# Diodes

#### aamirali

Joined Feb 2, 2012
412
I have connected a diode 1N4007 across 14.33 V supply & tested voltage as shown in figure in attached pdf. (File.pdf)The voltage was 14.08V. I was surprised how this happen as forward drop of diode is 0.7V at room temp at low currents . (this is what I know till now)

suddenly I realized my voltmeter has internal resistance of 10M ohm so i have connected a ammeter as shown & I noted the current 1.3uA. So drop at this current is 14.08V in next figure.

So I kept on adding 10M resistance as shown so overall resistance is 5M a shown in next figure.

In total when I=1.3uA, Vdrop = 0.25V
I=2.7uA, Vdrop = 0.31V
I=4.1uA, Vdrop = 0.36V

I also tested diode (whether its working right), by connecting across diode tester in voltmeter. It shows 0.535V in forward bias direction.
I thought also to check its current i.e when meter is used to check diode drop. So I connected ammeter in between:
New Vdrop= 0.568 (total across diode & ammeter)
I = 337.2 uA.

Help me to understand this & why diode drop is not 0.7V at low current, Diode forward voltage drop

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#### BillB3857

Joined Feb 28, 2009
2,564
Look at the data sheet for a 1N4007 and in specific, the Forward Voltage vs Current chart. It may help to explain what you are seeing.

#### mlog

Joined Feb 11, 2012
276
Two quick points.

1. Look at where your voltmeter is applied. Hint: It's not only across the diode.

2. You need a minimum current in order to effect a forward diode drop in the neighborhood of 0.7 volts. For this diode at room temperature, about 50 mA would be about right. That means you need a 270 ohm resistor in series with the diode and voltage source. Beware that the resistor will dissipate about 3/4 watt.

#### Mark_T

Joined Feb 7, 2012
47
To measure the Vf of a diode you connect the meter across the diode not in series. As above you will need a load resistor, try 10k where the meter is.

#### MrChips

Joined Oct 2, 2009
26,790
Think about it, if the diode current is zero, the voltage drop across the diode would be zero. How would the voltage drop eventually rise to 0.7v? It has to get there gradually starting at 0V.

#### Wendy

Joined Mar 24, 2008
22,988
Post #2 said it best. The amount of drop across the diode is dependent on current flowing through it, it is not a constant. At 1A it will be around 1.0V, at 1µA it could be around 0.6VDC.

#### crutschow

Joined Mar 14, 2008
30,438
Diode voltage drop is actually logarithmic with current. It's just that at the design operating current, the drop is about 0.7V for most silicon diodes and that is the value that is typically mentioned. Thus a silicon diode voltage drop at say 1μA forward current is only about 0.35V.

#### Audioguru

Joined Dec 20, 2007
11,248
Here is a datasheet graph of the forward voltage of a "typical" 1N4007 rectifier diode when its junction is 25 degrees C and the pulse width is small to avoid heating:

#### MrChips

Joined Oct 2, 2009
26,790
Diode voltage drop is actually logarithmic with current.
True, but usually we plot I as the dependent variable (Y-axis) and V as the independent variable (X-axis). So the relationship of I vs V is exponential.

#### aamirali

Joined Feb 2, 2012
412
I have seen that graph of I vs V earlier. In it u can seen Vdrop starts from 0.6V at 10mA. Does that mean dode drop can fall below 0.3V also at lower current.

Any technical reason explaining this?

#### Audioguru

Joined Dec 20, 2007
11,248
Yes, the forward voltage is very low at very low currents.
Look at the datasheet for the 1N4148 diode:

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