DIY Diode Clipping Circuit: +10/-10V Supply

[vk6kro]

Indirectly, I guess.

Isn't that how Hydroelectric power stations work? Something like water has a potential energy and this is used to produce motion and the resulting kinetic energy rotates a generator and produces electricity, which you could use to drive a motor.

I think you probably mean to use gravity directly. In that case, no, I don't think so.

 

[raxAdaam]

I just wrote 3 replies & in the process (think I) understood what you were getting at: all one need consider is the behaviour of the voltage divider once the diodes are conducting because we choose the value of the resistor s.t. when the voltage that is dropped across it is maximal (i.e. 5.4) the output is *not more* than 5 - one needn't worry about what happens for V < 5.4 because we know V_{out} will definitely be less than 5 anyhow...

Thanks so much for the help. You really cleared up a major conceptual road block for me. Out of curiosity, how do you think of an AC voltage source? I find I am confusing myself because I don't really know how one should think of them (vs. DC)? I find I imagine 'one side' of the circuit varies its voltage while the 'other side' effectively remains at 0V the entire time - but this inevitably leaves me confused.

I tried to include a simple AC-source - 'backward diode' - 1k resistor - circuit in ASCII, but the spacing is lost in the post, so I won't bother - but hopefully you can understand what I mean? If not, I'll draw a circuit & scan it in tomorrow.

Thanks again.

 

[vk6kro]

The lower voltages do get reduced by the voltage divider. eg if the input was 2.7 volts, the output would be 2.5 volts by voltage divider action.

However, the main aim is to use the components allowed to ensure that no more than 5 volts is passed to the output. There are better ways of doing this with other components.AC signals. Well, I usually attach one end of them to the common side of a circuit and then the other end has a voltage that varies in a sinewave fashion going positive and negative relative to the common lead.
It isn't often that you can't attach one end of the AC source to a common point in a circuit.

Usually, input signals for amplifiers are AC signals and these often come into the circuit via a coaxial cable and the shield of this can be "earthed" or connected to the negative supply of the amplifier.

Drawing circuits etc.
I use Microsoft Paint which comes with Windows. Use the magnify function and try to get the lines straight.
Do any lettering in a blank part of the screen and then position it with the select function (rectangle with dotted lines)
You can attach it to a Forum post with the paper clip thing at the top of the edit screen.

 

[raxAdaam]

Thanks for the thoughts on thinking about AC-circuits.

To clarify, in the circuit attached, V_o would be 0 for V_{in} &gt; 0 and would be V_{in} - 0.7 for V_{in} &lt; 0, right?

What is confusing me is that if I think of the top line as grounded and V_{in} is connected to the bottom line, then I'm unclear as to how the diode is coming into play . . . wouldn't current simply pass through the resistor and continue to V_o? Or does current have to flow through the diode for it to flow through the rest of the circuit as well? If so, why?On a not-exactly-related note, I built a full-wave rectifier & noticed that while the ripple is improved, the voltage is more than an additional diode drop lower (the diodes I'm using drop roughly 0.6 & the full-wave rectifier output is roughly 1.3V lower than a similar half-wave rectifier). Any thoughts as to why this would be the case?

Cheers,

RaxA

ps. how does one embed an image from one's comp. directly into a reply? It asked for the url of the image, do I have to post it online somewhere first (as you can tell, I'm very new to all of this).

 

[vk6kro]

It is usual to connect the AC source so that the wire that goes straight through to the output (without any components in it) is the common.
This would be the lower input connection in this case.

So, regard the top input connection as varying.
Any positive input won't get past the diode.
Any negative input less than 0.6 volts won't get past the diode.
Any negative input greater than 0.6 volts will be passed but with 0.6 volts subtracted from the input voltage.
I think this is what you said.

You probably used a bridge rectifier. These have two diodes in series with the output for either input polarity. So, you get two diode drops.

You can put an attached thumbnail in the text. Use the little arrow next to the paper clip at the top of the edit screen. This will put the thumbnail to where your cursor is.

To put a big image in with the text, you do have to store it somewhere else and give a URL for it.
Some of these sites let viewers browse through all your pictures. I like one called Dropbox which limits viewing to the one picture you want viewers to see.

 

[raxAdaam]

You probably used a bridge rectifier. These have two diodes in series with the output for either input polarity. So, you get two diode drops.

Yeah - I did use a bridge rectifier, but I observed more than one *additional* diode drop in the output. i.e. for the half-wave rectifier my output was c.10.9V and for my full-wave rectifier it was 9.6V - so it seems an extra ~0.7V are being eaten up somewhere . . .

& I really can't thank you enough for answering all these questions - should I be posting somewhere else, generally? I don't want you to feel obliged to answer my never-ending inquiries [though I do appreciate your lucid answers - which can be a bit of a rarity online!].

All the best,

RaxA

 

[vk6kro]

It sounds like you don't have a capacitor on the output of your bridge rectifier.
With no load, a capacitor will charge up to the peak values of the output from the diodes. This then drops with load, producing "ripple" on the output. Bigger capacitors give less ripple for the same load.
A typical capacitor for a small power supply would be 1000 uF at 25 volts.
Connect it between the + and - outputs of the bridge rectifier.


Yes, you should be posting in the Electrical Engineering section if these are not homework related questions.
https://www.physicsforums.com/forumdisplay.php?f=102
We have to answer questions differently in the homework section compared with the Electrical Engineering section.
Everyone is very friendly there and welcoming to beginners willing to learn.