dropping voltage with diodes in series

Electroplater

Member
so if u have an activation voltage with a diode at about 2 volts, is this activation voltage taken off my voltage as I pass through the diode?

Id say it is, but in falstad it doesn't put it in the simulation, but I've seen circuits where people drop the volts by passing through diodes.

Thanks for any help on the issue.
 
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Done it many times, if my supply voltage is higher than the needed voltage, the use of series diodes will reduce the voltage by 0.7V each silicon diode added.

+12V-----|>|------|>|-----|>|-------|>|-------+9.2V [0.7Vdrop x 4 Si diodes= 2.8V less]

+5V---------|>|---------|>|---------|>|-------+2.9V [0.7Vdrop x 3 Si diodes=2.1V less]

+2V----------------------|>|-------------------+1.3V [2V minus 0.7V trough 1 Si diode = 0.7V less]

Just make sure to use any needed heat sink and not exceed current ratings of the diodes.
 
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Note that 0.7V is a nominal value at a few mA of current.
It varies with the current in a logarithmic manner.

Below is the simulation for a small diode:

 
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The voltage drop of diodes decreases with increasing temperature.

Current flow increases temperature which decreases voltage drop which increases current flow.....
 
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You can paste Falstad links after File>export> make shortcut > copy.
All logic IC's are ideal in Falstad's with 0 Ohms out, so you must add R depending on logic family determined often by Vdd max e.g. Diodes also have bulk resistance which depends on size. The default diode can be modified if you know how.

74' logic family
5.5V max (~50 ohms) or 3.6V max (~25 Ohms)
 
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shortbus= said:
Or beer
Click to expand...
Never heard of it

However Google informs it was a very popular American beer, now long defunct.

It also informs me that there's a local micro-brewery in Derby, only about 20 miles from me - never heard of that either, but micro-breweries have become very popular in the last few years.

Home

www.falstaffbrewery.co.uk
 
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