# 1N4148 not giving correct voltage drop

I am trying to drop the voltage coming from a USB power line to 3.3V. I actually plan on using resistors to create a voltage divider (which I may have questions about later), but for now I was trying to experiment with the 1N4148. As I understand from the data sheet, it has a forward voltage of about 0.70V. Well, what I did was connect one to VCC, and checked what the voltage was. Prior to attaching the diode, I was getting about 5.42V, and after connecting, I was getting about 5.0V. Why is the drop not close to the forward voltage?

Thanks. If a diagram is needed, please let me know (although this is quite a simple circuit).

EDIT: Please find a diagram below • The diagram would indeed help. Aug 22, 2012 at 20:45
• @NickAlexeev There you go, added. Aug 22, 2012 at 21:07
• put a 1K resistor at the output of your circuit as load, measure the voltage on the resistor and you will obtain what you want. Jul 3, 2016 at 9:27

Actually it sounds like you're getting about what you should. The voltage drop accross a diode is a non-linear function of current. It is not a fixed voltage source, but is around 600-700 mV for most normal currents. If the only load thru the diode is a voltmeter, which are designed to draw as little current as possible, it may look like the drop is only 200 mV or so. You are seeing 420 mV drop, which is certainly plausible for some small amounts of current.

The basic problem is that you are trying to use a forward biased diode as a fixed voltage source. That does work in some cases when the situation is well enough known or controlled, but this is not the best way to achieve what you want. What you do want is a linear regulator. There are many to chose from, but something like the 3.3 V version of the MCP1700 is a cheap chip that would supply a few 100 mA and should work for your case.

Another issue is that your USB voltage is high. I don't remember the exact full range of valid USB voltage as can be seen by a device, but 5.42 V sounds out of spec. Still, most 3.3 V fixed linear regulators, including the MCP 1700, will be fine with that.

• Yes, that 5.42 V is out of spec. Vbus maximum is 5.25 V. Aug 23, 2012 at 6:41

You have placed the diode the wrong way (if red wire is VCC and black is GND). With your diagram you are not drawing any current, hence, the diode actually will have no voltage drop, so I assume your connection diagram is a bit wrong.

The cathode (line on diode) should be placed towards ground, having the anode side on VCC. These diodes does also come with different quality depending on where you buy them. I think that the voltage drop you see is actually a leakage through the "blocking" direction of the diode, and not at all what everybody else are answering.

You could also try setting your multimeter to diode test, connection the leads to anode and cathode, getting a measurement of how big the voltage drop is over the diode.

• That was just a mistake in the diagram, I don't actually have it connected like that. Sincerest apologies. Aug 22, 2012 at 21:17
• Ok :) Do you have any load on the circuit?
– chwi
Aug 22, 2012 at 21:20
• Unless you consider the multimeter a load, nope. I think that's the problem. Aug 22, 2012 at 21:23
• I just recently had the same "problem". If the diode does not carry any current, you will have no voltage drop over it (ohms law). Got my head twisted for a while, but look at the voltage characteristic of the diode and you see that with just a tiny bit of current the voltage will quickly rise to ~.7V
– chwi
Aug 22, 2012 at 21:26

On page 4 of the datasheet there is a graph, Figure 3, which shows the relationship between forward voltage drop, current, and temperature. It shows that there could be a fairly wide range of voltage drop depending on those factors, and 0.42v drop is not outside of what the part is spec'd at.

Another thing that graph shows is two lines, one for "typical" and one for "maximum". So even at the same current and temp there could be some variation from part to part.

In short, the forward voltage drop will vary and cannot be relied upon to be a constant value.

Since you are not drawing any current, the voltage drop will be ~0. Have a look at the current/voltage characteristic of a diode.

The forward voltage drop for a given load current can be accurately calculated using the Shockley diode equation. This can be tedious and monotonous, so just keep the following in mind.

A silicon diode's knee voltage is soft - at very light load currents, the forward drop is less that what it is at the rated current for the device. If you look at the curves on page 4 of the datasheet, you can see that the diodes are soft below 100mA or so.

I would argue that your application is using much less than 100mA, so you're in the soft region, hence the drop is lower than your expectation.

You won't get any voltage drop until you draw some current. Put a resistor across the leads where you measure the voltage, and see what the result is.