Author Topic: LAB PSU Schematic Explaination (Read 15452 times)

xoom · « **on:** March 19, 2016, 11:25:53 am »

Hi,
im trying to fully understand this schematic, but im stuck:)

most confusing part is that PSU positive rail is as ground.. and all my understanding disappears

I colored some rails for better visual understanding and marked some control circuits..

My questions is:
1. Control circuit that switches relay for other transformer tap. How it senses voltage on output. How to calculate what threshold is?
2. As i understand for current limiting control R22 is shunt resistor.. but again im confused how it measures voltage across it.

and there is more questions but for now i just want to understand this concept.

My point it to make Power supply with higher output voltage (30V@3A) and some more improvements.. like more stable voltage ref. (maybe some digital controls)

mmagin · « **Reply #1 on:** March 19, 2016, 01:04:11 pm »

Wow. That's like the "how not to draw schematics" example in The Art of Electronics.
And the way the voltage feedback circuit involving the 741 has a +12/-6 volt supply is extra confusing.
I think I'm not awake enough this morning to do more than observe a couple things:

1. that relay control circuit probably can be thought of as a comparator with hysteresis?
2. The shunt for the current limit is R22+R15. It provides a voltage to the - input of U2B, and the front panel pot provides a user-set voltage to the + input (referenced to 'ground', which is the positive output!). So if the voltage across R22+R15 exceeds the voltage from VR2, the output goes low, and through the 'diode-OR' current is pulled until the pass transistor brings the output down to the limit.

You probably cannot increase the output voltage without a different transformer (or rewinding the T-5/6/7 winding). I assume the T-8/9/10 winding is probably much lighter gauge. I don't think the stability should be horrible for a cheap supply, as long as that TL431 clone isn't that bad (could try replacing it with the real thing) and R3/VR3/VR4 and the 741 aren't too affected by temperature.

amspire · « **Reply #2 on:** March 19, 2016, 01:13:52 pm »

To understand the circuit, notice that there are two different ground symbols used. The 0V "ground" that of the control circuit is just that - the control circuit 0V and nothing more.

The chassis ground, or the mains earth ground is totally separate and is isolated from the power supply output.

From outside the power supply, the control circuit ground means nothing to you at all.

Now to explaining the circuit a little.

First, the control system has its own power supply. That makes design much easier - you can change the voltage at will to to main supply without any affect on the control circuit. You can measure the voltages across a current sense resistor and have the voltage of the sense resistor right at the 0V potential of the control circuit (there is a +/- supply for the control circuit, so 0V is in the middle).

Secondly, the thing that the control circuit is doing is controlling the base current of a NPN power transistor that is at a potential of about 0.7V above the output potive rail voltage. So if you choose the positive output rail as the 0V point of your control circuit, driving that pass transistor is simple.

Thirdly, if the control circuit has its own independent supply, and is has its 0V at the positive output rail potential, it is easy to scale the supply to any output voltage you want. This way, they can take this design and tweak it a bit to be a 30V supply, a 60V supply, a 100V supply - whatever they want. You will probably find that this supply is part of a family, and if you get the schematic of the 30V version, you may be able to just change the components of your supply to make it 30V. That will include needing a new transformer, new mains filter capacitor, perhaps a new power transistor and the output current will be half that of the current supply. You probably have to make a new meter scale ..... Is it worth it? Probably better of selling your supply on eBay and buying the model you want.

Now to your questions. The pre-relay circuit is a comparator with input connected to the positive output. The other input goes to a voltage divider that has one end connected to 12V above the positive output, and the other end to the negative rail. When the voltage across R35 is less then the supply output voltage, the relay turns on halving the voltage from the transformer. This will happen when the supply is below 15K * 12V/24K = 7.5 volts out.

The current control has one end connected to the more positive end of the shunt. The other end of the shunt is of course control circuit ground. The other current control opamp input goes to voltage divider that has one end at control circuit 0V (the output) and the other end at a reference that is 2.4V above the output. So when the voltage across the shunt equals the voltage on this divider, current limiting starts.

Why did they pick the positive rail for control circuit ground? They actually could have picked the negative rail and used a PNP power transistor or a P-Channel power MOSFET. They chose a NPN power transistor and the positive rail because NPN power devices and better, cheaper and more plentiful then PNP power transistors. That is the only reason.

Kleinstein · « **Reply #3 on:** March 19, 2016, 01:16:59 pm »

This supply is a more or less classic circuit using the floating regulator. So the regulating circuit is powered by the upper transformer. Ground for this is tied to the positive output terminal, and the output stage is running as a common collector circuit.

The voltage controlling amplifier looks a little odd: Normally one would have have a resistor from ground to the inverting input of the OP and a pair of antiparallel diodes instead of C1. They also seem to use one of the zero adjustment pins for compensation. In this respect the circuit is unusual and depends on the OP type used here.

homebrew · « **Reply #4 on:** March 19, 2016, 01:39:50 pm »

Interesting, I get a different tripping point. But maybe I'm confused now also ...

Let's see:
In the end this thing works as any other voltage comparator. R5 and R35 form a voltage divider. Try to imagine that we would select ground to be the negative output rail. Then the voltage on top of the divider would be V_out + 12V. Hence the output of the divider would be v = (R5 / (R5 + R35)) * (V_out + 12V). When this falls below V_out, the relay will energize. Hence solving this equation leads to a tripping point of Vtr = (12V / (1 - R5/(R5+R35))) and hence to 19.5V.

Hopefully that is not complete bullshit. If so, feel free to correct me ...

xoom · « **Reply #5 on:** March 19, 2016, 01:44:02 pm »

Thank you for replay.

@mmagin
Maybe that's why its so hard to understand for me

I will use different transformer with .. 13.2V X 2 @3.2A Maybe its not reach 30V DC but its ok.. problem is that it don't have second secondary for powering control circuits.. and using same secondary for controls and power i think is bad.. as its have 'ground" on positive rail.. also i use MAX6043 for ref. voltage (2.5V) so TL431 is no needed any more..

Other question what about voltage control.. its reference to negative rail of PSU. How it behaves?

@amspire

Thank you for more explaination. Im not owning this Power supply. I want to make it my self and using this schematic as reference which i want to modify to my needs

Now just need to read again your post and try to understand more better

P.S. Sorry for my bad English

homebrew · « **Reply #6 on:** March 19, 2016, 01:44:18 pm »

Yeah, forget it ...
I mixed up the resistors. When swapping R5 and R35, my equation also leads to 7.5V...

amspire · « **Reply #7 on:** March 19, 2016, 01:56:43 pm »

Quote from: xoom on March 19, 2016, 01:44:02 pm

@amspire

Thank you for more explaination. Im not owning this Power supply. I want to make it my self and using this schematic as reference which i want to modify to my needs Now just need to read again your post and try to understand more better

As I said, just find the schematic of the 30V version of the same supply. Then you can copy it directly. If it is an HP supply, there will be other models with the same basic circuit, but with different output voltages. You have just picked the circuit of the 15 volt model. when you building it, don't try too hard to find a transformer with all the windings. Get one big transformer for the power circuit, and a small transformer for the control circuit. The WL431 reference IC is more commonly known as the TL431 reference - very common. Might even find one in scrap PC or USB power supplies.

xoom · « **Reply #8 on:** March 19, 2016, 02:11:42 pm »

That's what im thinking to do with second transformer.. I sticked to this schematic becouse it looks simple (but still little confusing for me becouse in analog circuits im not so good) And yes i know TL431 but i have MAX6043 voltage ref. chip, so i want to use it. Also opamps i have ISL28207 so thinking to use them

And want to challenge my self a little to make this PSU by modifying this schematic and calculating new values for resistors

Its more learning then need of this power supply

Also as here is negative supplies regulated only by zeners i want to change them to voltage regulators and thinking about that -6V rail to get rid of it(i guess i could do it). and just supply only + - 12V

amspire · « **Reply #9 on:** March 19, 2016, 02:29:35 pm »

The opamps in the circuit look like LM324 quad opamp, and the uA741 which comes from the 1970s. A bit too crusty, I think. The compensation circuits are designed to suit the chosen opamps, but your Intersil opamps may actually work. It is possible the supply control circuit may oscillate if you substitute parts.

xoom · « **Reply #10 on:** March 19, 2016, 06:10:32 pm »

Looks like im really stupid

.. trying to redraw that comparator part in more understandable way, but allways come to a dead end

that negative rail comes like mystery ..

R19 and R20 are for hysteresis (im right?) so left with R5 and R35 which is divider as reference for comparator.
I'm looking at amspire formula but not understand why its calculated like that

when im looking at homebrew formula i see more familiar formula for divider Uout = (R2/(R1+R2))*Uin

saturation · « **Reply #11 on:** March 19, 2016, 07:07:44 pm »

This is the schematic for a widely made clone "Mastech" 1803D, well discussed on this forum. ~$50 new or a $30 used. Overall, its fairly reliable 1980s era tech. You can save time and maybe money buying a full working unit and mod it from there.

http://www.ebay.com/itm/MASTECH-HY1803D-VARIABLE-LINEAR-DC-POWER-SUPPLY-REGULATED-T2-C5-/291679603533?hash=item43e9758b4d:g:0WMAAOSwnLdWr8hs

Kleinstein · « **Reply #12 on:** March 19, 2016, 07:47:30 pm »

The divider is between the reference and output. Its easier to see it like comparing currents thus the trip voltage is at 12 V * R5 / R35.

This type of circuit needs the second transformer or extra isolated winding. Depending on the OPs used the auxiliary supply can have other, preferably lower voltages. It's similar to the system used in many HP supplies. Its well possibly to use this type as a basis for a own design with digital control - the control part also powered from the extra transformer. However the plan still has a few points that should/could be done better (e.g. current reading more accurate, better compensation for voltage control, add output enable).

Power supplies can be rather cheap - so even buying the part can be more expensive than getting a cheap Chinese one and modify it to a lower current to make it more durable. So building your own supply is not about saving money but more something to learn unless you have parts to reuse.

Iwanushka · « **Reply #13 on:** March 19, 2016, 11:34:26 pm »

Quote from: Kleinstein on March 19, 2016, 07:47:30 pm

Power supplies can be rather cheap - so even buying the part can be more expensive than getting a cheap Chinese one and modify it to a lower current to make it more durable. So building your own supply is not about saving money but more something to learn unless you have parts to reuse.

In our country anything that has label "LAB" is very expensive, this clone is at around 50+euros 30v goes for 100e (single channel) or something like that, and we have plenty of "free" transformers, so that thing can be made for 1/5 -> 1/10 and so on... of the retail price, and shipping from other countries would be like buying second PSU... if not two of them ;/

Kleinstein · « **Reply #14 on:** March 20, 2016, 10:11:38 am »

The self build solution can still be cheap, if you can start with some used / recycled part. For a lab supply something like an old (broken) Hifi amplifier would be a very good starting point. It just gets expensive when you need to buy things like the case, transformer and heat sink.

Today the price for the cheap Chinese parts you get from ebay should not be so different in most parts of the world. Just a few less developed countries have difficulty in delivery or high customs duties and a few more developed ones insist on proper CE/UL/FCC markings so you may not get all crap send directly from china there.

xoom · « **Reply #15 on:** March 20, 2016, 10:19:44 am »

I have transformer. Have heaps of heatsinks.. its more about learning.. i want to fully understand this circuit.. I tryed to emulate in not very good simulator (crocodile physics) but i cant.. it not works like that

If i put zero volt rail or ground rail i cant lift it like here.. only thing i can think of is to put negative voltage on negative rail and change it .. if anyone can redraw that part (comparator) in more understandable way (like for beginner) i would really appreciate it

xoom · « **Reply #16 on:** March 20, 2016, 11:19:59 am »

Ok done some progress i guess

Here ground symbols are same as in schematic. And Red probe is on Negative rail becouse only this one i can vary with signal generator. So i configured it to go from 0V to -30V.
And here is what i get:

At about -9V (which would be +9V in normal operation) relay kicks in.. and when the voltage comes back to 0V at about -6V (+6V) it de energize relay.

If im doing it right

)

BigEarl619 · « **Reply #17 on:** May 16, 2017, 10:51:52 pm »

Hello everyone,

I found this topic on the mastech power supply. What is the point of the DW3 zener diode that is connected to R10 then R10 to +12 volts and anode is connected to the OR circuit, seems like it is missing a connection from cathode to base of Q1 so it doesn't go above the zener voltage of 2V.

If not, what is the purpose of this zener diode if not used a ref volt/voltage stabilizer,

Thanks.

Kleinstein · « **Reply #18 on:** May 17, 2017, 04:24:41 pm »

The DW3 zener in the HY1803D circuit is there to prevent the output to turn on when the regulated 12 V supply is not yet high (e.g. about 4,x V) enough to allow the OPs to operate. So it is more to prevent / reduce a spike on turn on.

FabioLucca · « **Reply #19 on:** August 14, 2017, 01:41:22 pm »

This scheme is very similar (it seems almost a copy) of a source scheme of the December 1982 issue of Elektor (which I am assemble), read the article http://www.retro.co.za/zs1ke/projects/PrecisionPowerSupply/PrecisionPSU-Elektor-Dec-1982.pdf that will be clearer for you, I also took time to understand the operation and to make the source work, (I had to insert an extra resistor in the power circuit).
Here https://dicasdozebio.com/2014/06/12/projeto-fonte-de-alimentacao-cc-profissional-0-40v-0-4a/ has the process of assembling this source (Elektor) by a Brazilian.
My assembly worked (schematic Elektor magazine), in the Proteus simulator the circuit works but I want to improve the circuit with digital control and measurement with ADC so I need to study a lot. Obviously it's cheaper to buy the ready-made source or a DC-DC converter but what fun does it have? I'm in for learning. I apologize for the bad english of google translate.