[sdiy] Voltage regs + diode drops.

Neil Johnson neil.johnson97 at ntlworld.com
Sat Feb 21 20:07:41 CET 2009

Hi Dave,

Dave Kendall wrote:
> Is there any good reason not to use diodes in series after the  
> output terminal of a positive voltage regulator to get odd  
> voltages, eg. 2 x 1N4001 after a LM7809 to get just below +8V?
> It seems it *should* work, but I haven't seen it done anywhere yet.  
> Are there any potential problems that anyone can think of?
> FWIW, It would be for a CMOS logic circuit, and with a single OP- 
> amp or transistor level shifter on the output.

My own opinion: avoid if possible.  Voltage drop across the diode  
depends on current flow.  At low currents the voltage drop reduces,  
so increasing the output voltage.  It _can_ work if you can guarantee  
a reasonable current (a few mA or depending on the diode in question)  
through the diode to bias it well into conduction.  Same applies to  
zener diodes.

For example, some quick results from the bench with various diodes in  
series with a digital multimeter with 10M input resistance (i.e.,  
very low current, a uA or so):

Supply: 15.11V

1N4148 forward biased: 14.80V

= a drop of 0.31V (not 0.6V)

BZX55C3V6 zener forward biased: 14.62V

= a drop of 0.49V (not 0.6V)

  -- same --     reverse biased: 14.03V

= a drop of 1.08V (not 3.6V)

This shows that for both diode types what the simple model says  
("0.6V" or "the zener voltage") about the voltage drop is nowhere  
near at very low currents (e.g., CMOS circuits at low frequencies or  

In your case your single op-amp may consume enough static current to  
bias the diodes well into conduction.  But check.

Also note that rectifier diodes tend to have higher forward voltage  
drops, about 1.0V.


More information about the Synth-diy mailing list