Hopefully a SMPS...

Thread Starter

Allenph

Joined May 27, 2015
76


This is what I've come up with for a SMPS I need to power a large 3D printer I'm building.

It doesn't say on the diagram, but here we're outputting 25V AC on the transformer.

Intended Operation:
1) Transformer steps down AC to 25V.
2) BR1 rectifies AC into full wave DC.
3) C1 evens out full wave DC from BR1.
4) Voltage divider R1-R2 provides 5V to the astable multi-vibrator, with a max current draw from the vibrator of 5mA.
5) The next bit in the middle is an astable multivibrator with added diodes for improved rise time, and a P-Type signal mosfet substituted for one of the timing resistors. M1 recieves feedback from the load, and changes resistance based on the feedback, thus changing the duty cycle.
6) The bit on the right is a buck converter, converting the 25V into 12V, regardless of load.
7) D4 is a zener diode which acts as a feedback channel for the oscillator. It will only allow current through if the load exceeds or is at 12V. If it isn't at 12V M1 will have very little resistance, and the duty cycle will increase until it is at 12V.

I'm sure there are MANY fundamental mistakes in this circuit. But, this is what I've come up with. I think it would be best to get critique instead of starting electrical fires.

Thanks1
 

Dodgydave

Joined Jun 22, 2012
11,285
Q1 is the wrong way round, M2 gate is at 0v should be on Q2 collector ..how much current will you need to drive the printer, upto 3 amps you can use a LM2596 voltage regulator to drop the 25v to 12v,,if you need extra current use a transistor buffer.
 
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Alec_t

Joined Sep 17, 2013
14,280
I'm sure there are MANY fundamental mistakes in this circuit.
You're right!. Here are a few, at a quick glance:
Q1 is upside down,
Neither Q1 nor Q2 has a collector load,
M2 gate is not being switched,
If M2 is N-channel (unclear from the image) its gate voltage will never be high enough to turn M2 on.
 

AnalogKid

Joined Aug 1, 2013
10,987
The R1-R2 voltage divider to power the oscillator needs a filter/decoupling capacitor at a minimum, and would be better replaced with a 3-terminal regulator. Is this a home project, homework, for a commercial product, or what? There are many buck regulators on ebay that will do this for under $5, and many app notes available if you want to DIY. While there is absolutely nothing wrong with just jumping in and reinventing the wheel for the learning experience, you'll get better guidance if you let us in on the reason for the project.

ak
 

Thread Starter

Allenph

Joined May 27, 2015
76
The R1-R2 voltage divider to power the oscillator needs a filter/decoupling capacitor at a minimum, and would be better replaced with a 3-terminal regulator. Is this a home project, homework, for a commercial product, or what? There are many buck regulators on ebay that will do this for under $5, and many app notes available if you want to DIY. While there is absolutely nothing wrong with just jumping in and reinventing the wheel for the learning experience, you'll get better guidance if you let us in on the reason for the project.

ak
Thanks for the advice all.

The transistor being upside down was a rookie mistake, essentially a typo.

I didn't add resistors to the collectors of the oscillator BJTs because the voltage divider already regulates us to 5mA, which is a safe collector current. I didn't want to go much higher than that because then I'll end up using bigger than 1/4W resistors.

As to the question of what happens when Q1 or Q2 turn on...nothing that doesn't happen in a normal oscillator? What's wrong here?

Why won't M2 ever be switched on? And yes, it is an n-type.

I didn't use a voltage regulator because the machine I've designed uses three devices in addition to 3 high power stepper motors. In addition to those steppers I have a CNC motor, a plastic extruder, and a laser. The three "bits" are interchangeable, so only one will be on at once. But, if it's the CNC machine at startup all four of those high power motors coming on could cause some serious current spikes which I don't think would be okay to dump on a regulator, especially with these spikes happening often during the start-stop process of printing. Not to mention, printing can take hours, and this would waste power AND make my circuit kind of heat prone.

In addition, the printer utilizes a TI wifi integration IC. It's VERY sensitive to ripples and heat changes. I was planning on putting a regulator on the output, just to be double sure, without killing myself in the power dissipation area.

I wouldn't say it's commercial, but I also wouldn't call it a home project either. I've gotten custom PCBs made, I'm working on custom chasis, etc. It may not ever get sold, but the idea is to make it look like I bought it, and then maybe sell it later. So far, I've brought down a 1' X 1' X 1' 3 in 1 printer to under $150 w/o power supply which is...amazing.
 

Thread Starter

Allenph

Joined May 27, 2015
76
Oh, I see why M2 isn't switching. I understand that the gate needs to be collected to the collector not the emitter.
 

AnalogKid

Joined Aug 1, 2013
10,987
As to the question of what happens when Q1 or Q2 turn on...nothing that doesn't happen in a normal oscillator? What's wrong here?
Look at each transistor's emitter and collector connections, and imagine either one as a dead short from collector to emitter (it's effective saturated state), and see what happens.

ak
 

tcmtech

Joined Nov 4, 2013
2,867
So why the overly complicated multi step part linear part buck process to get a 12 volt 3 - 4 amp capable power supply that could be bought new online for under $20?
 

Thread Starter

Allenph

Joined May 27, 2015
76
Look at each transistor's emitter and collector connections, and imagine either one as a dead short from collector to emitter (it's effective saturated state), and see what happens.

ak
Every ounce of that 5mA has a path to ground, so the capacitor doesn't charge through the charging resistors...or at all actually. I need to limit it so that the current running through the transistor in the on state leaves enough current for the charging of the capacitors.

Is that what it is?
 

Jony130

Joined Feb 17, 2009
5,487
D1 and D2 serves no function because they are short by a wire. Also your feedback will not work as it should be. In fact all circuit will not work.
Try something like this
10.PNG
 

Thread Starter

Allenph

Joined May 27, 2015
76
So why the overly complicated multi step part linear part buck process to get a 12 volt 3 - 4 amp capable power supply that could be bought new online for under $20?
It's a hobby. I want to make everything from scratch; that's the whole point. Not to mention if I ever did sell it "PSU Sold Separately" wouldn't be an option.
 

Thread Starter

Allenph

Joined May 27, 2015
76
D1 and D2 serves no function because they are short by a wire. Also your feedback will not work as it should be. In fact all circuit will not work.
Try something like this
View attachment 87354
D1 and D2 serve functions. It's a pretty stereotypical way to increase rise times in this kind of astable oscillator. It isolates the collector current from the capacitor.

I also don't understand why my feedback channel wouldn't work. If the zener is not reverse biased (There's not 12V at the load) then the equivalent resistance of M1 is negligible, resulting in a rise in duty-cycle, resulting in a higher voltage. If it's above 12V, then the equivalent resistance of M2 increases until the duty cycle only produces 12V for the specified load.

OOPS! That's another mistake on the diagram. That zener should have the polarity reversed. And it's reverse threshold should be 12V.
 
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tcmtech

Joined Nov 4, 2013
2,867
It's a hobby. I want to make everything from scratch; that's the whole point. Not to mention if I ever did sell it "PSU Sold Separately" wouldn't be an option.
I can relate to the hobby part and the learning part but I still can't follow the logic of making something unnecessarily complicated just for the sake of it. Especially when it's a power supply that has to work at a fairly trivial power level and relevant efficiency.
 

Thread Starter

Allenph

Joined May 27, 2015
76
I can relate to the hobby part and the learning part but I still can't follow the logic of making something unnecessarily complicated just for the sake of it. Especially when it's a power supply that has to work at a fairly trivial power level and relevant efficiency.
I was just under the impression that using a linear supply like you suggested wouldn't work well because there would be some near constant current spikes for hours on end...isn't that a fire waiting to happen?
 

AnalogKid

Joined Aug 1, 2013
10,987
Is that what it is?
Q1 and Q2 - both emitters go to GND, both collectors go to R1 as the collector load. When either transistor comes on, it shorts out the other which should be open. The circuit cannot oscillate as drawn. A standard multivibrator has independent collector loads to a common power source.

ak
 

Thread Starter

Allenph

Joined May 27, 2015
76
Q1 and Q2 - both emitters go to GND, both collectors go to R1 as the collector load. When either transistor comes on, it shorts out the other which should be open. The circuit cannot oscillate as drawn. A standard multivibrator has independent collector loads to a common power source.

ak
Ahh. Got it. But, I'm unsure how to set this up. I chose the resistor that's on the oscillator specifically because when combined with the 5k total resistance of the divider it would produce a base frequency of 100kHz. (If the MOSFET was replaced with a similar resistor.)

I'm kind of at a lose as to what resistor values to isolate the collectors with. Now that we're getting down into splitting 5mA I'm wondering how much current is required on the output to switch the power MOSFET M2. As far as I understand, they're voltage controlled devices...right?

I guess I'm getting confused here. Do I need to supply a certain amount of current to the gate of M2 to get the full potential of 20A flowing? I know transistors have gain, but I also thought that they were voltage controlled devices...I.E. it's just ON if you apply a voltage to the gate.

In addition, I noticed how you only specified R1 as the common power supply for the collectors. I realized that when current goes through the transistors to ground I'm not really supplying current through the voltage divider, only R1. I could fix this by removing R2 and adding it to the emitter of each transistor, correct?
 

tcmtech

Joined Nov 4, 2013
2,867
I was just under the impression that using a linear supply like you suggested wouldn't work well because there would be some near constant current spikes for hours on end...isn't that a fire waiting to happen?
Linear is way more forgiving of current and power spikes than SMPS based systems are. As far as heat goes it's only a problem if you designed the circuit wrong and used way too small of rated devices and heat sinks.

If the primary load is almost all motor driver power then the voltage and current filtering is even less of an issue and a simple transformer to bridge rectifier to good old heavy capacitor bank is more than sufficient and clean enough and the dedicated more sensitive logic circuits can be ran off their own independant regulated power supply so they are isolated and for the most part immune to motor driver related noise.
 

AnalogKid

Joined Aug 1, 2013
10,987
https://en.wikipedia.org/wiki/Multivibrator

This is about as clear as it gets. In Figure 1, each collector and each base have independent power connections for total control over each half of the circuit. I suggest that you start here, and add your feedback MOSFET in parallel with R2 or R3. When Q2 turns off, R4 is the impedance that discharges the M2 gate capacitance to turn it off. The faster this happens the better the circuit efficiency. I don't know what you mean about the two diodes isolating collector current. The collectors supply current to the capacitors every half cycle. Even without the drafting errors in your original schematic, I say get rid of the diodes and get the circuit to function in the traditional manner, then start messing around.

ak
 

Thread Starter

Allenph

Joined May 27, 2015
76

There. Hopefully this will work now.

It's similar in nature, it's just that now each transistor, diode, and capacitor in the oscillator has its own source. I don't have to mess with high wattage resistors because each branch only handles a few mA, and I don't have to worry about figuring out how much offset is taking how much current at whatever point in time, because they're all independently sourced.

What was someone saying about using filtering capacitors?

I don't see the need because all the components in the oscillator can handle some wobble, and the output is either high or low, so whether or not it's 4V or 6V makes relatively little difference. At least I think...


https://en.wikipedia.org/wiki/Multivibrator

This is about as clear as it gets. In Figure 1, each collector and each base have independent power connections for total control over each half of the circuit. I suggest that you start here, and add your feedback MOSFET in parallel with R2 or R3. When Q2 turns off, R4 is the impedance that discharges the M2 gate capacitance to turn it off. The faster this happens the better the circuit efficiency. I don't know what you mean about the two diodes isolating collector current. The collectors supply current to the capacitors every half cycle. Even without the drafting errors in your original schematic, I say get rid of the diodes and get the circuit to function in the traditional manner, then start messing around.

ak
Seems I already kind of figured out the independent branching thing.

http://www.learnabout-electronics.org/Oscillators/osc41.php

That's explains why you use the diode. I've built a couple test oscillators on a breadboard, and it does DRAMATICALLY increase rise time. Without these, it's not really a crisp square wave. I believe this is called the Rosner Modification.

I'm not sure why I would add the MOSFET in parallel to another resistor. Isn't that just over complicating?
 
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