Lab questions!

[confused17]

So I did a lab in school where I had a 2.2k resistor and a forward biased diode in a series circuit where I increased the voltage across the diode 0-.75V(by .5) and measured the current through the diode. For the required results of the lab I am supposed to plot the diode curve current versus voltage, I have done that but than it asks me what type of curve it is. I'm not sure what that is supposed to mean, like a exponential growth or something? The current through the diode increased from 0 to .0004mA when the voltage reached 0.25V (which is the knee voltage right?) Then I am asked to calculate the forward resistance of the diode at 4mA and 15mA from the curve and to explain the difference. How do I find the resistance at 4mA when I only went to
.75V?
Any help please? I'm really confused!

 

[confused17]

Sorry for the double post but can anybody help me?

 

[cowana]

What was the current when it was 0.75v?

Do they say forward voltage, or forward resistance?

 

[Gary B]

I believe that there is a typo in the question. If you plot the curve at increments of .5 as stated, you only have two points (0 and .5) .75 is half way to the next test point. I suspect the correct sample size should have been .05.

Yes, they are looking for linear, geometric, logarithmic, or other curve type. So this is a curve fitting problem. What equation defines your data points? That equation also defines the curve. I am surprised that you were not also asked to state the slope of the curve at one or more points as well.

 

[MrAl]

Hi,

What kind of diode are you using? I find that a 1N4001 diode biased to 0.75v causes a current flow of 134ma in spice. A 1N4148 does 29ma.
Also, make sure you measure the voltage right across the diode, not at the power supply output jacks and dont use the built in meter (if it has one).

 

[crutschow]

So I did a lab in school where I had a 2.2k resistor and a forward biased diode in a series circuit where I increased the voltage across the diode 0-.75V(by .5) and measured the current through the diode. For the required results of the lab I am supposed to plot the diode curve current versus voltage, I have done that but than it asks me what type of curve it is. I'm not sure what that is supposed to mean, like a exponential growth or something? The current through the diode increased from 0 to .0004mA when the voltage reached 0.25V (which is the knee voltage right?) Then I am asked to calculate the forward resistance of the diode at 4mA and 15mA from the curve and to explain the difference. How do I find the resistance at 4mA when I only went to
.75V?

If I understand correctly you went to 0.75V across the resistor and diode. Your were supposed to go to 0.75V across the diode. The voltage across the resistor and diode will be much more than that. For example to get 15mA through the resistor will require greater than 33V.

If you do that then you will be able to measure the voltage directly across the diode (not including the resistor) at up to 15mA.

To calculate the equivalent diode resistance, just divide the diode voltage by it's current at each point.

Zener diodes have a "knee" but regular diodes do not. It all depends upon the scale when you do a linear plot of the voltage versus current. At the normal range of operating currents it typically appears to be about 0.6-0.7V for standard silicon diodes.

 

[confused17]

The current at .75V was .0928mA , yes im sorry it is .05 not .5 and the diode used was 1N540, I think I made the mistake of using power supply instead of checking the voltage across the diode.

Edit- I tried making the voltage across the diode .75 using multisim but you need a power source of more than 40 volts which is the max at my lab. Im not sure what it is now. Btw thanks for the answers everyone!

 

[MrAl]

Hello,

Why do you think you need a power supply of more than 40 volts? A few volts should do it as you only have to up to three quarters of one volt.

 

[crutschow]

Hello,

Why do you think you need a power supply of more than 40 volts? A few volts should do it as you only have to up to three quarters of one volt.

Because of the 2.2k ohm resistor in series with the diode. But he could go with a lower value resistor.

 

[RCinFLA]

Curve is exponential.

I = Isat * (e^(Vd / nVt)-1)

Diode modelling - Wikipedia, the free encyclopedia

On your measurements, be aware of temperature. Silicon diode voltage will change at -2.5 mV/degC

Know the Isat and temperature and you can predict the curve. (Isat is a very small number)

If you look up diode Spice models you can find the approximate Isat for your part number. Isat does vary lot to lot a little for a given part number due to manufacturing process variations.

 

[Roff]

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