Sinclair ZX Spectrum clone (CIP-03) power supply - Page 1
Forum section: electronics / general

author: NewBeginner
author_rank: Regular Contributor
timestamp: July 31, 2012, 08:10:20 pm
content: Hi,I decided to restore my old Sinclair ZX Spectrum clone (called CIP-03 -http://www.old-computers.com/museum/computer.asp?st=1&c=700).I intend to start with the power supply unit. I managed to get a schematic (attached image) of the original power supply.But I noticed that the overvoltage protection elements are missing from my unit.They appear on the PCB's silkscreen but the PCB was not populated (probably to reduce costs) with this elements.Since I am restoring the power supply I would also like to include the overvoltage protection elements.I have a few questions in order to make sure that I correctly understand what is going on in the schematic.What is the role of R2 and P1 resistors?Is it to lift or lover the output voltage in order to allow adjusting the sensitivity to the overvoltage protection? Lifting the output voltage will bring it closer to the protection triggering point.Or does it have another role?Since the BM323 (LM323 equivalent) doesn't seem to require an external minimum load in order to operate properly, what is the role of R1 resistor?My main concern is about the D3 zener diode value (5.6V). If I understand it correctly the protection triggering level would be at the Vz + Vbe of the PNP transistor.Since the output of the LM323 (equivalent of the BM323) can go as high as 5.25V, the total output at point (6) will be 5.25V + the voltage over the R2 and P1 resistors.My concern is that if the zener will start to conduct bellow 5.6 (nominal value) the protection circuit will trigger under normal operation when it shouldn't.How can I calculate the minimum voltage at which the zener will start to conduct? Can Vzk of the zener be calculated based on datasheet values?I would really appreciate if somebody could help me understand how can I calculate more precisely the worst / best case scenarios for the overvoltage protection circuit.Thank you

author: Bored@Work
author_rank: Super Contributor
timestamp: July 31, 2012, 09:39:53 pm
content: R1, R2, P1 are all needed to trim the output voltage upwards +6%. That should be enough to trim most, but not the extreme underperforming (< 5V) 323s to 5V. What they can not do is to trim a 323 delivering more than 5V down to 5V.Not being able to trim down maybe the reason why they skipped the protection circuit. Maybe it was too delicate. But you have to study the zener and PNP transistor data sheets in detail, and do some calculations for the worst case to get an idea if the protection circuit is too delicate or not.

author: NewBeginner
author_rank: Regular Contributor
timestamp: August 01, 2012, 09:04:46 am
content: Thank you for the clarification.QuoteWell, this is exactly where I need help. Unfortunately I couldn't find much info about zener diodes other than the basic usage.As I said in my previous post I don't really understand how to do this calculations. I would really appreciate is someone could help me in this area.Thank you

author: PA0PBZ
author_rank: Super Contributor
timestamp: August 01, 2012, 09:37:30 am
content: Quote from: NewBeginner on July 31, 2012, 08:10:20 pmThe way I see it is that it's just the 5.6V of the zener. As long as it is not conducting the Vbe of T1 will be zero, but when the zener starts conducting (at 5.6V), T1 gets enough Vbe to open and trigger Th1.

author: Balaur
author_rank: Unknown
timestamp: August 01, 2012, 09:52:52 am
content: Unless you are going for a full restoration, I cannot see why you would like to keep that heavy black communist box instead of modern capitalist imperialism multi-amp 5V switching PSUs.I have very fond memories about myHC85and I'm always hoping to repair it (mysterious and sudden fault) but I cannot be arsed about this because I know that I'll lose interest 5 seconds after repair.Cheers and good luck,Multe salutari,Dan

author: NewBeginner
author_rank: Regular Contributor
timestamp: August 01, 2012, 10:47:55 am
content: @PA0PBZAre you saying that the trip point for the protection circuit is given only by the zener voltage (5.6V)? Because I don't think that is quite correct.I think that the output from the LM323 needs to be at about Vz + Vbe for triggering the protection circuit.Could you describe it in more details?Here are some reference circuits I found related to this:http://axotron.se/index_en.php?page=26http://www.maxim-ic.com/app-notes/index.mvp/id/760My main concern is that those values are typical values and they can vary, and I don't understand yet how to calculate the worst case scenarios.@BalaurWell, I do intend to do a full restoration(not really sure what you mean by that). I too have lots of good old memories about my CIP-03.Also, since I am a beginner in electronics, I think that the task of restoring it may help me learn more about electronics.I too will probably lose interest for it after restoration is finished. Playing on it is not my main goal.I consider it more like a "piece of history"so I would like to keep the original design as much as possible.Thank you,Multumesc

author: PA0PBZ
author_rank: Super Contributor
timestamp: August 01, 2012, 11:34:07 am
content: Quote from: NewBeginner on August 01, 2012, 10:47:55 amAh yes of course, while the zener will trip at 5.6V you still need a voltage drop of Vbe over resistor R4, my badSo it is safe to say that the protection would turn on somewhere between 6.15 and 6.2 volts, and that is probably what you want.

author: SeanB
author_rank: Super Contributor
timestamp: August 01, 2012, 03:45:59 pm
content: The idea of the trip is that it will hopefully work if the regulator dies and goes short circuit input to output. Thus if it only operates if the output is over 6-7v then it is fine, because the input to the regulator is probably around 10V. Only way to test is to replace the fuse with a 10R resistor, and use a bench power supply via another 10R resistor to feed the output and pull it up till it trips. TTL though is pretty tough, it will handle overvoltage very often with no problems, so 7V is not going to be likely to kill it.

author: NewBeginner
author_rank: Regular Contributor
timestamp: August 02, 2012, 09:08:52 am
content: @PA0PBZ Thanks for the clarification@SeanB Yes, but is there a way to calculate the worst/best case scenario? I don't know yet how to calculate this.The input to the regulator is about 13V.Thank you

author: amyk
author_rank: Super Contributor
timestamp: August 02, 2012, 10:32:30 am
content: Isn't D6 backwards in the schematic?Quote from: SeanB on August 01, 2012, 03:45:59 pmThe CPU is NMOS which isn't quite as tolerant as TTL, although still tougher than modern CMOS. The Z80 datasheet lists 7V as absolute maximum.

author: NewBeginner
author_rank: Regular Contributor
timestamp: August 02, 2012, 11:12:30 am
content: Yes, D6 LED seems to be connected backwards.

author: SeanB
author_rank: Super Contributor
timestamp: August 02, 2012, 03:39:04 pm
content: 5.6V across the zener, about 0.6v Vbe on the transistor. Generally the zener has some leakage, thus the low value of R4, so as to hold the transistor off until the zener is conducting around 1.8mA. Then the transistor turns on and then it will supply enough current ( more than 20mA is needed, and this will flow with a base current of 200uA or less) to turn on the SCR. Thus it will operate at least between 5.6( worst case on the zener and a really hot transistor) and 6.5V ( mid winter in Siberia and the highest tolerance zener) in all cases. Can be a problem only in hot places or if the unit is too hot, when it will blow the fuse and shut down gracefully.

author: Zad
author_rank: Super Contributor
timestamp: August 02, 2012, 05:20:20 pm
content: I wonder why they changed the circuit so much, it is nothing like the original ZX Spectrum's power supply:http://www.1000bit.it/SUPPORT/SCHEMA/zxspectrum/spectrum_sch_2.gifSo far as I remember, inside the PSU block was just a transformer+ rectifier / cap. Certainly no LED, that must have been a high spec Russian edition!

author: SeanB
author_rank: Super Contributor
timestamp: August 02, 2012, 05:53:42 pm
content: Probably were trying to get rid of the hot spot on top of the original units, where the poor 7805 sat on a tiny bent plate heatsink under the keyboard membrane..

author: NewBeginner
author_rank: Regular Contributor
timestamp: August 02, 2012, 07:12:22 pm
content: @Zad Actually the CIP-03 is a Romanian clone of the ZX Spectrum.@SeanB Could you please describe it in more details how did you calculated those values? I still have trouble understanding the calculations.Thank you

author: SeanB
author_rank: Super Contributor
timestamp: August 02, 2012, 07:49:49 pm
content: Zener is 5v6 nominal, and with a 5% tolerance ( typical unless you pay for a tighter selection) this means it will be from 5.4 to 5.9V voltage drop. Higher or lower and it would be binned as a different voltage band, 5.1V or 6.2V respectively. Transistor Vbe is around 0.6V, and can be considered to be reasonably constant. Calculated the current at 0.6v drop across the resistor R4 at around 1.8mA, so this is the current roughly that must flow through the zener ( actually a little lower but not by much) before it will start to turn the transistor on. The transistor will turn on the SCR when it is able to supply enough gate current, which is typically 20mA, and the gate resistor will mean you need a little more, but not much more, it is there because the transistor will be leaking a little current all the time, and more if it is hot, so the resistor is just to ensure that the transistor has to provide a lot more than leakage before tripping occurs. Thus the trip point is somewhere between 5.4+.2V( really hot and leaky transistor that is already passing a lot of leakage and just needs a small push) to 5.9+0.6V( Siberian winter with cold power supply) before it trips.Just remember this trip does not protect under all faults. A shorted rectifier diode will not blow the fuse, but will burn out the transformer in short order. It does though protect against the common fault of a shorted regulator quite well, by blowing the fuse. It can be defeated if the fuse is replaced with one of too high a current ( also known as a nail) which can burn out the SCR, though they either fail short or open circuit, depending on the current and time.

author: Balaur
author_rank: Unknown
timestamp: August 02, 2012, 08:14:17 pm
content: Quote from: NewBeginner on August 02, 2012, 07:12:22 pmA very simple (and idealistic) approach to the issue is the following:- If for some reasons the output voltage goes over the 5V6, the Zener starts conducting- The voltage on the R4 resistor is the difference between the output voltage and the 5V6- If the R4 voltage > 0.6V, the transistor T1 becomes active (BE junction is forward biased, BE voltage = 0.6V)- The T1 collector will go at a high potential- The thyristor Th1 will trigger and latches until the fuse Si 2 gives up- User replaces the fuse Si 2 with a nail and continues to play Dizzy V happily- Magic smoke comes out from the transformer and the unit is replaced with a 24V transformer stolen from industrial equipment and a junk regulator based on IPRS-made 2N3055s- User is happy and finally finishes R-Type before moving to better and bigger games such as Robocop or Saboteur IIThus, in this very simplistic scenario, the threshold voltage from which everything goes south is 5.6V (on the Zener) + 0.6V on R4 = 6.2VLayers of complexity can be added:- 5.6V is not nailed down, you should look at the I-V curve of the Zener and also take in account its tolerance- R5 will steal some of the voltage from the T1 BE junction. Not that important here anyway- The thyristor Vgk indicates the required voltage on R6. From here you can compute the current through the resistor, which is provided by T1- The T1 transistor has some finite beta factor that will dictate how much current you need through the base (and R5) to reach the required Ice current.In all, a peculiar and communist implementation choice.Cheers,Dan

author: NewBeginner
author_rank: Regular Contributor
timestamp: August 03, 2012, 08:45:34 am
content: Thank you guys.@SeanB I think I understand nowOff topic:@BalaurI don't see how this can be a "communist implementation". I mean it is a linear power supply with a crowbar over-voltage protection circuit. As far as I know linear power supplies are still in use today.Maybe they are not so widespread as they used to be but they are still in use.Yes, the CIP-03's power supply design is a 20 years old design, but I don't think that this design was specific to communist countries.Anyway, as I said before, my goal is to restore the unit and hopefully learn some electronics in the process.Thank you

author: Balaur
author_rank: Unknown
timestamp: August 03, 2012, 09:34:05 am
content: Quote from: NewBeginner on August 03, 2012, 08:45:34 amEh, most of my communist remarks were told jokingly. Of course there is nothing wrong with a linear power supply; it's a very useful concept and the application domains are endless.As for the protection circuitry, why not. Too bad that most of the time (always?) this section was not populated.(Still joking) I developed a (healthy in my cynic opinion) disdain for our beloved past, because I imagine that I know too much of the communist electronic industry & academy: mass production of electronic engineers, 50% of very competent teachers, 50% of copy-paste ones, 90% of students that gave no f***s about their profession; good intentions but very poor implementation; competition between the several electronic companies in communist Romania was a joke, leading to vastly similar and poorly performing products, centrally-taken terrifically bad decisions leading to zero personal responsibility and involvement and so on. There have and there are wonderful people and professionals, but I fear that the general environment towards the end of the regime was strongly disparaging. Sadly, the echoes of this epoch are still with us today.Smilingly cheers,Dan

author: SeanB
author_rank: Super Contributor
timestamp: August 03, 2012, 02:55:58 pm
content: I used to work on equipment that had similar crowbars, but these omitted the transistor, and used a 6V2 diode straight. A lot of times the thyristor quietly failed open circuit, and the one time it was needed on a unit it was silently faulty, and was happy to allow the unregulated 14V rail through the pass transistors. Funny enough the box of TTL ran more or less normally at this voltage, there was only one SN5403PM3 that objected and popped it's lid. Funny in being the only chip with open collector outputs and they were guaranteed to 30V on the output.

author: NewBeginner
author_rank: Regular Contributor
timestamp: August 05, 2012, 05:03:59 pm
content: @Balaur I understand you@SeanB I never worked with thyristors until now. Are they so unreliable in general?Thank you

author: SeanB
author_rank: Super Contributor
timestamp: August 05, 2012, 05:22:15 pm
content: Generally no, but these were sitting for years with power applied, and were subject to multiple cycles and vibration. Thus any failures that were open or poor sensitivity never were weeded out, and, as they were in a stud package made in the 1960's, they mostly died from old age. Generally reliability has improved with process changes and die shrinks over the years, and the diffusion processes are now a lot better understood and controlled.

author: NewBeginner
author_rank: Regular Contributor
timestamp: August 06, 2012, 08:51:29 pm
content: I see. That's a relief.Thank you