Q3 and Q4 are power transistors rated for high voltage and high current. In your circuit they get 10V at almost no current then are not needed since the output of the LM393 can drive the Mosfet if R3, R9, R11 and the 7805 and 7810 are removed. The regulator for the LM393 should be a 7812 since the IRF710 Mosfet needs a Vgs of 10V to fully turn on. The current determining resistors need to be re-calculated.

 

I'm surprised the mods are letting this one through... it's a transformerless power supply or have the rules changed while I've been away.

 

ectification then transfers you into the DC realm, whereby you apply filtering (capacitors) to smooth the pulse to a DC voltage. You can then control your current level with a very simple current regulator using 2 PNP BJTs and 2 resistor

Q3 and Q4 are power transistors rated for high voltage and high current. In your circuit they get 10V at almost no current then are not needed since the output of the LM393 can drive the Mosfet if R3, R9, R11 and the 7805 and 7810 are removed. The regulator for the LM393 should be a 7812 since the IRF710 Mosfet needs a Vgs of 10V to fully turn on. The current determining resistors need to be re-calculated.

thank you for a helpful answer. I will work on changes.

 

I'm surprised the mods are letting this one through... it's a transformerless power supply or have the rules changed while I've been away.

I don't know whom on here is really an electrical engineer or not. But every LED driver in your house, open them up and look, are working without a transformer. If my question is "blocked" then people are not living in today's reality ? What rules? Are we not allowed to ask questions?

 

Hello,

Are you going to use isolation from the mains?
This is a must on this site, otherwise I will close the thread.

From the user agreement:


Bertus

 

can we put idiotic designs on the list?

 

Hello,

Are you going to use isolation from the mains?
This is a must on this site, otherwise I will close the thread.

From the user agreement:
View attachment 223199

Bertus

Hello,

Are you going to use isolation from the mains?
This is a must on this site, otherwise I will close the thread.

From the user agreement:
View attachment 223199

Bertus

I'm using an isolation transformer with a 300mA fuse. Without this isolation my oscilloscope would ground the circuit to the house's ground. So yes the circuit is isolated.

 

@farhadsaberi

Stop (and step back for a moment). You don't know enough. No disrespect. You've done a lot, but don't be in love with your design. Instead, to achieve what you need, you should be looking at simpler, safer circuits to achieve it with AND understand it.

AC power is handled in a couple of ways- using field devices, as opposed to thermal. In this case, capacitors instead of resistors to get that current into a _safe realm_. Resistors can then drop the voltage to a safe realm prior to rectification.

Rectification then transfers you into the DC realm, whereby you apply filtering (capacitors) to smooth the pulse to a DC voltage. You can then control your current level with a very simple current regulator using 2 PNP BJTs and 2 resistors:

Constant Current:

View attachment 223197

Disregard resistor values, you need to calculate those to control output current.

Learn about zener diodes, how to use them in conjunction with BJTs etc as well. There are many ways to control current and voltage- just remember they are never independent of one another (unless manipulated to be so).

Get this book, it will get you up to speed, including BJTs faster than almost anything else:

Title: Understanding Basic Electronics, 1st Ed.
Publisher: The American Radio Relay League
ISBN: 0-87259-398-3

Hello. I am very interested in hearing more about how to lower voltage using capacitors. The only way I know, and to pass 300mA at 60Hz mains outlet, I would need a non-polarized 10uF cap that will be the size of my fist. If you know of specific articles on this subject please point me there.

The constant current idea snippet you attached is an idealized one. In reality It will vary too much with temperature and cannot be used with current sensitive devices such as simple LED's. Those BJT's will also get very hot and require huge heat sinks. My solution keeps a constant current at any temperature and without using any heatsinks. The circuit I have so far I believe is on the right track. Unless someone can tell me specifically why it is so bad and provide a solution as well, not just criticisms without any solution ideas.

In fact it is not my solution, but pieces of information i've gathered and put together while studying. Thank you.

 

can we put idiotic designs on the list?

Hi. Please provide your solution. Thanks.

 

Hi. Please provide your solution. Thanks.

Solution to what? Still missing things like what does the circuit do, what are the technical requirements. what is "L", etc.

ak

 

Hello. I am very interested in hearing more about how to lower voltage using capacitors. The only way I know, and to pass 300mA at 60Hz mains outlet, I would need a non-polarized 10uF cap that will be the size of my fist. If you know of specific articles on this subject please point me there.

The constant current idea snippet you attached is an idealized one. In reality It will vary too much with temperature and cannot be used with current sensitive devices such as simple LED's. Those BJT's will also get very hot and require huge heat sinks. My solution keeps a constant current at any temperature and without using any heatsinks. The circuit I have so far I believe is on the right track. Unless someone can tell me specifically why it is so bad and provide a solution as well, not just criticisms without any solution ideas.

In fact it is not my solution, but pieces of information i've gathered and put together while studying. Thank you.

Yes, the constant current circuit is simple, and being in linear region it can get warm. It works fine with LEDs, which don't use a lot of current, unless your talking high-energy (multi-watt) LED setups. How warm it gets is a function of the current you require. It was just one concept to show that there are simpler ways.

As a rule of thumb, use simple discreet circuits to control signal/current parameters, and then scale output with other devices (like OpAmps, transistors, multipliers, etc.

Capacitors are your most powerful to way to control current in the A/C environment. Look in the book I pointed out to you to learn about capacitive reactance. Here is another link:

https://www.electronics-tutorials.ws/capacitor/cap_8.html

You can rework the equations to determine the capacitor size. Engineering is always a trade-off of constraints. Figure out which constraints you can't change, and which ones you can.

 

Yes, the constant current circuit is simple, and being in linear region it can get warm. It works fine with LEDs, which don't use a lot of current, unless your talking high-energy (multi-watt) LED setups. How warm it gets is a function of the current you require. It was just one concept to show that there are simpler ways.

As a rule of thumb, use simple discreet circuits to control signal/current parameters, and then scale output with other devices (like OpAmps, transistors, multipliers, etc.

Capacitors are your most powerful to way to control current in the A/C environment. Look in the book I pointed out to you to learn about capacitive reactance. Here is another link:

https://www.electronics-tutorials.ws/capacitor/cap_8.html

You can rework the equations to determine the capacitor size. Engineering is always a trade-off of constraints. Figure out which constraints you can't change, and which ones you can.

When I started this project six months ago capacitive resistance with AC was the first method I looked at. It is definitely not the way to do this. Say I want 200mA for a large LED bulb at 40 V (this is what I actually have).I would need a *huge* capacitor to begin the circuit with. And then still use a comparator with hysteresis to achieve any constant current I want that will not be affected with temperature changes.

Turning the knob of a BJT half way will also make it way too hot and never provide an exact current. Vhfe (Beta) value varies too much. The only way must be using a mosfet with a clean square wave at its gate with feedback to a comparator, so it achieves constant current and also doesn't get hot. Unless somebody can show me otherwise.

I discovered something last night about why Q3 won't turn on. Q3 not turning on is the reason I came here for help. Once I finalize this I will share it. Thanks.

 

@farhadsaberi
I found an appropriate automotive capacitor on mouser that is only 2.25" long (largest dimension), hardly 'fist-sized'. You would be surprised how capacitors and inductors are used in substations to control much much larger currents and voltages. Your incumbent utility uses them all the time.

It isn't capacitive resistance, it's reactance. There is a difference. Virtually no thermal/heat dissipation with the capacitor. My goal wasn't to force you to a solution, it was to reconsider what you've done and simplify it. What you've done is far too complex, IMHO.

If the book I pointed out to you seems too simple for you, then try this one:

The Art of Electronics 3rd Ed.
Author(s) Horowitz & Hill
ISBN-10: 9780521809269

Wishing you well on your endeavor. If you still have trouble figuring out why Q3 isn't firing, perhaps the first book I suggested would be worth your time.

 

When I started this project six months ago capacitive resistance with AC was the first method I looked at. It is definitely not the way to do this. Say I want 200mA for a large LED bulb at 40 V (this is what I actually have).I would need a *huge* capacitor to begin the circuit with. And then still use a comparator with hysteresis to achieve any constant current I want that will not be affected with temperature changes.

Turning the knob of a BJT half way will also make it way too hot and never provide an exact current. Vhfe (Beta) value varies too much. The only way must be using a mosfet with a clean square wave at its gate with feedback to a comparator, so it achieves constant current and also doesn't get hot. Unless somebody can show me otherwise.

I discovered something last night about why Q3 won't turn on. Q3 not turning on is the reason I came here for help. Once I finalize this I will share it. Thanks.

Could you please explain what the circuit is supposed to do, and why do you need to run it on an high voltage , as said in other posts Q3 won't turn on because you're trying to pulse it with a lower voltage .