negative PSU schematic, how does it work?

Hello, I have seen this dual rail capacitor charge psu.
Could you please explain how the capacitor charge mechanism works for both rails?

It works only for the positive rail.

When the 555's output is low it's at 0V (the Earth in your diagram). The output labelled "negative voltage output" produces nothing.

Syncopator said:

It works only for the positive rail.

When the 555's output is low it's at 0V (the Earth in your diagram). The output labelled "negative voltage output" produces nothing.

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You mean it works only when the 555 outputs a positive pulse and not zero?

The negative portion works, I have tested it and it gives a negative voltage of -5v to -11v at 70-10mA proportionally. The output pulse of the 555 is about 0v and +12v at 50% square wave

I am confused of how it actually works. A guess is this:
When a positive pulse comes out from the 555 this charges the positive rail input cap, which charges the output cap through the series diode. but what about the negative rail?

I need to figure out how the circuit works and at what 555 pulses, positive or negative or both?

Ok, I must admit I took only a swift look at it. I shall think again. :???:

Hmmm, now I'm a bit confused myself, but about the positive output works. Additionally, I don't see the necessity for it. Why not use the positive rail to the 555 directly?

Let's see what's happening. I've shown the output of the 555 as a switch which connects pin 3 to either the supply rail or to 0V.



With the 555's output high, C3 will charge via D3 to almost the positive supply value.

I suppose C2 will too, but I don't think passing reverse current through C1 is a clever thing to do.

When the 555''s output goes to 0V, the positive plate of C3 is connected to 0V , so, as it is charged to the voltage of the supply, its negative plate will now be at -Vs and it will charge C4 via D4 to almost the same voltage.

I don't understand the purpose of C1 and D1.

There is an animated simulator that can clarify how this circuit operates:

www.falstad.com/circuit

Here is a screenshot of my simulation:



It is able to generate a link (below) which will open Falstad's webpage, load my schematic, and run it on your computer. (Click Allow to load the Java applet.)

https://tinyurl.com/az4yrkt

You can change values at will. Right-click a component to bring up an edit window.

These posts are so helpful but so problematic... now I need to install java, slowing down even more the lab computer just to see how this circuit works?
There must be an explanation somewhere

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Syncopator said:

Hmmm, now I'm a bit confused myself, but about the positive output works. Additionally, I don't see the necessity for it. Why not use the positive rail to the 555 directly?

Click to expand...

If an opamp is needed to be powered by this circuit, the two rails must have equal opposite voltages. Since this circuit's voltage will degrade as current increases, one can never maintain a similar voltage to the negative one, if taking that voltage directly from the vcc. an identical positive rail to the negative one must be used, to maintain voltage balance. I can see it this way.

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I think the answer is this
both cycles are needed. the positive cycle charges the series cap and the negative discharges it, generating the -v
that is for the negative rail.

the positive rail charger the output cap at both cycles. when the 555 is at 0v, the output cap takes its energy from the input cap as well.
but then, why is thee shunt diode needed at the positive rail? maxim does not include this diode.

If you need more power, this schematic uses the same concept, except you replace the 555 with two transistors in totem-pole configuration.



Its effective impedance is much less than the 555.

To get a +8.2 V supply, install the corresponding positive network.

Notice that any bipolar supply has some internal impedance. I have trouble picturing a type where you draw high current from one polarity, without some imbalance happening.

Syncopator said:

Hmmm, now I'm a bit confused myself, but about the positive output works. Additionally, I don't see the necessity for it. Why not use the positive rail to the 555 directly?

Let's see what's happening. I've shown the output of the 555 as a switch which connects pin 3 to either the supply rail or to 0V.

I don't understand the purpose of C1 and D1.

Click to expand...

I think D1 anode should be connected to +ve to get a more +ve voltage