Voltage Drop of 1N4007 Diode 0.3 V ?

[pkshima]

So I am measuring the voltage with the multimeter and diode in series with a 9V battery.

The reading is 9.01 V whereas with multimeter directly across the battery, the reading is 9.41 volts.

So the voltage drop comes out to be 0.3 V.

But shouldnt it be 0.7 V for 1N4007 is a silicon diode ?

 

[MikeMl]

Do you have a load between the cathode of the diode and the neg terminal of the battery (other than the internal resistance of the voltmeter)? The meter resistance is high, probably >20megOhm, so the current being drawn through the diode is likely less than 400nA. Try again with currents between 1mA and 1A flowing through the diode.

 

[crutschow]

But shouldnt it be 0.7 V for 1N4007 is a silicon diode ?

That's a generality for normal operating currents where the forward drop is typically about 0.65-0.7v, but it varies from that with current (look at a data sheet for curves showing this). Ignoring the diode resistance, its drop is actually logarithmic versus current. Since a typical multimeter has a 10MΩ input impedance, the diode current for your measurement is only about 0.9 micro-amps, and at that low current the diode apparently only has a forward drop of about 0.3

 

[pkshima]

Thanks MikeMl and crutschow ! It surely was that.

I get expected reading if I put a diode and 220 Ohm resistor in series with the battery and measure the voltage across the diode. It was exactly 0.7 volts.

I wanted to know this so I can use a 6V battery with PIC by using a diode in series.

Thanks again !!

 

[joespaz7]

So I am measuring the voltage with the multimeter and diode in series with a 9V battery.

The reading is 9.01 V whereas with multimeter directly across the battery, the reading is 9.41 volts.

So the voltage drop comes out to be 0.3 V.

But shouldnt it be 0.7 V for 1N4007 is a silicon diode ?

you need a resistor in seris with it

 

[MikeMl]

you need a resistor in seris with it

Not if the OP's load draws sufficient current...

 

[Hero999]

Post the datasheet for the PIC - most PICs can take 6V anyway.

 

[crutschow]

you need a resistor in series with it

A resistor in series has little effect on the diode forward voltage drop. It's only to the extend that it changes the current through the diode.

 

[pkshima]

Post the datasheet for the PIC - most PICs can take 6V anyway.

Its a PIC16F628A I/P (DIP package). Max it takes is 5.5 V.

 

[Hero999]

Its a PIC16F628A I/P (DIP package). Max it takes is 5.5 V.

According to the datasheet, the absolute maximum rating is 6.5V (see page 131) so running at 6V isn't exactly good for it but it shouldn't do any harm especially if it's for short periods and 5.7V is safer still.
https://www.electro-tech-online.com/custompdfs/2010/03/16f628a.pdf

If you're intending to use four AA cells, I say stick the diode in series and it'll also serve as an idiot diode as well as a voltage dropper.

 

[pkshima]

According to the datasheet, the absolute maximum rating is 6.5V (see page 131) so running at 6V isn't exactly good for it but it shouldn't do any harm especially if it's for short periods and 5.7V is safer still.
https://www.electro-tech-online.com/custompdfs/2010/03/16f628a-1.pdf

If you're intending to use four AA cells, I say stick the diode in series and it'll also serve as an idiot diode as well as a voltage dropper.

Quite true. Thanks !! I guess I will go with that.
Its for my bot which I intend to run on a 6V 4.5 AH SLA battery.
So I will have the diode in series for the uC portion so that it gets 5.3 volts.
The 78L05 I have been using was forcing me to use a 9V + battery.


Thanks again everyone !!

 

[mvs sarma]

Its a PIC16F628A I/P (DIP package). Max it takes is 5.5 V.

be safe to protect the chip
you can afford one LM1117-5 LDO regulator for 6V input.( provided 6V is stable .

though the package is different, it can be adopted on solder side.
Otherwise work with 3 dry cells at 4.5V
PIC works well at 4.5 or even 3.8V till the cells discharge.

 

[Hero999]

Quite true. Thanks !! I guess I will go with that.
Its for my bot which I intend to run on a 6V 4.5 AH SLA battery.
So I will have the diode in series for the uC portion so that it gets 5.3 volts.
The 78L05 I have been using was forcing me to use a 9V + battery.

Unfortunately the voltage of a 6V SLA can be as high as 7.2V when hot off the charger so you'll need at least a couple of diodes to drop the voltage to a safe level.

A LDO regulator is a better option but more expensive.

 

[mvs sarma]

Unfortunately the voltage of a 6V SLA can be as high as 7.2V when hot off the charger so you'll need at least a couple of diodes to drop the voltage to a safe level.

A LDO regulator is a better option but more expensive.

as we do testing and make a project ( not a bulk manufacturer), a one off would always help, instead of killing a microchip.
somehow the series drop methods are worth LM2931 series also have 5V with

 

[MikeMl]

Another nit: If you are using the ADC or any other parts that require a fixed, and known Vcc, the regulator will keep the Vcc within a few mV, while all the schemes using dropping diode(s) will cause the Vcc to vary by at least 1.5V as the battery discharges.