Power Supplies for Home Construction

Electronic circuits, whether they be valve or solid state (transistor, for example) all require one or more sources of DC electricity. Valves in addition require a filament supply, which is usually an AC supply. Whilst it is almost always possible to provide the required supply voltages by wiring dry batteries (primary cells) in series for the high voltages needed to operate valve circuits, or to use dry or rechargeable batteries batteries for the low voltages needed for solid state projects, it simply is not always practical and is very expensive. This article describes how to make power supplies for your projects, with examples of both valve and solid state power supply circuits.

In South Africa, the mains voltage is 230 volts AC at 50 Hz. The supply voltage and frequency may be different in your country, so please bear this in mind as you study this article.

Safety

When dealing with electricity, please bear in mind some safety rules. Your power supply should have a double pole switch between the mains input and the rest of the circuit and the metal chassis/case should be grounded, if it uses one. There should be a suitable fuse between the line and the transformer and there should be a pilot light to indicate whether or not the power supply is switched on. There should be adequate precautions to prevent accidental touching of mains or HT supplies. Use heat-shrink tubing over switch terminals and fuses. Large  resistors at high potential (such as those connected to a voltage stabiliser) should have insulated leads. “Table top” style transformers should have a cover over the terminals.

If you are making a high voltage (anything over about 100 volts) power supply, it is a good idea to place high value resistors across the power supply capacitors (we'll come to those) because inadvertently connecting yourself across a charged high voltage capacitor will cause you to jump, in turn causing some other more serious accident as you knock something off the workbench!

When working with power supplies, wear rubber soled shoes and, if possible work with one hand in your pocket. The worst possible electric shock is when the supply is connected from one hand to the other so the voltage can affect your heart muscle.

Finally, always use a mains transformer. This will isolate your project (and you) from the mains supply. There are usually plenty of local transformer winding companies that will be willing to provide you with a transformer to your specification at a reasonable price.

Notation

In this article, I will be writing all equations the same way as they would be written in a computer program written in a language like BASIC, rather than the mathematical way with superscripts and subscripts. It means you can cut and paste the equations into a program and it means that I don't have to download a large number of images for each equation. I will recant if the community prefers the original equations. If the equations occur inside an image, I will try to remember to use both.So:

becomes E0 = EPEAK*SIN(2*PI*F*T)
PI = 3.14159... or 355 / 113 in fractions...

Recalling the Basics

An AC voltage of 230 volts is a varying voltage that changes from zero to approximately 325 volts then down to zero then to -325 volts and back to zero again 50 times per second. Thus, the difference between maximum negative and maximum positive voltage is about 750 volts (which is what it feels like if you get yourself across it.) This can be expressed mathematically as:

E0 = EPEAK*SIN(2*PI*F*T) ;where

E0 = Instantaneous AC voltage
EPEAK = The peak output voltage measured from zero to the maximum positive voltage
F = Frequency in Hz (50 in South Africa, 60 in the US)
T = Time in seconds

In a DC circuit, the power dissipated in a resistance R with a voltage V applied to it is :

(V^2) / R.

In an AC circuit, we would like to have a similar simple relationship, so that when we calculate the power drawn by an electric kettle with an element resistance of R, we can express it in the form:

((k*EPEAK)^2) / R; where k is a constant.

Since E0 varies, we must take an average for the value of E0^2 over a full period of the AC waveform. We do this by calculating the area under the curve for E0^2, as shown in the following diagram:

(Beware!  - Squaring a sawtooth waveform, for example, does not result in equal cyan and buff areas, so the RMS value will have a different relationship to the peak. )

See also article on RMS measurements

In this article, we will be talking about the “load resistance” (RL) this is simply the device that is going to draw power from the power supply. (One valve radio, transistor amp, computer experiment etc.) The load will operate at a certain voltage (V)and draw a certain current (IL). The value of the load resistance will be :

RL = V*IL

Diodes

Diodes are devices that only allow the passage of electricity to flow one way. They may be semiconductor diodes, such as the ubiquitous 1N4007, or may be valves, such as the EZ81, 5Y3 etc. The valves mentioned have two diodes in one envelope. Semiconductor diodes also come in a bridge configuration, consisting of four diodes in one package.

Diodes have a maximum reverse voltage. In the case of the 1N4007, this is 1000 volts. The EZ81 has a “Peak Inverse Voltage” of 1.3kV – but don't even think about it.

They also have a maximum current, the 1N4007 can stand about 1 Amp, the EZ81 can deliver 150mA at 350 volts, but this decreases as the output voltage increases.

Diodes have a small voltage drop. Silicon diodes normally have a forward voltage drop of about .5 to .7 volts, whilst diode valves have a drop caused by their internal resistance of a few hundred Ohms.

Power supplies normally have a smoothing capacitor. Too big a value will cause too much current to flow through the diode, causing it to prematurely fail in the case of a valve and instantly and disastrously fail in the case of a semiconductor diode. Beware of too much capacitance with the 5Y3 rectifier.

The choice of whether to use a valve or solid state rectifier isn't just one of aesthetics. If you use a solid state rectifier with a valve circuit, the HT supply will be available long before the heaters have brought the cathode to the correct temperature. This can cause problems with some valves. Tektronix in their 1970's 'scopes always used a thermal delay switch to ensure the cathodes were hot before HT was applied. Valves are made nowadays for the high-end audio market, with JJ and possibly others, making a selection of rectifiers.

Smoothing Capacitor

In effect this is a reservoir of energy. The transformer and rectifier top the capacitor up with electric charge, while the load draws the reserve of energy down. This effectively smooths the DC supply and provides a reserve of energy in the case of audio amplifiers.

Almost always electrolytic capacitors are used. Modern electrolytic capacitors are a fraction of the size of the original electrolytics used in old radios. Whereas once upon a time, high-tension (HT) supplies for radios used choke coils and an additional capacitor for smoothing, we now almost always use one big capacitor. If we are making a valve power supply, then depending on the valve, we may be limited to a maximum value of 50uF or even less for the smoothing capacitor.

Electrolytic capacitors have an approximate value marked on the can. If you buy a 47uF capacitor, it can have a value of anything from 25uF to 75uF. They will also have a value of working voltage – be sure to have plenty in hand, for example, if you are making a 250VDC power supply, use at least 450 VW (Volt Working) capacitors.

Power Supply using Half Wave Rectifier Circuit

In what follows, it is assumed that the smoothing capacitor C will have a “large” value. The design proceeds by selecting a suitable transformer.

Suppose we need a power supply for the two valve superhet.
Supply voltage = 200 Volts
Supply Current = 20mA
Transformer required will be 200 * 20 / 1000 = 4 Watts
Filament 6.3 Volts * 0.6 A total = 3.78 Watts
Total 7.78 Watts
Allow a factor of safety approx 1.5 and use a 12 Watt transformer.

Now select a rectifier – I'm going to use a 1N4007, basically because they are extremely inexpensive and will handle 1.0 Amp, with a peak reverse voltage of 1000 Volts. Max surge current for this type is 30 Amps, which would give a value for Rs = 141/30 = 4 Ohms. The chances are that the transformer secondary resistance plus the effective primary resistance is way above this value, but its advisable to measure them all the same. You will have to multiply the primary resistance by (NS^2/NP^2), where NS/NP is the turns ratio of primary to secondary, or VS/VP – the voltage ratio.

The remaining component is the capacitor. For a radio receiver, we might want a ripple of about 0.5 volts. Note that this is 0.5 volts peak-to-peak. To get the RMS value of the ripple, we divide by 2.828 (or 2/SQR(2.0) ), this gives us 0.176. The % ripple is then (0.176/200)*100 or .08%. (Not everyone considers the % ripple this way, and I'm not sure whether this is correct, or whether simply taking 0.5 / 200 = .25% is the correct answer. For the purist, none of the above are correct, since the ripple approximates to a sawtooth waveform.)

Then C = ((10^6) * .02 ) / (50 * 0.5) = 800.0uF
I think we can agree that this is not a particularly happy answer and although readily available, this capacitor will cost around US$ 20.00.

What if we had decided to use an EZ81 (Strapping the anodes together) ? Well – if we look at the data sheet, we see that the maximum value of capacitor we can use is 50uF per anode, and that we also have to use an Rs of 150 Ohms per anode. This means that our ripple voltage will be around 4 volts peak-to-peak and we will lose 20 * 150 /1000 = 2.15 volts. The power rating for the resistor Rs would be (2.15 ^ 2) / 150 – very small. To arrive at the transformer secondary voltage, we need to add up all the voltage drops on the way to the load. These might be:

Diode drop (1N4007) = 0.7 volts
Transformer Internal resistance = 30Ohms *.02 + (230 / 200)^2 * 40 Ohms *.02 = 1.66 Volts
Ripple voltage = 0.25 volts (0.5 pk-pk)
Total = 2.6 Volts
So, we need 202.6 peak volts from the transformer = 143.3 volts RMS.

If you're using the EZ81, then there will be voltage drops in the series resistors (if used) and estimate the voltage drop for a load of  20 mA, using the valve characteristic curves. By eying the characteristic curve data, it looks like about 15 volts, for the two anodes strapped together.

EZ81 Drop  = 15 volts
Transformer Internal Resistance = 1.66 Volts
Ripple = 2.15 volts
Remainder of valve series resistance = 75 Ohms but Transformer Internal resistance = 82.9 Ohms
Total voltage, peak = 218.8
Transformer voltage = 154.7

Full Wave Rectifier Circuits

The half-wave rectifier circuit has little to recommend it now that rectifier diodes are so inexpensive. For one thing, the transformer only really does any work during the half cycle the diode conducts and it requires double the size of smoothing capacitor for the same ripple (as we shall see). Any savings made by cutting out 3 diodes will be totally annihilated by the additional costs of a capacitor.

The three circuits given all have their valve counterparts, but the one most commonly used is the half-bridge with the centre tapped transformer. For low voltage op-amp power supplies, transformers are readily available with two secondary windings. These may be connected in the correct phase so they can be used with the 3rd circuit.

Calculations can proceed as before by selecting a suitable transformer (or winding one), allowing for the voltage drop across Rs and the diode voltage drops. To be strictly accurate, half the ripple voltage should also be deducted, particularly if it is large. You might have to make this calculation quite carefully if you are planning on following your unregulated supply with a 3-terminal regulator. In this case it would be wise to take into account variations of 10% in the mains supply – because you don't want the regulator to “drop out”. (Where I live, the mains voltage can vary between 310 volts and 150 volts, which is rather harrowing.)

Suppose we want to design a 15 volt 500mA power supply using a 7815 three terminal regulator.
Then we need at least 18 volts to prevent drop out. If the ripple is 1 volt, then we need to add this.
There will be a further 0.7 volts drop in the diodes and a drop caused by resistance of the transformer windings of, say, 30 Ohms * (20^2)/(230^2) = 0.24 + 1 Ohm in the secondary, giving a voltage drop of .6 Volts. (Making a guess that the turns ratio is 20/230)

Total voltage required is thus (15 + 3 + 1 + .7 +.6 )*1.1 = 22.3volts.

This could be provided by a transformer with a 22.3*.707 = 16.8 volt secondary. The nearest transformer available off-the-shelf has a 17.5 volt secondary.

Now we calculate the value of the capacitor required to give a 1 volt ripple, remembering the ripple frequency is now 2*50Hz:

C = ((10^6) * .5) / (2*50* 1.0) = 5000uF (The nearest preferred value is 4700uF, but it might be better to play safe and use a 6800uF capacitor – remember the tolerance!). I would make it at least a 35 Volt Working.

Adding an output filter

The theoretical values calculated for the smoothing capacitors may have come as a shock. This will be particularly true for people used to working with old radios with 5Y3 power supplies having small capacitors of 8uF or so.
 
What always used to be done, was to add a choke in series with the output and a capacitor to ground:
Typically, the choke might have had a value of 10H and a resistance of  330 Ohms. The capacitor C would have had a value of anything from 8 to 270uF or more. The combination of the choke impedance and the capacitor acts as a voltage divider, on the ripple voltage so the ripple becomes much less.

Taking our EZ81 half-bridge with its 4 volts peak to peak ripple and a capacitor C of 100uF, then the choke impedance at 50 Hz (Its a half-wave circuit) would be:
 2*PI*F*L  = 2*3.14*50*10 = 3140 Ohms. In addition, it has a resistance of 330 Ohms.
The capacitive reactance would be:
10^6/2*PI*F*C = (10^6)/(2*3.14*50*100)  = 32 (Much, much smaller than the load resistance, which can be ignored)
So the ripple becomes 4 * 32 /(3140 – 32 ) = 0.04 Volts
This is all very interesting, but unless you are making a hi-fi valve amp, you probably won't want to use a choke and in any case, they are heavy and not readily available. Note that in theory it is possible to get L and C to resonate.

Toward the end of the valve era, most sets had done away with the choke. Even published Hi-Fi circuits just had a simple resistor-capacitor input filter. Usually, there would be a connection from the input capacitor to the final output stage, followed by an input filter feeding the remainder of the circuit – see for example the Mullard 3-3 amplifier, or Brimar VA10.

Suppose we simply were to use a 3300 Ohm 2Watt resistor and 100uF capacitor, then the ripple would be:
4 * 32 /(3300+32) = .038 Volts
BUT we would lose 66 volts across the 3300 Ohm resistor. Actually, a more sensible value for the resistor might be 330 Ohms, then we would lose only 6.6 volts.

Pilot Light

It is possible to run LEDs directly from the mains, but they require a bit more current than a neon pilot light. They operate from about 1mA onwards, but since you are going to use them in a half-wave circuit (effectively) you need to double that to get the equivalent light output. I tested a few red LEDs, and these all operated well with a 100k resistor in series. If you are designing a low voltage power supply, then you can use the DC voltage to drive the LED. Suppose you have a 5 volt power supply, and want to run the LED at 2mA, then the resistor needed in series = 1000*(5 -1.5)/2 = 1 750 Ohms – no harm in using a 1500k, it all depends on how bright you want the LED. Just don't do what a well known manufacturer did to my tape deck and put a blue LED running at (at least) 20mA as a pilot light. It hurts the eyes and makes everything In the room look blue.

Regulation

The supplies mentioned in this article all have a fair amount of  resistance in series with the load. This means that if you draw more current than designed for, the voltage will drop. If you draw less, it will rise. In addition, the ripple voltage will increase with increasing load. In addition, the output will vary according to the input mains voltage. You can see how well the power supply will perform by repeating the calculations for different load conditions. You can also build the power supply and apply last-minute tweaks to the series resistors.

Computer Programs

There are a number of computer programs that can be used. HAMCALC by VE3ERP has a power supply design program written in GWBASIC. There is also a nice little simulation program called “PSU Designer II” from Duncan's tube amps. Then there is  LTSpice available free from Linear Technologies - but it might need quite a bit more study than the other software programs, but is well worth it. Start with the filter circuits. I'm sure there are many more that I don't know about.

Conclusion

This short article came about because I felt rather guilty at the glib way I had dismissed power supply design in my article on the two-valve radio project. I will try to include some examples of well constructed power supplies as soon as I have done the things I should have done to the power supplies I have made.

References

Measurements on a small transformer – rmOrg
Horowitz and Hill “The Art of Electronics” - Cambridge University Press 1980
WW Smith “Electronics for Technician Engineers” - Hutchinson Educational 1970

Radio Communication Handbook - RSGB, 1994
BRIMAR Valve Data Book

 

 

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