# Not getting expected Voltage Drop with diodes in series

#### jellytot

Joined May 20, 2014
72
I have a wall adapter that converts AC power to 12V output. It powers a load that draws 300ma. I want to drop the voltage closer to 9V. I read that diodes have a voltage drop of between around 0.6V-0.8V per diode, so I hooked up 4 diodes in series and expected a new voltage of between 9.6V-10.2V, but my multimeter reads it's around 11 Volts.
Did I misunderstand how this works?
I do have a 9V regulator so I can use that instead but was wondering why this didn't work.
Previously, I remember I had the same issue when I tried to drop 3V to 1.5V using 2-3 diodes, but voltage was still higher than expected. 1.5V regulators are harder to get so I was hoping to use this diode method, if I can get it to work.

Joined Jan 15, 2015
4,913

Ron

#### jellytot

Joined May 20, 2014
72
Yes, in my 12V example, it powers a 300ma load.

#### Papabravo

Joined Feb 24, 2006
12,770
Yes, well the output voltage may not be regulated and the diodes may be dropping less than you expect at the 300 mA load current. If you have a regulated +12V output then your scheme will work a little better. Measue the unloaded output and tell us what diodes you are using. For all we know they might be Schottky diodes!

#### crutschow

Joined Mar 14, 2008
24,115
The diodes need to be loaded when you measure their voltage drop. It's not clear if you did the measurement under load.
What is the diode part number?

Joined Jan 15, 2015
4,913
When the wall wart is under a load, as you mention it is, just measure the voltage drop across each diode or all the diodes in series. Just remember this needs done while the wall wart is driving a load. Also, per crutschow, what are the diodes?

Ron

#### DerStrom8

Joined Feb 20, 2011
2,390
Don't forget to place high value resistors (~1-10Mohm) in parallel with each diode. Otherwise some will work harder than others and may burn out.

#### crutschow

Joined Mar 14, 2008
24,115
Don't forget to place high value resistors (~1-10Mohm) in parallel with each diode. Otherwise some will work harder than others and may burn out.
That may be needed if the diodes are reversed biased, but here they are only forward biased.

#### jellytot

Joined May 20, 2014
72
The diodes were under load when measured. The multimeter reads the load at 250ma, at 10.90V. Diodes are standard rectifier diodes 1N4001.

---(+power)----[+1N4oo1-]---[+1N4oo1-]---[+1N4oo1-]---/[Multimeter]\-------(-power)---

With the diodes removed, multimeter reads 11.9V
---(+power)------------------------------------------------------/[Multimeter]\-------(-power)---

Which means a voltage drop of ~0.33V per diode. Nowhere near 0.6-0.8V I read it's supposed to be.

#### joeyd999

Joined Jun 6, 2011
4,291
The diodes were under load when measured. The multimeter reads the load at 250ma, at 10.90V. Diodes are standard rectifier diodes 1N4001.

---(+power)----[+1N4oo1-]---[+1N4oo1-]---[+1N4oo1-]---/[Multimeter]\-------(-power)---

With the diodes removed, multimeter reads 11.9V
---(+power)------------------------------------------------------/[Multimeter]\-------(-power)---

Which means a voltage drop of ~0.33V per diode. Nowhere near 0.6-0.8V I read it's supposed to be.

#### bmuigai

Joined Sep 13, 2012
1
have you checked the datasheet for your diodes. Germanium should drop less voltage, although the 1N400x's i have used, ...not sure... but I think they are silicon n should drop according to your expectation. Also pardon me for asking stupidly, but if your arrangement is as indicated above, where is the load in this case?

#### MikeML

Joined Oct 2, 2009
5,444
Why using series diodes to drop voltage is a bad idea: Note the output voltage vs current to load.

#### DerStrom8

Joined Feb 20, 2011
2,390
That may be needed if the diodes are reversed biased, but here they are only forward biased.
The resistors are there to equalize the voltage drops across each diode, since they will not all be the same. It is still recommended for forward-biased diodes, for the same reason it is recommended when using multiple diodes in series for rectification to increase their overall voltage rating.

#### wayneh

Joined Sep 9, 2010
16,162
Why using series diodes to drop voltage is a bad idea: Note the output voltage vs current to load.
On the other hand, if you limit the x-axis to a practical 10X range of current, for instance in the 10-100mA range, your plot shows why diodes work fairly well. Poor compared to a regulator, but plenty good for things with steady loads.

#### crutschow

Joined Mar 14, 2008
24,115
The resistors are there to equalize the voltage drops across each diode, since they will not all be the same. It is still recommended for forward-biased diodes, for the same reason it is recommended when using multiple diodes in series for rectification to increase their overall voltage rating.
1-10 megohm resistors will have little effect on the forward voltage drop of each diode except perhaps at very low currents.
Those resistors are normally used when a reverse voltage is applied and you want to equalize the reverse voltage across each diode.
I don't see how any of that applies here.

#### crutschow

Joined Mar 14, 2008
24,115
According to my LTspice simulation you should be getting over 0.8V drop across each 1N4001 diode at 250mA, giving an output of about 8.6V from a 12V supply.

So to explain your results, here are some choices:
A. Your circuit is wired incorrectly.
B. You don't have a 250mA load.
C. The 12V supply is greater than 12V.
D. The diodes are faulty.

#### MrChips

Joined Oct 2, 2009
19,928
The diodes were under load when measured. The multimeter reads the load at 250ma, at 10.90V. Diodes are standard rectifier diodes 1N4001.

---(+power)----[+1N4oo1-]---[+1N4oo1-]---[+1N4oo1-]---/[Multimeter]\-------(-power)---

With the diodes removed, multimeter reads 11.9V
---(+power)------------------------------------------------------/[Multimeter]\-------(-power)---

Which means a voltage drop of ~0.33V per diode. Nowhere near 0.6-0.8V I read it's supposed to be.
You are performing your test incorrectly.
Replace the voltmeter with a 47Ω 5W resistor.
Measure the voltage across this resistor.

#### MrChips

Joined Oct 2, 2009
19,928
It would be better to use 9V 5W zener and a 10Ω 3W series resistor.

#### MikeML

Joined Oct 2, 2009
5,444
On the other hand, if you limit the x-axis to a practical 10X range of current, for instance in the 10-100mA range, your plot shows why diodes work fairly well. Poor compared to a regulator, but plenty good for things with steady loads.
On the third hand, and the reason I did the sim, is to show what happens if you load the diodes with only 10 or 20megΩ (the input impedance of the voltmeter)!