Basic , with Relay's Code Circuit Diagram...

[ljcox]

Firstly , Error Relay Coil parallel appr. 4700uF Condensator required.

Why do you need a capacitor that large?

What is the resistance of the relay coil?

 

[HD ELEKTRONİK]

Why do you need a capacitor that large?

What is the resistance of the relay coil?

If any Error Buttons press.For this is delay time required...
(appr.2200-4700uF--> For 5sn (appr.) delay time to get).

 

[ljcox]

Here is the circuit with additional error logic so that circuit will be reset if B1 or B4 are pressed out of sequence.

I have changed the name of relay E to RS as it now does a reset function even when the correct sequence of buttons is pressed.

The release time of RS (determined by C2) allows time for the release of any operated relays.

 

[Leftyretro]

What a mess relay logic can quickly become. I was always amazed that the telephone companies switching relay logic worked as well as it did. Seems to me I read an early 'computer' was built using relay logic during or right after WW2.

Relays are still very useful. They have great isolation (input/output) and can be sized for crazy voltages and current levels. So they are still very viable as a final control element, but as a logic solver?

Get real, a simple micro-controller driving the door solenoid via a Mosfet is so much simpler, better, cheaper and programmable that it's painful to even have to explain.

Lefty

 

[ljcox]

You're overlooking one distinct advantage relay logic has over electronic logic.

In its quiescent state, it consumes ZERO energy. Therefore it could be run from a battery that would only have to be replaced every 2 ~ 3 years.

How long would an electronic one run on a battery?

The CMOS one I posted above may last quite while (as it is a combinational lock, there is no oscillator running or other power consuming items), but I'm not too sure about a PIC version.

Besides, the op specified a relay version - I don't know why - perhaps it is a school project.

The first "computer" - the Colossos - was built in England during WW2 to assist in decoding German cyphers was partly electronic and partly relay logic.

 

[HD ELEKTRONİK]

1N4007 required...

**broken link removed**

 

[HD ELEKTRONİK]

Ok...
Now identically for X4 Relays delay time required.(approx. 1-3sn)...

 

[Nigel Goodwin]

The first "computer" - the Colossos - was built in England during WW2 to assist in decoding German cyphers was partly electronic and partly relay logic.

And was then dismantled and the parts returned to the Post Office - an amazing story!.

Britain had a massive world lead in computers and all from a single guy who worked for the Post Office, and 'stole' all the parts to build it!

I'm sure I read somewhere a bit back that they have built a reconstruction of it now?.

 

[ericgibbs]

And was then dismantled and the parts returned to the Post Office - an amazing story!.

Britain had a massive world lead in computers and all from a single guy who worked for the Post Office, and 'stole' all the parts to build it!

I'm sure I read somewhere a bit back that they have built a reconstruction of it now?.

hi Nigel,
He [Tommy Flowers] actually spent a £1000 of his own money as well as the 'borrowed' bits to build the first one,
which after many years the Govt paid him back..
The other 9 Collososii were built with Govt funding.

Mr W Churchill at the end of the war ordered all the 10 to be dismantled and the documentation to be destroyed..
he was concerned it would fall into the hands of the USSR.

The working reconstruction was shown working on a recent TV program...

 

[ljcox]

1N4007 required?

No

This is a standard technique used to increase the release time of relays.

But a diode is not used in that position as it is not necessary.

 

[ljcox]

Ok...
Now identically for X4 Relays delay time required.(approx. 1-3sn)...

X4 is operated for the time that the B4 button is pressed, so I don't see why you need to increase its release delay.

 

[ljcox]

hi Nigel,
He [Tommy Flowers] actually spent a £1000 of his own money as well as the 'borrowed' bits to build the first one,
which after many years the Govt paid him back..
The other 9 Collososii were built with Govt funding.

Mr W Churchill at the end of the war ordered all the 10 to be dismantled and the documentation to be destroyed..
he was concerned it would fall into the hands of the USSR.

The working reconstruction was shown working on a recent TV program...

The reconstruction was driven by Tony Sale. Some years ago I looked at their web site and obtained some circuits on a CD that Tony sent to me (there was a modest cost).

They produced 2 input AND gates by connecting the inputs to the grid and suppressor grid of a pentode valve. And XOR gates using 2 AND gates and a wired OR.

The German teleprinter cypher (code named Tunny) was transmitted via the international standard teleprinter code but the text was encoded using relay XOR gates and switching wheels.

The code to be decyphered was stored on a loop of paper teleprinter tape (5 bit words). It rotated very quickly around a series of pulleys and was read by photo cells.

Apparently, the tape broke occasionally - with spectacular results.

 

[HD ELEKTRONİK]

X4 is operated for the time that the B4 button is pressed, so I don't see why you need to increase its release delay.

If B4 button 0,1sn pressed...
Then X4 0,1sn operated and DS 0,1sn active.
(DS --> 1-3sn active to be).
...
So ;
If B4 button 10sn or more pressed...
Then X4 10sn or more operated and DS 10sn or more active...
This is a problem for DS...(DS is break down !).
...
.............................................

Series resistor (82 Ohm) is only for charge required.

**broken link removed**

 

[ljcox]

If B4 button 0,1sn pressed...
Then X4 0,1sn operated and DS 0,1sn active.
(DS --> 1-3sn active to be). So you want to limit the time that DS is energised regardless of how long B4 is pressed. If so, then you will need to do more than just making X4 slow to release....
So ;
If B4 button 10sn or more pressed...
Then X4 10sn or more operated and DS 10sn or more active...
This is a problem for DS...(DS is break down !). Yes, DS is likely to over heat and burn....
.............................................

Series resistor (82 Ohm) is only for charge required. Yes, but it makes little difference to the release time. As I said previously, this is a standard technique used to increase relay release time.
I have never seen a diode across the resistor.
What you are missing is that the resistor increases the time constant and may in fact sligthy increase the release time compared to the case where the diode is included.

It is a complex situation. The initial voltage across the coil will be 12 Volt, not 11.3 as you have shown since, due to the coil inductance, the current does not change when the button is released.

In your second diagram, ie. where there is no diode, the voltage across the coil will be 12 Volt and the voltage across the resistor will be I * R where I is the coil current and R is the resistance of the resistor. For example, if the coil resistance is 600 Ohm, the current will be 12/0.6 = 20 mA.

If R = 82 Ohm, the voltage across it will be 20/1000 * 82 = 1.64 volt.

 

[ljcox]

This is what I mean

Oops I made an error. See the revised diagram.

 

[HD ELEKTRONİK]

**broken link removed**
**broken link removed**

 

[ljcox]

You did not look at my latest edit. The initial relay current is 20 mA.

My error was that I forgot to adjust the relay coil voltage.

It is 10.36 Volt which obeys Kirchoff.

I expect your next point will be - that is not 12 Volt, so the relay delay will be less.

But the magnetic force on the armature is proportional to the current, not the voltage.

My diagram shows that the current does not change (due to the nature of inductance) when the switch is opened. It decays from 20 mA.

So there will be little difference in the release delay. In fact it may be slightly longer since the capacitive time constant is a little greater due to the 82 Ohm resistor.

It is not possible to do this by intuition. You need to solve a second order differential equation and include the relevant initial conditions.

See the attachments. Start at the heading "RLC Transient" near the bottom of page 247.

This gives you an idea of what I'm trying to say. In your case, the initial conditions are:-
1. the capacitor has an initial charge of 12 Volt
2. the inductance of the relay coil has an initial current of 20 mA (assuming a 600 Ohm coil resistance).

So these need to be included in the mathematics.

 

[HD ELEKTRONİK]

Your Theory ...

**broken link removed**

Please check again...

 

[ljcox]

Please check again...

You have added the capacitor and resistor voltages.

The arrows show the polarity. Therefore the voltage across C1 is 10.36 + 1.64 = 12 Volt. As I said previously.

The point you're missing is that the back EMF of the coil is 1.64 Volt

I have simulated the circuit using Switcher CAD III.

The relay coil is simulated by R3 and L1. I have assumed that the inductance is 1 Henry. I have use a transistor to simulate the switch.

The Red curve is the current through C1. It is negative when the capacitor is charging.

The Green curve is the transistor base voltage.

The Blue curve is the current through the coil.

The Pink curve is the collector voltage.

Note that, while the transistor is on, the coil current is 20 mA and the C1 charges to 12 Volt.

When the transistor is turned off, the coil current decays from 20 mA.

 

[ljcox]

Here is a diagram showing the relay coil voltages at the instant that the switch is opened.

For simplicity, I have not shown the capacitor and 82 Ohm resistor that are connected across the coil.

It shows that the back EMF is 1.64 Volt and this is driving the current. The back EMF is of such a value as to keep the same current flowing that was flowing just before the switch opens,

The coil can be considered to be an inductance in series with a resistance.

EDIT:
Here is a more accurate wave form chart. The transistor was not switching fast enough, so I had to change some parameters in the simulation.

Notr that the C1 current is equal to the L1 current immediately the transistor is turned off - as expected. They both decay from 20 mA.