in an earlier post. The simplest "fairly efficient" way to do it is to use the output of the 555 to drive two current amplifiers, one inverting the other not, then use their outputs to drive a center tapped transformer in push-pull configuration. It's the easiest way to ensure saturation without adding more than a few components.

Brian.

 

[Easy peasy]

a cmos 555 will drive mosfets pretty well - esp at 50/60 Hz, so I imagine a bipolar one will too ... they just don't last due to the inherent shoot thru current spike in the o/p stage ( bipolar ones ). Driving two low side mosfets in push-pull also allows for an Ron and pnp turn off xtor - giving some dead time - which is usually very useful ...

 

[Audioguru]

The old inverter circuit in this thread was designed long before Cmos 555 ICs were even dreamed about.

My electricity is very reliable so I have never bought or made an inverter. On the few rare occasions when the power fails we all cheer and have a party in the dark.
Most of my electrical products rely on the higher peak voltage of a sinewave and do not work properly on the lower voltage of a squarewave like in this old project. The lower peak voltage of the squarewave was designed to light old incandescent light bulbs or heaters. I don't think squarewave inverters are sold anymore, most inverters use a stepped "modified sinewave" that is actually a modified squarewave. This circuit is the only inverter circuit I have seen that tries to smooth its squarewave into a sinewave with the LC parts. If the LC is tuned properly then it might boost the peak voltage a little.

 

[Easy peasy]

In the OP's ckt there is no attempt to modify the square wave into a sine wave, the DC blocking cap biases itself to 6V under normal operation - and the Tx sees +/- 6V ( ideally ) from a 12V source.

a series resonant approach would not work here - as the load too variable

if all you want is a square wave to drive some other equipment then this ckt might suffice, but it is a poor choice in the main.

 

[Javid.zare.s]

Is the NE555 a fake from ebay?
Did you measure the resistance of the 100 ohms resistor? If it is shorted (ebay?) then the 555 gets hot when the circuit has a load.
If the transistors are fake (ebay?) and are simple base-emitter diodes instead then again the 555 gets hot when the circuit is loaded.

-it was the cheapest so definitely fake, but operating fine just like original.
-yes I did measure 100 ohm res and is 100 ohm
-do they really do such a thing? I mean replacing transistors with diodes?!! is there a way to check that???

- - - Updated - - -

yes, either partially dead 555 or the xtors are very low gain & nearly dead - or the wiring is wrong somewhere - again - you need higher gain xtors if this circuit is ever to supply a load ...

I can say I checked the wiring for many times, and I will attempt to try with new parts for next 2 or 3 days, new timer ic and new transistors

- - - Updated - - -

That old circuit is begging for Mosfets instead of ordinary low gain transistors.

do you think mosfets can it resolve my problem ? I've never worked with any of them are there any common mosfets that you can name, so I can order online?
will I need to rewire the circuit for mosfets?

- - - Updated - - -

try NJW0281 and its complement NJW0302 gain approaching 150 for good ones ...

I've searched and this part (which I don't know what it is), is not available on the website that I buy my parts

- - - Updated - - -

The gain of output transistors is needed in a linear amplifier. But this circuit is not a linear amplifier, its transistors are saturated on-off switches.
The (Japanese?) transistors recommended have a 250V rating that is not needed in this 12V circuit. Their saturation specs are only slightly better than the American TIP41 and TIP42 transistors used in this wrongly designed project.

thanks, so I should not expect this circuit with Tip41c/42c to be operating at all.
this project was a big failure for me, I should have done more research before starting it

 

[Easy peasy]

try ali-express for NJW0281 & 0302 ... 1000's there ...

but yes the schematic has never been built and tested - otherwise the originator would know how bad it is ...

email me for a better one EP@pwrtrnx.com

 

[Audioguru]

ebay is Chinese and sells cheap fake or defective Chinese electronic parts.
AliExpress is also Chinese and sells cheap Chinese parts. Fakes? Made by Chinese companies I never heard of.

ON Semi is American and sell the high voltage, high current and high gain transistors. I trust ON Semi.

The Original Circuit will work but with a reduced output voltage and more voltage drop when loaded if it uses a normal American NE555 or LM555 and normal American TIP41 and TIP42 transistors. The 555 will never get hot because the 100 ohms resistor allows its maximum output current to be 1/4 of its maximum allowed current.

 

[Easy peasy]

the original ckt will be very power limited, at 15VDC in you will be lucky to get +/- 4V across the Tx pri at no load - the volt drop across the 100E resistor limits emitter o/p voltage quite severely - as the ckt is an emitter follower, i.e. the emitter follows the base voltage less 0.6 volts

the push pull is the better way to go here - tons of ckts on line ...

As to Ali-express - yes some vendors need to be avoided on this platform, but for the 4-star plus vendors, we have bought many 1000's of dollars of parts with no issues - they are always tested by our inwards goods though ...

 

[Javid.zare.s]

the original ckt will be very power limited, at 15VDC in you will be lucky to get +/- 4V across the Tx pri at no load - the volt drop across the 100E resistor limits emitter o/p voltage quite severely - as the ckt is an emitter follower, i.e. the emitter follows the base voltage less 0.6 volts

the push pull is the better way to go here - tons of ckts on line ...

As to Ali-express - yes some vendors need to be avoided on this platform, but for the 4-star plus vendors, we have bought many 1000's of dollars of parts with no issues - they are always tested by our inwards goods though ...

no I don't buy stuff from ebay ali.... it is not cause I don't want to ,actually they are unavailable here in Iran, all for government and economical sanctions...

 

[Audioguru]

the original ckt will be very power limited, at 15VDC in you will be lucky to get +/- 4V across the Tx pri at no load - the volt drop across the 100E resistor limits emitter o/p voltage quite severely - as the ckt is an emitter follower, i.e. the emitter follows the base voltage less 0.6 volts

It could be worse if the TIP41 and TIP42 transistors have their minimum hFE and the supply is a normal 13.2V battery:
1) If the output power wanted is only 13.2W then the transistors conduct 1A. Their minimum hFE at 1A is 20 so their base current will be 50mA. The maximum voltage loss of the 555 at 50mA is a total of about 2.75V. The voltage loss of the 100 ohms resistor at 50mA is 5V. The maximum base-emitter voltage of the TIP41 and TIP42 transistors at 50mA of base current is a total of 2.2V. So the total of the voltage losses is 2.75V + 5V + 2.2V= 9.95V!
The primary of the transistor must be 13.2V - 9.95V= 3.25V and the output voltage will rise like crazy when the load current is less.

 

[Easy peasy]

Correct - the effect of the 100 ohm base current limiting resistor is to make the emitter follower ckt impractical - a properly driven transformer with a 71% PWM duty cycle on a 280Vpk output gives 230Vac rms with the least harmonics - otherwise known as modified sine wave, even better if the Tx is shorted during the 29% dead time ...