This is most common arrangement of clipping diodes that you will see. For some reason, I'm looking at this and thinking that D2 is doing nothing. Imagine your favorite boost running on 9v is at the input. The signal flowing from input to output is from 0v to 9v centered on 4.5v. When the signal is greater than 0.7v D1 conducts and the signal gets clipped. It seems D2 does nothing. Since the max voltage of 9v is not enough to reverse bias the D2 and make it conduct. The data sheet gives a the 1N4148 a reverse bias of 75v.
(http://www.super-freq.com/wp-content/uploads/2010/01/picture-2.png)
People have been doing this for years I see this in so many places I must be missing something. I can't believe we have been wasting wasting so many diodes!
Your signal is AC. One diode clips the "top hafl" and the other diode clips the "bottom half"
That's what I always thought to. But the signal is still in the range of 0v to 9v. So if you have 0v to 9v flowing past the top of the diodes and 0v at the bottom of the diodes. I can see where D1 clips. But why would D2 conduct?
If ground were 0v and the AC signal was +4.5v to -4.5v I could see D2 doing something.
Quote from: soggybag on January 22, 2010, 01:10:11 PM
So if you have 0v to 9v flowing past the top of the diodes and 0v at the bottom of the diodes.
The GND on your picture has to be the ground on your circuit. If you are using a split supply then GND is actually 4.5V, NOT 0V. You are making a virtual ground to get around the fact that you don't have a split power supply - you cant go negative obviously because you have no negative voltages from your power supply, but you can go between 0 and 9, so you divide that by half and make the middle a fake ground - now you can go 4.5 BELOW ground (fake) and 4.5 ABOVE ground (fake). It replicates a split supply, such as a true -4.5V, GND, and 4.5V supply. So your AC signal goes around 4.5V, and when it is 3.8 (0.7 less then 4.5) D2 conducts to the Vref (reference voltage, virtual ground, etc) and when it is 5.2 (0.7 more than 4.5) D1 conducts to Vref.
Why do so many effects have these two diodes in this same configuration going to -9v? Take the Rat for instance. Seems the second diode is not doing anything. Same thing with the Distortion Plus. I could find a hundred more.
It seems like this is a really common practice that really doesn't do anything. speaking of D2 of course.
The Analogman King of Tone, on the other hand, connects the two diodes to Vr which would get diode 2 into the action.
In the rat and Distortion plus, there's a cap in series with the signal before the diodes and a resistor to ground after them. This references the signal to ground, do it swings above and below ground, the diodes then work the same as in JKowalski's example.
Quote[People have been doing this for years I see this in so many places I must be missing something. I can't believe we have been wasting wasting so many diodes!
Unfortunately, you don't have a clear picture of what's going on. If you don't think it's having any effect, then remove D2 and see what happens. Clearly it will sound vastly different, and you will observe that there is a function to having D2 there.
Then re-read Kurtlives comment above. Google diode clipping and read Aron's wiki articles.
I'm sure it's me. I'm just trying to understand what is happening here. I'm one of those remedial students.
Looking at the Rat schem at GGG. I see both D1 and D2 are connected to ground. I want to call this 0v. On the left is C7. The opposite side of which is connected to the op-amp which is biased at 4.5v. On the other side is C9 which connected to Q1. The C9 Q1 connection is biased at something like 2.8v.
What am I missing? It seems the lower end of both diodes is at 0v and the top side of both diodes should always be greater than 0v?
Quote from: soggybag on January 22, 2010, 02:21:51 PM
I'm sure it's me. I'm just trying to understand what is happening here. I'm one of those remedial students.
Looking at the Rat schem at GGG. I see both D1 and D2 are connected to ground. I want to call this 0v. On the left is C7. The opposite side of which is connected to the op-amp which is biased at 4.5v. On the other side is C9 which connected to Q1. The C9 Q1 connection is biased at something like 2.8v.
What am I missing? It seems the lower end of both diodes is at 0v and the top side of both diodes should always be greater than 0v?
No, lookk up decoupling capacitors for an explaination. Capacitors will NEVER pass DC. They are two plates without any physical connection between them. All they do is pass AC signals. Now, say you have a signal at 4.5V. You put it through a decoupling capacitor. On the other side, it couples to whatever DC signal it can find. If you had a resistor to 0V on the other side of the capacitor, it would couple to 0V, and form a RC high pass filter (only passes high frequencies, rolls off to infinite attenuation at DC. If you had a resistor to 9V on the other side, it would couple to 9V, etc. This is how you move your signal from one "ground" to another "ground". In your schematic, the first capacitor couples the signal to 0V from being biased at 4.5V. Now it's at 0V so the signal will swing between some negative voltage and some positive voltage depending on the strength. If this "some voltage" is far enough away from ground to get over the diode drop it will conduct.
The second capacitor couples the signal that is "biased" at ground to the required voltage that the transistor needs to be biased at to function correctly.
It's just switching grounds, to the right ones that the circuit needs to operate at. You have to not think of 0VDC as the definition of "Ground", ground can be any voltage, it just depends on how you use it in the circuit, really. GROUND is arbitrary, not absolute.
EDIT: Took a peek at the RAT schematic and there is no resistor to ground after the first capacitor, which may confuse you. In this case the diodes act as the link to ground instead of the resistor mentioned.
Thanks for the explanation. I get the idea, and I've seen this type of clipping all the time. This is pretty easy idea.
I was having a hard grasping the idea that at one point the circuit could be producing a signal centered around 4.5v (0 to 9v) and at another point acting as if this signal centered on 0v (-4.5 to +4.5v). I had overlooked that capacitor and it's function.