Some issues with voltage drops ???

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Suppose that you connect only one end of a resistor to the positive
terminal of a 9V (9.23 actual) battery and use a DVOM to measure the
voltage drop between the other end of the resistor and the negative
terminal of the battery. Is the voltage drop across these two points
always going to be 9.23V regardless of the size of the resistor?
Using a 1K and 2K resistor my DVOM measured 9.23, but measured 8.39
with a 1M resistor.

I also did some experimenting with a silicon diode and a 741 op amp.
I am using two 9V batteries to make a +/- 9V power supply. I have a
voltage divider supplying 4.07 (couldn't adjust to exactly 4.0) volts
to the non-inverting terminal and a silicon diode connected between
Vout and the inverting terminal (cathode on the non-inverting side).
On paper, I was expecting to have about 4.67V at Vout and a few micro
volts below 4.07V at the inverting terminal. What I ended up getting
in the real world was 4.26 at Vout and 4.07 at the inverting
terminal. The diode is dropping about 0.2 volts instead of the
typical 0.6 ???

Doing some more experimenting I made up a different circuit using just
a singe 9V battery, voltage divider (POT), and the same diode.
Connecting the anode side of the diode to the output of the voltage
divider and measuring the voltage drop between the free end of the
diode and the negative terminal of the battery produced values ranging
between .20 and .29 volts. Why am I not getting voltage drops around .
6V?

If I make a complete circuit putting the diode in a series with a
resistor, I can get a voltage drop of 0.6 across the diode so I thing
that the diode is working ok.

Any help would be greatly appreciated. Thanks

 

[neon]

get some books and read the writeing don't look at the pages. quit experimenting if you don't even now the wheel of fortune. a diode has an exponential curve. it never is .6 .7.3 v dependes on the current flow. .6v .7v is an assumable impirical number. when you measure less voltage in series with a res. because your meter has a voltage drop because of its internal inpedance. read book quit experimenting and then ask basic questions.

 

[Eeyore]

Suppose that you connect only one end of a resistor to the positive
terminal of a 9V (9.23 actual) battery and use a DVOM to measure the
voltage drop between the other end of the resistor and the negative
terminal of the battery. Is the voltage drop across these two points
always going to be 9.23V regardless of the size of the resistor?
Using a 1K and 2K resistor my DVOM measured 9.23, but measured 8.39
with a 1M resistor.

Because the DVM draws a small input current. In fact the above shows it has
an input resistance of 10 M ohms.

Graham

 

[Rich Webb]

Suppose that you connect only one end of a resistor to the positive
terminal of a 9V (9.23 actual) battery and use a DVOM to measure the
voltage drop between the other end of the resistor and the negative
terminal of the battery. Is the voltage drop across these two points
always going to be 9.23V regardless of the size of the resistor?
Using a 1K and 2K resistor my DVOM measured 9.23, but measured 8.39
with a 1M resistor.

That's because your voltmeter has a high -- but not infinite --
resistance on its input. From the numbers you gave, it's about 10 meg
ohms. You can check the math by figuring a voltage divider with 9.23 V
"on top" dropping across a 1 meg and then a 10 meg resistor. What do you
get for the voltage between the two resistors?

I also did some experimenting with a silicon diode and a 741 op amp.
I am using two 9V batteries to make a +/- 9V power supply. I have a
voltage divider supplying 4.07 (couldn't adjust to exactly 4.0) volts
to the non-inverting terminal and a silicon diode connected between
Vout and the inverting terminal (cathode on the non-inverting side).
On paper, I was expecting to have about 4.67V at Vout and a few micro
volts below 4.07V at the inverting terminal. What I ended up getting
in the real world was 4.26 at Vout and 4.07 at the inverting
terminal. The diode is dropping about 0.2 volts instead of the
typical 0.6 ???

The forward voltage drop across a diode varies with the forward current
and also a few other things. See: The Diode Equation.

Doing some more experimenting I made up a different circuit using just
a singe 9V battery, voltage divider (POT), and the same diode.
Connecting the anode side of the diode to the output of the voltage
divider and measuring the voltage drop between the free end of the
diode and the negative terminal of the battery produced values ranging
between .20 and .29 volts. Why am I not getting voltage drops around .
6V?

See above. It can be a very instructive exercise to actually plot the I
vs V curves for a few diodes. Try this with different types (small
signal, Schottky, rectifier). How do the curves differ? Why?

Note: This is fairly easy to do if you have two multimeters and set one
on a low current scale (in series with the diode) and one on a low
voltage scale (across the diode). Use your pot to vary the voltage and
stop at convenient current readings to write down the I & V values.
First, though, sketch out the circuit and calculate the expected values
for current through the meter and diode (assume a low value for the
voltage drop). You don't want to burn up the meter ...

If I make a complete circuit putting the diode in a series with a
resistor, I can get a voltage drop of 0.6 across the diode so I thing
that the diode is working ok.

Again, see above. Once you've plotted the I-V curves for a few, I think
you'll see the answer.