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### Help with simple diode circuit ???

Posted: Tue Dec 18, 2001 12:06 pm
Suppose that a diode is connected in series with a resistor to a battery. Assuming
that the values for Vsource, R, and the voltage curve for the diode (some function
f(I) ) are known, how can the current I (current) be calculated? Also assume that
f(.020) = 1.2 volts . Also lets assume that the diode begins to conduct at around .6
volts.<p>I believe that the following formulas are true:<p>Vdiode = f(I)
Vr = Vsource - Vdiode,
I = Vr/R,
Vr = Vsource - f(I) .<p>I also believe that if I (current) were to somehow increase, Vdiode would increase
which would mean that Vr would decrease. However if Vr decreases, then this
would mean that I (current) would have to decrease.<p>Any help would be greatly appreciated. Thank You

### Re: Help with simple diode circuit ???

Posted: Tue Dec 18, 2001 2:40 pm
Im not sure about what your trying to achieve? Are you looking for the "Internal Happenings" of the diode, or just the general OHMS law, external to the diode? The voltage through a diode and its passage in general consideration, don't change or vary in that they break down at the given voltage, like point seven or point three volts regardless of other general* external factors. However because other factors do exist within the diode it self, they [diodes] can be made to "pop" on and off like a pressure cooker weight by passing a voltage/current that drops off as soon as it passes through, closing and opening the current passage within a diode at very high speed. Because this factor follows the recovery time of the stated diode, this effect can be utilized as a microwave signal, a timing circuit, and many other functions can also be utilized by this action.

### Re: Help with simple diode circuit ???

Posted: Tue Dec 18, 2001 4:42 pm
I also believe that if I (current) were to somehow increase, Vdiode would increase
which would mean that Vr would decrease. However if Vr decreases, then this
would mean that I (current) would have to decrease.<p>This seems to be the hangup. YOu have to define the "somehow". In order for the current to increase either the applied voltage would have to increase or the voltage across some element would have to decrease.<p>Think of the diode as a fixed voltage at the junction with a built in series resistance. If the applied voltage increases then the current will increase as well as the voltage across the diode and the resistor. <p>If the voltage across the diode were to decrease say from a change in temperature then the current would increase and the voltage across the series resistor would increase.

### Re: Help with simple diode circuit ???

Posted: Tue Dec 18, 2001 6:56 pm
You are absolutly right, therefore the current cannot increase (or decrease) unless something is changed in the circuit. If the diode drop that was assumed in the beginning was wrong, then the calculated current may have to increase in order that all the drops around the circuit add up to zero. But, once the correct current is found, it will not change unless the diode temperature changes, or some other change is made.

### Re: Help with simple diode circuit ???

Posted: Wed Dec 19, 2001 5:36 am
<blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr>Originally posted by Chris Smith:
Im not sure about what your trying to achieve? Are you looking for the "Internal Happenings" of the diode, or just the general OHMS law, external to the diode? The voltage through a diode and its passage in general consideration, don't change or vary in that they break down at the given voltage, like point seven or point three volts regardless of other general* external factors. However because other factors do exist within the diode it self, they [diodes] can be made to "pop" on and off like a pressure cooker weight by passing a voltage/current that drops off as soon as it passes through, closing and opening the current passage within a diode at very high speed. Because this factor follows the recovery time of the stated diode, this effect can be utilized as a microwave signal, a timing circuit, and many other functions can also be utilized by this action.<hr></blockquote>

### Re: Help with simple diode circuit ???

Posted: Wed Dec 19, 2001 5:39 am
<blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr>Originally posted by john albers:
I basically just want to know how to calculate the current through the wire given R, Vsource, and the curve for the diode.<hr></blockquote>

### Re: Help with simple diode circuit ???

Posted: Wed Dec 19, 2001 11:17 am
In a simple circuit where there are two fixed components like one resistor, and one simple type diode, and for simplistic understanding, ohms law applies and you can equate the values as if the diode were a on/off switch rather than a diode. The diode is either on, or off, not varying in resistance from one value to another, but on or off. The curve mentioned for diodes show that the actual curve in time is rather instant than gradual and snaps on or off in terms of "Normal" time. As I mentioned this curve in time can be made to "hover" in this region and be turned into a microwave signal because of its speed of transition, but for the common circuit it is on, or off, and basically doesn't exist in terms of a variable. The resistor also behaves in this same manner in that it also doesn't vary its value of resistance during a current flow, or a current change. Resistors are fixed and with the exception of overheating they remain fixed in value. To consider how a resistor works, think of it like a water valve. Increasing the pressure in a water pipe doesn't change the opening size, but more water will flow with a pressure increase. The diameter remains the same in any valve even though more water flows with a increase in pressure. The diode for the most part also does the same with only two choices, open or closed. The diode however reacts like a pressure relief valve in that it remains closed until a given pressure is achieved and then its open, and remains open as long as the threshold pressure or voltage is met. It does have a given intrinsic value of resistance in the open state, and infinite resistance in the closed position, for argument sake. However for argument sake, these values don't change in that once again, It is either open or closed. Standard ohms law applies to find the current, voltage drop, and all other values. The diode to be measured will have a value drop of the point 7 volts after the diode, and this point 7 wont vary once the diode "opens" and will remain fixed for any increase in voltage or current flow. Ten volts in will yield nine point three volts out with point seven drop at any voltage above the starting point. Increasing the voltage will increase the current flow through the diode, but still point seven volts will drop through the diode. This plus the above posts should cover all of you possibilities or answers? The "curve" in most diodes turns sharply into a straight line upwards, again for argument sake making it on, or off.

### Re: Help with simple diode circuit ???

Posted: Wed Dec 19, 2001 12:14 pm
<blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr>Originally posted by john albers:
[QB][/QB]<hr></blockquote><p>If you have the diode V-I curve, you can use a "load line" to determine the diode's operating point for a given supply voltage and series resistor. Simply add a line to the graph containing the diode curve, as follows:<p>One end point is on the horizontal (voltage) axis, and is equal to the supply voltage. The other end point is on the vertical (current) axis, and is equal to the current which will flow through the resistor if the diode is shorted out (I=V/R). Connect these two points, and the place where the resulting line crosses the diode curve is the diode's operating point.<p>For example, if you have a 5 volt supply and a 1 kohm resistor in series with the diode, place a point at 5 volts on the horizontal axis, and place another point at 5ma (5 volts divided by 1 kohm) on the vertical axis, then connect the two points with a straight line. The intersection of this line and the diode curve will be the operating point. As Chris pointed out, in this case you will get, for normal silicon small-signal diodes, about 0.7 volts and 4.3 ma. For large rectifiers, the diode voltage will be a little less.<p>If you are interested in the equation which defines the diode curve, search the web for "diode equation".<p>Ron H<p>PS I obviously don't understand how to post a quote. <p>[ December 19, 2001: Message edited by: RonH ]</p>

### Re: Help with simple diode circuit ???

Posted: Wed Dec 19, 2001 1:17 pm
Ron, ...what do you mean by quote? You can "copy and paste" Address lines and other info as if they were plain text, and photographs / drawings have a "photo" insert section below under "image"? I haven't tried the image yet?

### Re: Help with simple diode circuit ???

Posted: Wed Dec 19, 2001 2:22 pm
Well, I clicked on the "" (reply with quote) at the top of John Albers' last post, naively assuming it would place the contents of his post in mine. Obviously, I was wrong.
Regarding images - the FAQs say you can't include files. Is there a way to include an image without including a file?<p>Ron H

### Re: Help with simple diode circuit ???

Posted: Wed Dec 19, 2001 2:39 pm
I need to do some experimenting also, but I would say that if the files are already within the JPG, they will be considered to be a "Photo" in it self, regardless of the data. If something has two parts to it, as in a separate "JPG" and a separate "TXT" file, they would have to be separated by you, or combined by you, into a separate entity called or labeled as a single "JPG"? Scanners can combine both photos and text into a single "JPG" file. I think the "image" section can only accept extensions of "jpg, gif, bmp, etc"? "Reply with Quote", I haven't tried it yet?<p>[ December 19, 2001: Message edited by: Chris Smith ]</p>

### Re: Help with simple diode circuit ???

Posted: Wed Dec 19, 2001 3:16 pm
John Albers knows how to post a quote. You there, John?<p>Ron H

### Re: Help with simple diode circuit ???

Posted: Thu Dec 20, 2001 5:01 am
<blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr>Originally posted by RonH:
John Albers knows how to post a quote. You there, John?<p>I clicked on the post with quote button.<p>Ron H<hr></blockquote>

### Re: Help with simple diode circuit ???

Posted: Thu Dec 20, 2001 5:03 am
<blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr>Originally posted by john albers:
[QB][/QB]I mean reply with quote button. I think that you also have to type your message between the quote tags.<hr></blockquote>