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

[novop]

Homework Statement

Hi. I'm being asked to make a circuit to take the input waveform, on the left, and extract the output waveform on the right. In my circuit, I can only use basic diodes, resistors, and DC supplies of +10 and -10V.

The Attempt at a Solution

 

[berkeman]

If you can assume 0.6V drop across a diode, then just stack them up to make 10V one way and 10V the other way -- you wouldn't need the power supplies in that case. Effectively you make a 10V bi-directional clamp with the two diode strings...

 

[novop]

I only added the 10V supplies because I can't make 10V using 0.6V drop diodes, the closest I can get is 10.2V

 

[berkeman]

I only added the 10V supplies because I can't make 10V using 0.6V drop diodes, the closest I can get is 9.6V.

Can you assume 0V forward drop diodes? If so, you can use a slightly modified version of your solution circuit. Look at the direction of the diodes compared to the polarity of the suppllies...

 

[novop]

I can only "use" 0.6V drop diodes. That being said, is my first solution the closest I can get?

 

[berkeman]

I can only "use" 0.6V drop diodes. That being said, is my first solution the closest I can get?

To be honest, I'm having trouble understanding how your circuit would work. Have you tried simulating it?

 

[novop]

I don't have any of the software or the know-how to do that. I got the idea for my circuit from this:

http://www.allaboutcircuits.com/vol_3/chpt_3/6.html

The third circuit, as far as I can tell, is essentially what I'm trying to do with my circuit.

 

[berkeman]

I don't have any of the software or the know-how to do that. I got the idea for my circuit from this:

http://www.allaboutcircuits.com/vol_3/chpt_3/6.html

The third circuit, as far as I can tell, is essentially what I'm trying to do with my circuit.

But in that circuit, the power supplies keep the clipper diodes reverse biased until the voltage across the diode + power supply is enough to forward bias the diode, causing clipping at that voltage.

In your circuit, I think I see that you are trying to subtract 10V from the 15V total drop across the diode strings, but I don't think that will work (I could be wrong).

I think I know how they want you to do this, so I'll try a hint to see if it clicks for you.

Look at that 3rd circuit, and imagine what you would get if the diode reverse-biasing supplies were 10V, and the diodes were 0.6V (they use 0.7V in their circuits, but whatever). What would that circuit give at its output? How far off would the output be from what your problem is asking for? Then, can you think of a way of adjusting the output voltage to get it to match your problem?...

 

[novop]

If the supplies were 10V then the circuit would clip the signal at +/- 10.6V (even though the input in this particular problem is +/- 10V). So, I'm 5.6V off from where I want to be. The last part is where I am stuck, I can't think of a way to adjust the output voltage!

 

[berkeman]

You're only 0.6V off in each direction...

 

[novop]

In the diagram in the first post, the output is clipped at at 5V/-5V.

 

[vk6kro]

You need to clip at +/-5 volts.
You could get close to this with 9 diodes for positive clipping and 9 for negative clipping. This would give you (9 times 0.6) or 5.4 volts. Or you could use 8 diodes to get 4.8 volts.

You have to have a series resistor to avoid excess current flow while the clipping takes place.
Something like 330 ohms.
Clipping occurs when the input exceeds 5 volts in either direction.

To get exactly 5 volts, you could take the 5.4 volts and arrange a voltage divider. This is two resistors in series from the output to ground. They need to be large compared with the 330 ohm resistor.

Clipping with diodes would give a "rounded top" type of clipping, which is useful for some applications.

 

[mmmboh]

So there is no place one can put a capacitor to fix the clipping voltage problem? And how did you get 330 ohm's for the resistor?

 

[novop]

I don't understanding where to make the voltage divider in this case. If, say, we take 9 diodes in each direction, for 5.4V clipping, how can I reduce this to 5V given the circuit?

 

[vk6kro]

[PLAIN]http://dl.dropbox.com/u/4222062/clipper%203.PNG

You would put the resistors across the output as shown above.

Since this is the homework section, I have not shown the value of the lower resistor (although I did calculate it. :) )

The 330 ohm resistor was a bit arbitrary. I just guessed a value of 15 mA peak for the diodes.
This is a value I often use for LEDs.

 

[berkeman]

Good solution, vk6kro. I did have a math error in my proposed solution.

 

[novop]

Thanks very much vk6kro. If I understand this correctly, the resistor R's value is arbitrary, it's purpose is a voltage sink after the diodes start conducting, right?

Assuming the above to be true, the bottom resistor should have a value of 12.5k ohms to extract the right waveform. Am I making sense?

 

[vk6kro]

Yes, that is right. In practice, you would know the load that existed after the output and could design around that.
In this case, the only load is the voltage divider.

R isn't entirely optional. You need a substantial current to ensure that the voltage drop of the diodes is about 0.6 volts.
If you chose 15 K and kept the same voltage divider, the diodes would never turn on because the voltage across them would not be greater than 4.7 volts.

Yes, I agree about the 12.5K. Good.

 

[novop]

Thanks very much.

 

[skeptic2]

I programmed the circuit into PSPICE and ran a simulation. The attachment shows the circuit and the simulation input and output. Also included is an alternate circuit with its simulation run. The second circuit is what I thought you were thinking about when you were talking about using a 10 V supply.

 

[novop]

For the first simulation, did you place the output over the 12.5k resistor? The output is peaking at 5.4V, so I don't think the voltage divider is working.

 

[skeptic2]

Yes I did. If you look at the bottom left of the black part of the plot, it indicates what points are being plotted. The first is V(V1:+) which is the plus side of V1 and shows the triangle wave. The second reference is V(R3:2) which is the top side of R3 or the 12.5 k resistor.

Actually diodes are not very good voltage references. Their voltage varies with current and inversely with temperature. The turn-on voltage is somewhat gradual which gets magnified by having 9 diodes in series.

The circuit could be made much better using an opamp but that wasn't part of the problem.

 

[berkeman]

I programmed the circuit into PSPICE and ran a simulation. The attachment shows the circuit and the simulation input and output. Also included is an alternate circuit with its simulation run. The second circuit is what I thought you were thinking about when you were talking about using a 10 V supply.

Very freaking clever, skeptic! I did not see that solution. Very cool.

 

[vk6kro]

LT Spice models the 1N4148 as having a forward drop of 0.717 volts at 15 mA. So, if you chose this diode, you would model about 6.4 volts from 9 diodes in series. 7 would be better.

Since no diode was specified, the standard voltage is 0.6 volts per diode.


That other clipper circuit shows considerable rounding of the tops of the waveforms because the extra current in these resistors when clipping happens causes a higher voltage drop across them. LT Spice gives a voltage of 5.3 volts peak on the right hand side of the 10 K resistor.

Changing the 1 K resistors to 100 ohms improved this, but then there was a greater drop across the diodes in series with the DC supply and about 4.7 volts out. Substituting Schottky diodes an all positions brought the voltage back up to 5 volts and with a much improved clipping effect.

 

[skeptic2]

Can you post the plots please?

 

[vk6kro]

[PLAIN]http://dl.dropbox.com/u/4222062/clippers.PNG



The top one is like the one posted. The bottom one has the 1000 ohm resistors replaced by 100 ohms and Schottky diodes substituted to compensate for the extra drop in diodes in series with the power supply. Note the superior clipping.
The output on the top one is 5.25 volts.

As there is no adjustment available, you take what you can get for clipping voltages.

The model is probably unfair to 1N4148s. I didn't think they were as bad as that.

 

[raxAdaam]

Hi -

I'm pretty new to circuits & diodes in particular & worked through this as an exercise, but get my lower resister value c.13.75k ohms [I plugged in R = 100 ohms instead of the 330 used above], but even when I use the suggested 330 ohms, I get that the lower resister of the voltage divider should be even larger (~16k). When I go through the calculations this seems to work - so what am I missing?

Just to explain my approach: I imagined what was happening in the circuit just as it approached 5.4V (i.e. before the diodes conduct), say 5.399..., then my R_eq = (100 + 1000 + R_2).

V_in = I*R_eq
I = V_in / R_eq = 5.39/(1100 + R_2)

Now, we want the voltage drop across R_2 (i.e. V_out) to be 5V so:

V_2 = V_out = I*R_2 = [5.39/(1100 + R_2)]*R_2 = 5 *aside: no chance this forum is latex enabled?!*

=> (5.39-5)R_2 = 5500
=> R_2 = 5500/0.39... =~13.75k


From the above, it seems clear that a higher value of R (100 -> 330) would yield an even higher value of R_2.

What am I missing?

& thanks for all the above. Ver ver enlightening.

~ Rax

 

[vk6kro]

Yes, you can use Latex.

Ignoring the diodes and just consider this as a voltage divider.

You have two resistors in series with 5.4 volts across them. The top one is 1 K.

The bottom one has 5 volts across it, so the top one has 0.4 volts across it.

By Ohm's Law, the 1K has a current of 0.4 volts / 1000 ohms or 0.4 mA flowing in it. OK so far?

Now, this current can't go anywhere else, so it also goes through the bottom resistor.

This resistor has 5 volts across it and 0.4 mA flowing in it so it has a value of 5 volts / 0.0004 amps or 12500 ohms

 

[raxAdaam]

Alright, that all makes sense but I still don't see what exactly is wrong with the logic I applied above? Is the discrepancy in our answers simply a matter of accounting for the 1st resistor?

& why *is it* justifiable to ignore that 330 ohm resistor?

Thanks a ton -


Rax

 

[vk6kro]

The 5.4 volts is only across the diodes and therefore across the two output resistors. The diodes are certainly conducting when the input voltage is greater than 5.4 volts and it is this 5.4 volts that we want to reduce to 5.0 volts.

There is a much larger current flowing in the 330 ohm resistor and most of this goes into the diode strings and then back to the bottom end of the supply. It has no effect on the voltage divider. The diode current peaks at about 15 mA.

You get to Latex via the SIGMA character on the advanced reply screen. It is at the extreme right of the second row of symbols at the top of the screen.
You get to that screen when you click "NEW REPLY" rather than "QUICK REPLY".

 

[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 $$V_{in} > 0$$ and would be $$V_{in} - 0.7$$ for $$V_{in} < 0$$, 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.