Really Bad Design.

From time to time, really bad examples of design are seen. This is a new section of the site to draw attention to such things, and may include magazine projects, things seen on the internet, and commercially made designs.


1. "Television" magazine's colour TV project.
Spread over issues from 1972 to 1973, was a series of articles on how to build your own colour TV. The article which drew attention was Part 10; the power supply, described in January 1972.

Alarm bells rang when I read the description of the heater circuits for the valves and CRT. Let's look first at the valves, and see where the designer has gone horribly wrong:

Wrong! Wrong! Wrong!  The voltage from the diode dropper is not the average at all. As we have seen from the detailed analysis in the diode dropper article here, the voltage will be 170V rms, when the supply is 240V. The introduction of the diode halves the power, not the voltage. Looking at the heater string, there are three valves in series which require 91V at 300mA. With the diode in circuit, and our supply now 170V rms, the resistor should be (170 - 91) / 0.3 = 263 ohms. A far cry from the 100 ohms specified! For the 100 ohm resistor to be correct, the heater string would have to add up to 210V. Incidentally, the correct 263 ohm resistor will dissipate 24W. Turning next to the CRT, the designer has made just as much of a blunder:


We have a 24V transformer secondary which is to feed the 6.3V 900mA CRT heater. Why a 6.3V winding was not specified is just bizarre. Instead, the designer has worked out ways to get rid of the excess voltage with a diode, a thermistor, and a resistor. A fuse is erroneously included in case the diode shorts. So much extra work (and inefficiency) when a simple 6.3V winding would have avoided all this. Now to the calculations:
24V AC fed into a diode dropper will result in 17V rms output; not 12V. The 1 ohm thermistor will drop 900mV, which means a total of 7.2V needs to be supplied to the CRT + thermistor. With 17V rms to start with, the dropping resistor should be (17-7.2) / 0.9 = 10.9 ohms.
This is actually close enough to that specified, so in this instance the CRT will be OK, although maybe just a fraction over-run, depending on the actual heater current. However, the designer's calculation to get to this value is incorrect. 900mA will result in 9.9V drop across the 11 ohms, consisting of the thermistor and 10 ohm resistor. 9.9V + 6.3 = 16.2, not the 12V assumed. He's got the power rating of the resistor right though.

Colour CRT heaters are not designed for series heater use, so there may be some slight variation in current, and the resistor should really be selected on test. The fuse to protect the heater in case of the diode shorting won't do much good. The maximum short circuit current with just the 11 ohms of resistance in circuit will be 2.18A. A 1A fuse will certainly not blow immediately. But, the current with the CRT heater in circuit will be quite a lot less than this, and it could be hours before the fuse blows. "What a nice bright picture!" exclaim the viewers, as the CRT heater is being fed something closer to 12V, happily stripping the cathodes.  Why didn't the designer just put a reverse biassed shunt diode across the heaters to protect them?

On to the other interesting features of this power supply:

Wrong Again!
It seems readers did notice the error, because in Part 13, April 1973, mention was made of their concerns. Alas, the designer seems taken aback that his theory could possibly be taken as incorrect. Yet again, he is insistent that the rms voltage is simply halved. He keeps going on about mean voltage, instead of power. He has got it right in that the area under the curve is what heats the valves, and that removing one half cycle removes half the power. The error is in assuming that half the power across a given resistive load equates to half the rms voltage. Let's see how's he's fallen into this common trap.

An example: A 12R resistor is connected across a 12V supply. It draws 1A and the power is 12W. These values are the same for DC or  AC rms.
Now, our designer thinks that if the power across the resistor was halved, so would its voltage. So, for 6W dissipation, what will the voltage across the 12R resistor be? (The designer would think 6V, but we'll see in a moment that is not correct).

P = V^2 / R. Therefore, 6 = V^2 / 12.
This means that V^2 = 72. The square root of 72, and thus V, is 8.49V.
We see that the voltage is not halved at all!

And guess what? 8.49 / 12 = 0.707. If you remember from the diode dropper article, you will recall that the voltage fed into the heater chain was 0.707 times the supply voltage. Here, I've arrived at the same figure, but by a different means.

Put simply, the mistake is in assuming a linear relationship between power dissipation and voltage. He's forgotten all about I^2 x R. Power dissipation vs. voltage is not linear.

As an example of how to do it correctly, have a look at the calculations for the Thorn R2M television. A set made in the thousands by a reputable manufacturer. And the valve heaters all work at the correct voltage!

There's just one more little mistake to wrap up his analysis of the heater string. Regarding the drop across the heater rectifier diode; "about 0.1V at 300mA". Oh dear...not terribly brushed up on the characteristics of silicon diodes are we? Fortunately, whether it is 0.1V or 0.7V will have no practical effect in this application, but just that statement shows a lack of knowledge.

Also mentioned was the CRT heater supply. He did admit the 1A fuse was futile by saying, "This was dropped because of favourable operation without it". But his 10R heater resistor was justified on the basis that the mains voltage might be 10% high, and the tolerance of the mains transformer might also be of 10% tolerance, thus "the transformer might rise to 29V instead of 24V". He actually thinks the CRT heater is being underrun "in order to offer some protection". Actually, under running the heater on a CRT is more detrimental, due to cathode poisoning, to say nothing of reduced cathode emission. Does he think the heaters will burn out? Well, actually heaters of indirectly heated valves are quite forgiving of excess voltage. Take car radio valves for example; operating on closer to 7V than 6.3V. No, the CRT heaters aren't going to go open circuit with normal mains and transformer voltage variation. Think of CRT brighteners; it's not the heater burning out that eventually kills the tube, but no emissive left coating on the cathode. However, with all that said, it was just fortuitous that his erroneous theory did end up with close to the correct operating conditions, although ironically, slightly overrun!

Correction at last.
In August 1973, (20 months since the circuit was first published) we finally see the designer has realised the error; or more likely, notification of the error continued to come in, and notice had to be taken. Perhaps some readers were disturbed by valves glowing like light bulbs. In the last sentence for the article on the receiver project, it simply reads; "The correct specification for R501 is 260 ohms 25W".
Then in September 1973, readers are reminded again of the correction, and this time an RS components part is specified.

However, there's still no word on the error of the original calculation.

The matter is finally put to rest in April 1974, with an article titled, "The Diode Dropper". This went into the calculations in some detail, and somewhat coincidentally, the colour TV project was used as an example. It's quite likely that the author was one of those who questioned the original design.


2. Headphones Connected to the Mains.
For many years, circuits have been shown in magazines and on internet sites of radio receivers using a transformerless power supply for the B+. What the designers appear to be unable to understand is the lethal earphone/headphone connections.

This example was from a Japanese site and shows a one valve VHF receiver. While some of the design aspects are questionable, the important point here is that the crystal earphone is connected directly to the chassis, which in turn connects to the 100V power mains via the bridge rectifier. The earphone will always be live, regardless of how the mains plug is inserted.


Earphone live at 100V AC.
 

The book, "Radio for the Millions", is a compilation of projects which have appeared in Popular Science. This publication presents a pot-pourri of deadly circuits, typical of those often described in U.S. magazines. The following is a typical example:


Would you be comfortable with 120V on your head, let alone the front panel?

In this set, we see the aerial connected to the 110-120V power mains with no isolation. Furthermore, the metal panel which forms the front of the set is connected directly to the mains. The headphones connect straight to the B+, which in turn connects to the mains supply via a half wave rectifier. Regardless of which way the mains connected, the headphones will always be connected to the mains, either through the rectifier, or the 25B8 pentode. More than enough lethal current can flow by either of these paths.


3. Non-isolated Power Supplies.

Also from "Radio for the Millions", this circuit invites the constructor to connect their portable radio to the mains without any isolation.

What of the external aerial connection that the radio might have, exposed screws, or other metallic parts connected to the chassis?



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