# Diode and Capacitor [Basic Questions]

#### Clyd

Feb 19, 2014
3
Hi I'm a first year student and currently studying AS Electronics. I don't quite understand how capacitors charge and how diodes clamp in an astable circuit, although I know quite well how these components function independently or at least their basic characteristics. I've also done a lot of research, watched youtube videos and used educational websites and I'm now getting some of the idea, however there are still gaps being left since we touched down on this topic.

So my questions are:

• At which side does a diode clamp and how?

Does voltage without current charge a capacitor and vice versa?

Sorry, these questions might seem very stupid, maybe I just have very low technical understanding regarding circuits.

#### Arouse1973

Dec 18, 2013
5,178
Do you have an example circuit. There are many astable circuit configuration.

#### BobK

Jan 5, 2010
7,682
Current charges a capacitor. The voltage across it will change as it charges.

A diode limits the forward voltage across it to something like 0.6V. In an astatable multivibrator circuit built from two bipolar NPN transitors, it is used to keep the voltage on the base from going far enough negative to damage the transistor.

Bob

#### Clyd

Feb 19, 2014
3
Ok so I'll just bring my notes here:

On the first note, I don't understand why Vc is clamped only at 0.7V when you have 5V as feedback coming from the schmitt inverter, cause on the last part, the input became ( 5 - 0.7 = 4.3V) only ACROSS the forward biased diode.

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#### BobK

Jan 5, 2010
7,682
Did not notice that there is a resistor between the output of the Schmitt trigger and the anode of the diode? How does that effect the voltage on the anode?

Bob

#### Clyd

Feb 19, 2014
3
But when I follow the Ohm's law at least from what I understand is that there is no voltage drop across a single resistor which means the voltage are gonna be the same as the supply? Since there's no specified resistance, let's say we've got 2 Ohms

I = V/I
5/2 = 2.5A

V = IR
2.5 x 2 = 5V

now how does the diode affect the 5V when it is forward biased?

and what will be the effect? would it become 4.3V?

Sorry if things are bit shady...

#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
At a theoretical level, the diode does not affect the resistor. The resistor will continue to operate as per ohms law. In fact (ignoring the charge on the capacitor) the current flowing through the diode when it is forward biased comes from the output of the gate, through the resistor, and then through the diode.

The diode has a constant voltage drop (theoretically), the gate has a constant output voltage (theoretically). This works out close enough in practice provided R is large

#### BobK

Jan 5, 2010
7,682
But when I follow the Ohm's law at least from what I understand is that there is no voltage drop across a single resistor which means the voltage are gonna be the same as the supply? Since there's no specified resistance, let's say we've got 2 Ohms

I = V/I
5/2 = 2.5A

V = IR
2.5 x 2 = 5V

now how does the diode affect the 5V when it is forward biased?

and what will be the effect? would it become 4.3V?

Sorry if things are bit shady...
Yes, the diode will allow enough current to flow through the resistor to drop the other 4.3V, as a first approximation. Of course if the current required is too high, it something will burn out. If the current is very low, (say it is a 1 Meg resitstor) the current will be below the knee of the diode's IV curve, and the voltage drop across the diode will be lower than the nominal 0.7V.

Bob

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