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Sunday, July 16, 2017

Considerations about the Hiland Adjustable DC Regulated Power Supply Kit‎

My "lab" power supply was a homemade one I made/upgrade some years ago. It was constructed around a LM350K TO-3 regulator and worked very well. But I really missed one function: adjustable current limit.

The LM350K regulator has a fixed current limit around 2-3 amps, but usually that is just too much current for a prototype. Many times, when something went wrong, one or more components went burnt.

The Hiland Adjustable DC Regulated Power Supply DIY Kit‎

A friend send me a link to one regulable power supply with current limitation in kit form. I googled a bit about it and I found the schematics and many information about the kit. It fits perfectly into what I was looking for and it was really cheap (around 6€ including shipping) so I ordered one. The kit is known as the Hiland Adjustable DC Regulated Power Supply DIY Kit‎. There are some versions of the kit, one 100% analog and another one controlled by a microcontroller. I went to the 100% analog way.



The Kit

This is what you will get

The power supply comes in a really nice (but incomplete) kit form. There is a nice red double sided PCB, a bunch of resistors, a few capacitors, three ICs, some transistors and diodes, two potentiometers, a few screw terminals... It does not come with the power transistor heatsink neither the obligatory fan: this parts must be obtained separately.



How it works

The Hiland power supply schematic



The heart of the power supply is U1 and D8. D8 is a 5.1 volt zener and U1 in that configuration creates a 10.2 volts reference voltage at its output pin (pin 6). This reference voltage is used both to set the output voltage and the current limit.

The 10.2 volts reference voltage is applied to P1 who works as voltage divisor. Its output is applied to U2, who is configured as a x3 voltage amplifier ( 1 + R12 / R11 ). This means the theoretical maximum output voltage of this power supply is 10.2 x 3 = 30.6 volts. Q2 and Q4 work as a current buffer for U2.

The 10.2 volts reference voltage is applied also to the voltage divider formed by R18, P2 and R17. The output from the divider is compared with the voltage at the shunt resistor ( R7 ). If the voltage across the shunt is larger that the one set by P2, the output of U3 goes low, and thanks to D9, the input of U2 also is reduced, producing the current limitation. When U3 output is low, Q3 turns on the current limiting LED. With the supplied values of R17, R18 and P2 the current limitation goes as low as 2mA and as high as 3A.

The group of components in the bottom left corner (R2, R3, C2, C3, D5, D6 and D7) creates an -5 volts isolated rectifier for the correct operation of the operational amplifiers. Q1 is a safety transistor: if the -5 volts voltage is not present, it switchs off the power supply output. This is the reason the power supply must be feed with an AC current from a transformer. It will not work with DC current.

As you can see it is a very classic design and really easy to modify to meet your needs.



Problems

If you make a quick search about this power supply you will find it is advertised with different specifications. In some places it is a 30V, 3A power supply, in other places it is advertised as 28V 2A power supply... in practice both claims seems to be very optimistic.

Let's do some numbers, but before that, lets name some points in the power supply schematics just for reference:

Nomenclature used in this article



This names will be used in the rest of the article so keep them in mind.

The power supply instructions says you will need a 24V, 3A transformer. A 24 volts transformer will produce around 33 volts between the positive rail and the ground rail: 24 times √2 minus 1.2 volts for both conducting diodes. There are 5 volts between the ground rail and the negative rail. This means there are 38 volts between the positive rail and the negative rail.

Now look how U2 and U3 are connected to the supply rails. Don't forget U2 and U3 are TL081 operational amplifiers and its datasheet says the absolute maximum supply voltage is +/-18V, this is 36 volts. They are out of specifications, so they will fail sooner than later.

The 3 amps output current is also very optimistic. With an output of 3.3 volts at 3 amps, Q4 will need to dissipate 89 watts as heat. The small heatsink, even with the fan will not be enough to keep this transistor at a reasonable temperature. Also, don't forget at higher voltage outputs the power supply will not be able to produce 3 amps because the ripple at C1.

In my implementation I used an old box from a power supply, with a 18 volts transformer and a big heatsink for TO3 transistors, so I replaced Q4 with a properly mounted 2N3055 transistor. In this way I get the full power supply specifications up to 15 or 16 volts output with all components in the safe area. Just enough for me.



Adding volt / amp meter

No lab power supply can be called lab power supply without a voltmeter and an ampmeter, so you will install them. Your options are basically two: analog or digital meters. If you choose analog meters the problem reduces almost to zero: Place the voltmeter in parallel with the output and the ampmeter in series. That's all.

But if you choose a digital meter things can be tricky. Probably you will be tempted by that myriad of cheap voltmeters / ampmeters in eBay, Aliexpress and many other places in Internet. They are cheap, and sometimes they are quite precise.

But to install one of these meter in this power supply first you must understand what are they and how they work



The cheap voltmeters / amp meters

These devices are microcontroller based. There are many different versions with different specifications, ranges, etc, but basically they are all the same. A microcontroller with two analog inputs reads the voltage using a resistor divider, and the current using a shunt and a DC amplifier constructed around an operational amplifier. All you need to connect is five wires. A crude schematic can be this:

Simplified diagram of cheap voltmeter / ampmeter



If you watch carefully the drawing you can spot the main problem with these meters: Both black wires are the same, and they are also the negative lead from the microcontroller power supply. How can this meter be connected to the Hiland power supply? It seems easy, but it is not.

At a first try you probably will do this:
First proposed wiring diagram



This will work fine, but has a small drawback: The current drawn by the meter goes through R7, the power supply shunt. This means the variations of current because the turn on and off of segments will modulate the current limiting function of the power supply. That's not a very desirable side effect.

There could be another problem with this configuration. Remember the voltage at the power supply positive rail? It can exceed the maximum supply voltage of the meter, and if it does, it will burn. Sellers advertise the maximum supply voltage of these meters as 28 or 30 volts. I found many of them with regulators who has a maximum input voltage of 24 volts. Because the meter draws a few mA, you can use something like a 7812 or 7815 voltage regulator to lower the positive rail to a safe value, as long as it is under 35 volts: the maximum input voltage for these 78xx regulators. You can also lower the voltage using a series resistor with the meter supply. and a small electrolytic capacitor to smooth the voltage, but watch out the resistor dissipation!

Because current limit modulation by the meter is not desirable in a lab power supply another configuration must be made, and this is when things become tricky. You must eliminate the meter supply current to flow trough R7. Why? A possibility can be this:

Second proposed wiring diagram



Now both shunts (the power supply shunt and the meter shunt) are now in series. This produces a very small effect on the power supply: the meter shunt is just a piece of wire, with much lower resistance than R7 so its effect is minimal and the current drawn by the meter now does not circulate through R7.

It seems perfect! But if you try this, you will notice soon the displayed voltage moves up with the current. Exactly 0.47 volts for each amp the load draws from the power supply. Why?

Because the Ohm's Law and the fact the signal reference of the meter is the ground rail of the power supply, not the negative output of the supply. Now the current flowing by the power supply shunt is added to the output voltage, so you will read the correct output voltage only with no load. Unacceptable for a lab power supply.

Fortunately there is a solution for this. You need to subtract to the positive output voltage (as seen from the ground rail) the voltage developed at the shunt. How? Using a differential amplifier:

Differential amplifier



With an operational amplifier and four identical resistors you can build a differential amplifier with unity gain. In this case Vout is equal to V2 minus V1. So make V2 the power supply output voltage, V1 the voltage drop at R7 and Vout will have the real power supply output voltage referenced to the power supply ground rail, this is, ready to be measured by the meter:

Third proposed wiring diagram



A little messy but it works! Keep in mind you need to use precision resistors for the differential amplifier. Don't use anything over 1%. And better, choose four resistors with the most similar values from a large lot. What values? Anything from 10 to 100k will work fine.

You can also use ready to use differential amplifiers. I used the INA105 with excellent results. You can also use the AMP03, or AD8276, or any other unity gain differential amplifier. Just keep in mind the differential amplifier must be supplied between the power supply positive rail and negative rail, so watch out the maximum supply voltage!

This solution works as long as your meter has a "low resolution" current meter. Why? If you look carefully at the power supply schematics you will find U1 current (and associated components) return to the ground rail through the power supply shunt, R7.

U1 is supplied through the shunt resistor



This means there is always a minimum current circulating through R7. How much current? 5 milliamps. So, if the meter can't see that small current, everything will work fine. If it can, you will have a annoying problem.


Idle current through R7



So, if you use one of those cheap volt / amp meters with low current resolution the proposed wiring diagram with a differential amplifier will work just fine. But if your meter has enough resolution (and a lab power supply really deserves it) you still have a problem.

At this point there are not too much options. Probably it is possible to develop another circuit to compensate those extra 5 mA but the increased complexity really calls for a simpler solution. Much simpler: The first proposed wiring diagram, but with the meter supplied with its own and independent power supply. For this task I used a small 9 volts 1.2VA PCB transformer, a small bridge rectifier and a 470 uF 16 volts capacitor. This produces around 12 unregulated volts, more than enough to feed the meter's regulator with confidence. In this way everything works as expected.


Fourth proposed solution in the case the third one does not work as expected because of idle current through R7



In this way the small transformer is connected in parallel to the main one, and both transformers are switched by the same switch. It is interesting to note this solution can be cheaper than a unity gain differential amplifier like the INA105, so in some cases it will be an acceptable solution even for low resolution meters.



Improvements

There are many improvements of this power supply in Internet. Because I am not using the power supply as is, only a small subset of them are useful to me, for example:

- Use 5 amps diodes: The rectifier diodes in the kit are 1N5408 with maximum current of 3 amps. It is probably a good idea to replace them with 5 amps ones, like SR5100, or similar.

- The reference voltage of the power supply is D8, a 5.1 volt zener. If you need absolute precision and minimize temperature drifting you can replace it with a LM431 plus a pair of resistors.

- Use a 4700 or even 6800 uF 50V capacitor for C1. This will improve high current output at higher voltages.

- Use multi turn potentiometers, specially for the current adjustment.



Conclusion

In this article I expose the Hiland regulated power supply and how those cheap volt /amp meters works so you can use them together or with other types of power supplies.

I see this kit not as a regulated power supply, but as a building block for your regulated power supply. It works just fine, and the current limit function (the function I really missed) works like a charm. Now I can use the power supply to charge all kind of batteries with great success. Limiting the amount of current a prototype can draw is really a must. Also, Did I mention the power supply goes down to zero volts? Not a every day need but definitely a good feature.

Definitely not bad for 6 euros including shipping.

Miguel A. Vallejo, EA4EOZ

2 comments:

  1. You could benefit in this circuit from having a low value resistor of 0.2Ohm in series with the emitter of the pass transistor.

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  2. Congratulations for your article, Miguel. It will save me a lot of valuable time when wiring a voltage and current meter in my Hiland power supply. I will go through the transformer solution.

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