Chess challenger overheated!

[microhenri]

The Fidelity SC9, with the 6502 processor, which also runs on 5 volts,
works fine on my regulated 6 volt adaptor
L

I did try, but my CC10c does not start up with a regulated (DC) Voltage. Also not on 9 Volt DC.
Does this computer also have a power supply circuit that uses the AC voltage to make -5V and +12V? I found a circuit diagram of the CC1 proto type that showed such a circuit.
Kind regards, Henri

[mychess]

Hello,

The µA7805 (regulator on CC10) need 7 volts DC for supplied 5V DC.
So the transformer needed for this 7 volts DC is (7+2*0.7)/2^(1/2)=5.9 AC, not DC (0.7V are lost for each diode, and 2 are always used in a Graetz bridge).
So a 6V AC will be working.

Cordialy.

[microhenri]

The µA7805 (regulator on CC10) need 7 volts DC for supplied 5V DC.
So the transformer needed for this 7 volts DC is (7+2*0.7)/2^(1/2)=5.9 AC, not DC (0.7V are lost for each diode, and 2 are always used in a Graetz bridge).
So a 6V AC will be working.

Hi All. Slowly with all information I get from you all I getting things more clear. This is what I like… Thanks.
Now I know a uA7805 is used, I did do some temperature calculations. When the heat sink is the same as the CC10B, then I did do the assumption that the temperature resistance of this heat sink is about 7 °C/W. With the information of the datasheet I can calculate the maximum temperature allowed inside the wooden case of the Chess Challenger. At Junction temperatures of the uA7805 higher than 150°C will reduce the current (and also the output voltage) and the system will become unstable.

Here the calculations:
Current consumption 0.85 A
Tj max 150 °C
θJP of the uA7805 3 °C/W
θHA of the heat sink 7 °C/W

I did calculate with input voltages of 6V (minimum), 9V (according to spec) and 10.5Vac (as I did measure in my situation).
Total Power consumption:
@6V = 5.1VA
@9V = 7.7VA
@10.5V = 9VA
Power consumption of the power regulator
@6V = 1.8W
@9V = 5.4W
@10.5V = 7.2W
Temperature increase of the uA7805 junction (internal chip temperature) so ambient temperature has to be added to get the real temperature.
@6V = 17.7°C
@9V = 53.8°C
@10.5V = 71.8°C
Maximum allowable air temperature inside the Chess Challenger casing before the uA7805 internal temperature protection is activated.
@6V = 132.3°C
@9V = 96.2°C
@10.5V = 78.2°C

This all means that increasing the input voltage the computer will consume more power and will therefore increase the temperature inside the casing of the computer. The power regulator has to dissipate this extra heat. Therefore the maximum allowable heat sink temperature is lower. The chance that the junction temperature (chip temperature) gets higher than 150°C is increased.

Conclusion
Increasing the input voltage works two ways on the temperature. It does increase the internal air temperature and does decrease the allowable air temperature inside the chess challenger casing.
Lowering the input voltage will work two ways in the right direction.
Worth to try…
Thanks all
Kind regards Henri

[microhenri]

Hi All. Here an update with new answers and new questions.

Previous posts where almost pure theoretical. It is now time to share my practical experiments with you that I did after some valuable tips from you guys.

First I found beautiful pads and added them underneath the CC10c. The size is perfect and the color suites the CC10c very nice. The pads are normally used underneath furniture to protect the wooden floor. When you did not know they were new, you would believe they have always been there.

Lowering the AC voltage did not work quite well.
I have an 18W DC adapter with a sliding switch to choose voltages for 3V up to 12V in steps of 1.5 Volt. The transformer inside this adapter does have a series of windings. With the switch you can choose another winding with a lower or higher voltage. After the rectifier and capacitor you get the wanted DC voltage on the output.

I did solder two wires on the two AC inputs of the rectifier. This is after the switch so I can choose an AC voltage from 3 to 12 Vac. This was the theory. In practice the voltages differ a little. I did measure the output voltages with a load of 850mA. With the switch on 6V the output voltage was 8Vac. With the switch on 7.5V the output voltage was 9.3Vac.

Now I did connect the CC10c. This is the hard part, because I do not have the special female connector.

With 9.3Vac input voltage the CC10c did start up. Great, that works.
With 8Vac the CC10c failed to start up!!! This was not what we expected!

I did look at the datasheet of the 8080A microprocessor. What I did find was that this uP needs beside the +5V also a -5V and +12V.

When I use the formula provided by “mychess” to calculate the rectified DC voltage inside the CC10c then I come to the next finding:
9.3Vac input voltage gives a rectified DC voltage of 11.75V.
8Vac input voltage gives a rectified DC voltage of 9.9V. This is way below the specified 12V +- 5% that the 8080A needs. I now think this is the reason the processor does not start at 8Vac input voltage and below.

What about the 10.5Vac of the standard transformer. This gives 13.5Vdc instead of 12V. This is too high for the 8080A or is there some kind of circuit that regulates this voltage.
Does someone have a schematic diagram of this chess challenger?

Kind regards,
Henri

[Fernando]

MAN, READING YOUR TECHNICAL POST I felt like reading a paper by Einstein

[mychess]

Hi all,

There is a 78L12ACDB regulator in the CC10c, with small heat sink.
This regulator has a 1.7 V dropout voltage.
For the -5V, probably a simple zener only, very small current is necessary (0.01mA for the 8080).

Happy new year for all readers.

[Brutus]

Replace the voltage regulator 7805 or LM7805 or LM7805C (this is the one with the three legs getting too hot) with this one from TRACO:
TSRN 1-2450 or TSRN 1-2450A
6.5–42V DC to 5.0V DC @1A

http://www.produktinfo.conrad.com/daten ... 1_2465.pdf

The unit will stay cool, it will not require the heatsink anymore.
New technology regards, Wolfgang

[microhenri]

Hi all
@ myChess: Great. Thanks again.
My CC10c works fine now. No overheating problems. So far so good!!!

@Wolfgang: Thanks for the tip.
When the problem occurs again in the future than I can try this “2017” solution!

Kind regards,
Henri.

P.S. Schematic diagrams or clear pictures of the Printed circuit boards of these early Chess Challengers are very welcome.

[Scally]

Hi all,

I've just been reading older posts on Chess Challenger Power Supplies as I tested all of mine yesterday.

All my 9v power supplies are supplying between 16.0 and 16.2 volts which seems very high...

1) Could this damage my Chess Challengers or is this a false reading as they are not under load ?

2) Could this 16v be produced as the power supplies have been simply fitted with a UK plug, we have twice the US voltage ?

3) is the Voltage regulated in the Chess Challengers ?


Thanks in advance

Al.

[microhenri]

Hello Alan

16Vac is indeed high. Did you measure AC or DC.

The voltage in the UK is 240Vac. In the Netherlands we have 230Vac. When I measure the secondary voltage of the transformer without load it is about 10.8 Vac (10.5 Vac with load). In the UK (at 240Vac) my transformer should have a maximum voltage of 11.4 Vac, not 16Vac.

Yes, the Chess challengers do have regulators but they are linear. This means that when you increase the input voltage, these linear regulators have to dissipate the extra power. This is not a big problem with the CC7 that only consumes 240mA. The older models do need more power, and will get hotter.

When you use an American power supply than you should use a 240-110 V transformer to lower the voltage.

Kind regards, Henri

[paulwise3]

Hi Henri,

Nice to see a fellow dutchman on the forum. Do you live near Eindhoven?
I do anyway

Best regards,
Paul

[Scally]

Hi Henri,

Thanks for your reply.

I assume I'm measuring DC, I'm using a multimeter switched to 20 volts. I used the same setting to test a new battery and it showed 1.6 volts, another 12v power supply shows 12.2 volts, yet all my CC power supplies show 16v.

I have an U.S to U.K converter plug as I'm aware of the differences, I was just wondering what would happen if an American plug was removed and a U.K. plug put on in its place before I bought them, as we're twice the voltage and 16 is nearly twice 9, could that be the problem?


Guessing,

Al.

[microhenri]

Hello Alan.
What is your adapter tell you about the output voltage:
When it is a DC adapter then it should show for example 9V= or you see a + or – sigh.
When it is an AC adapter then the inscription is for example 9V~ or 9Vac.
My guess is that you are using a DC adapter. Then the 16Vdc is normal. When I take the 11.4Vac as mentioned before and calculate the voltage after rectifying to DC then it is approximately 11.4*1.414=16.1Vdc.

Kind regards,
Henri

[microhenri]

Hi Henri,

Nice to see a fellow dutchman on the forum. Do you live near Eindhoven?
I do anyway

Best regards,
Paul

Hello Paul,
Yes, I live close to Groningen. (Small country )
Henri

[paulwise3]

Hi Henri,

Nice to see a fellow dutchman on the forum. Do you live near Eindhoven?
I do anyway

Best regards,
Paul

Hello Paul,
Yes, I live close to Groningen. (Small country )
Henri

Too bad . Almost as far as possible...
I sent you a PM with a few questions.

Best regards, Paul