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pulse output

eTech

Active Member
Now I’m also confused. How did “pulse output” turn into power supply design.(?)
Perhaps the OP should re-state what he/she is trying to accomplish?
 

gophert

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Most Helpful Member
Now I’m also confused. How did “pulse output” turn into power supply design.(?)
Perhaps the OP should re-state what he/she is trying to accomplish?
I'm also confused that the relay coil is external tot he relay and it looks like the case of the relay is conductive and shorting the ground with the positive snd negative rails.
 

arivel

Member
What are those bridge's "internal diodes"?
I was referring to one of the diodes in the rectifier bridge that you see immediately after the transformer

Why do you have current flowing backwards (orange line). Current flows FROM the "ground" TO the V-.

View attachment 128906
this is an absolute novelty for me :oops:. of course I disagree. if it were as you say there would be a short circuit on diode 1 in addition current would flow in the coil continuously through the internal diode (diode2) of the half bridge H or of the push pull darlington at this point the driving circuit becomes useless.

Now I’m also confused. How did “pulse output” turn into power supply design.(?)
Perhaps the OP should re-state what he/she is trying to accomplish?
I have already explained in post number 7. I thought I was clear enough but I see it's not. as soon as I can I put a more complete scheme

I'm also confused that the relay coil is external tot he relay and it looks like the case of the relay is conductive and shorting the ground with the positive snd negative rails.
the container that can be seen on the far right is not the relay casing but the driver circuit, it is also written.
half H bridge or push pull darlington.
the choice could fall on TC4420 / 29 or TIP120-TIP125.
What are those bridge's "internal diodes"?
 

Attachments

gophert

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at this point the driving circuit becomes useless.
Then how do you explain the diode on your LM79xx voltage regulator? That regulator would be useless if your orange line is right - current would just flow through that diode (circled in green) instead of being regulated by the LM79xx just above the diode. That greeen diode is there to block current just in case of an upset from inductive kickback. Shouldn't that diode be flipped

545C3F1D-8639-4E3F-8C16-2247C220E0FE.jpeg
 

atferrari

Well-Known Member
Then how do you explain the diode on your LM79xx voltage regulator? That regulator would be useless if your orange line is right - current would just flow through that diode (circled in green) instead of being regulated by the LM79xx just above the diode. That greeen diode is there to block current just in case of an upset from inductive kickback. Shouldn't that diode be flipped

View attachment 128927
If so, the one in the positive regulator should also be flipped.
 

gophert

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Most Helpful Member
If so, the one in the positive regulator should also be flipped.
And, not to further confuse the OP but, the "output" of an LM7912 regulator (-12VDC) is the "regulated" side. Allowing -35 to -14v input - Even though current is flowing into the output.
 

throbscottle

Well-Known Member
I'm sorry but the proposed circuit cannot work.
I make an example with the attached circuit which is the same as the one proposed. if the push pull or the half bridge H supplies the coil with negative voltage when the relay switches off it must discharge through the indicated path but is blocked by one of the internal diodes of the diode bridge. the same thing happens when the positive voltage is supplied to the relay coil but following another path. It should be noted that the darlington phus pull pairs or the half bridges H have inside them the diodes which are arranged as in the drawing in the box.
Possibly I am going to regret joining this thread, but...
Your statement makes no sense. How do you think a split supply works? This is a classic design used in various forms on production systems for decades. You are over-thinking this - just treat the power supply as a black box with +/0/- outputs and don't worry about where the current goes inside it, because you are getting confused. The + or - side can supply current and the return path is through the 0v line. They can supply current simultaneously and 0v absorbs the difference. Don't mix PSU design into your switching requirements until you've done the rest and decided the PSU needs to be somehow special.
 

arivel

Member
Then how do you explain the diode on your LM79xx voltage regulator? That regulator would be useless if your orange line is right - current would just flow through that diode (circled in green) instead of being regulated by the LM79xx just above the diode. That greeen diode is there to block current just in case of an upset from inductive kickback. Shouldn't that diode be flipped

View attachment 128927
to tell the truth it was you who proposed the scheme with the diode at the ends of the regulator. go see the post you wrote at number 36.
the scheme that I made is actually the same as yours, I simply joined the two grounding lines together and to do this you have to rotate the lower branch if you check you will see that it is so

And, not to further confuse the OP but, the "output" of an LM7912 regulator (-12VDC) is the "regulated" side. Allowing -35 to -14v input - Even though current is flowing into the output.
Are you kidding me ?

Possibly I am going to regret joining this thread, but...
Your statement makes no sense. How do you think a split supply works? This is a classic design used in various forms on production systems for decades. You are over-thinking this - just treat the power supply as a black box with +/0/- outputs and don't worry about where the current goes inside it, because you are getting confused. The + or - side can supply current and the return path is through the 0v line. They can supply current simultaneously and 0v absorbs the difference. Don't mix PSU design into your switching requirements until you've done the rest and decided the PSU needs to be somehow special.
I honestly did not understand what you wrote
 

throbscottle

Well-Known Member
0v for your purposes is ground, or grounding line. It's the return path for all signals. Please explain if any other part of what I wrote doesn't make sense.
 

gophert

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Most Helpful Member
0v for your purposes is ground, or grounding line. It's the return path for all signals. Please explain if any other part of what I wrote doesn't make sense.
"Ground" is the "return path" for the positive supply. It is the current source for the negative supply.

The negative supply can be the "return path" for the positive supply.

"Ground" is simply an arbitrary reference point of voltage potential in a system.
 

throbscottle

Well-Known Member
True, but it's easier to think about my way around. One supply is positive, the other is negative. Or maybe it's just easier for me to think about, my way around...
 

gophert

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Most Helpful Member
True, but it's easier to think about my way around. One supply is positive, the other is negative. Or maybe it's just easier for me to think about, my way around...
Your way causes confuses which way a diode should be placed on the negative supply and implies no current would flow if the relay coil is disconnected in the drawing. Current WILL flow from V+ to V- .
 

throbscottle

Well-Known Member
But them I'm having to think about a + supply which is 0v, and a higher voltage + supply, which is V+.

I don't think of 2 positive current sources, I think of a positive one and a negative one, which produces the correct result anyway.
 

gophert

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Most Helpful Member
But them I'm having to think about a + supply which is 0v, and a higher voltage + supply, which is V+.

I don't think of 2 positive current sources, I think of a positive one and a negative one, which produces the correct result anyway.
There are two sources above V- and you should be aware of both as you design. A bulb connected across V+ to V- will have double the voltage than a bulb connected from V- to ground or V+ to ground. You cannot expect all current t return through ground.
Even in an output of an audio amplifier's push-pull design, the positive rail charges the output coupling capacitor when the npn (upper transistor) is activated and the pnp (lower transistor) discharges the cap to the negative rail. The other lead of the capacitor is connected through the speaker to "ground".
 
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gophert

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Most Helpful Member
I think the complexity has spiraled out of control snd we are now teaching as we solve his solution rather than solvi his original problem, after rereading the earlier posts, it looks like he is trying to drive a latching relay.
He has yet to tell us exactly what kind of latching relay.

the easiest option would be a dual coil so we don't need to change direction
A0542137-D280-44DC-9811-D46AB902B294.jpeg

then we can add a circuit like this...
2BCFB54C-366E-490D-A104-C3D9F6C41FC6.png

and, if the OP insists on a single coil, this seems tobe the preferred method - resistor valves need to be adjusted depending on current draw of coils and coil voltage.
1F49982D-ED4C-4291-9DF9-C62B977C4520.jpeg
 

arivel

Member
I had already written which relay it was in post number 7.
this is what I wrote:
"I need it to drive single coil bistable relays.
in these relays the current flows in both directions. ".
what does this sentence mean?
"snd we are now teaching"
I didn't know I was in school :D, who is the pupil ?
the first circuit you propose is made for double coil bistable relays with a common terminal.
I don't like the second circuit. when the coil of the relay has to discharge it is found in front of two resistors in series which in total make 94 ohm more when the transistor is in saturation most of the current goes on the resistance.
I have only one driving channel coming from the cd40193 and not two, otherwise I would have used a full H bridge which are specially designed for this purpose.
if I insist on the single coil bistable relay there is a reason.
 

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