Clamping Diodes

[juantravel]

I'm new to electronics and I'm trying to understand how do diodes work as voltage limiters. Ive seen schematics that have two diodes with one p region facing the input and another with n regions facing the 0v (negative).

0v
|
__|__
/ \
/ \ diode
/___\
|
__input______|______________output
|
__|__
/\
/ \
/___\
|
|
0v

 

[(*steve*)]

Your diagram got little broken. But the purpose is to clip the input signal to approx 0.7 volts either side of 0v.

This is typically done where the signal is expected to be small in amplitude (say no more than 0.2 V RMS) to prevent overload to the following stages if an unexpected spike appears on the input.

Another common thing you'll see is diodes to either supply rail. This is used to limit the input voltage to a diode drop more than the supply rail. This is used to protect inputs from excessive voltages.

 

[juantravel]

reply

I do understand that the two diodes are set up to clip the voltage to .7 which is the voltage needed for the electrons to flow over the junction in silicon. What characteristic of the diode allows that. I'm guessing that if a input is higher then .7 it causes and avalanche effect... correct? but if that is the case wouldn't that just fry the diode and cause the high voltage to go into the amp circuit. The circuit I was looking at had a op-amp. The input had 2 diodes one with n type to the input and another with the p type to the input. Then the input continued to the op amp.

 

[(*steve*)]

It depends on the impedance of the input signal. The impedance acts like a series resistance and limits the current. Even small signal diodes can handle currents of many tens to several hundreds on mA.

 

[juantravel]

reply

input impedance? Is that the op amps impedance? So what I'm thinking is that the diode becomes forward bias when voltage gets higher then .7 and the diode becomes low impedance and allows the voltage to flow out through the diodes instead of flowing over into the input? Is this type of circuit used for voltage spikes or voltage limiter? I understand that current will flow through the least resistive place. wouldnt that cause no voltage to the op amp?

 

[(*steve*)]

Sorry, the impedance of the signal SOURCE, the impedance of the output of whatever is providing the input TO this stage (this stage is the one with the back to back diodes across it).

Yes, this limits the input voltage swing to around 0.7V either side of the 0V rail.

No, any signal less than 0.7V will pass through to the input of the op-amp (actually it's a bit more complex than that because diodes don't suddenly switch on at 0.7V, but let's just pretend they do). Any signal greater than 0.7V will have that portion above 0.7V clipped off and the remainder will be passed to the op-amp.

Current does NOT flow through the least resistive place. If you're taking about resistances, it flows through all resistances in an amount inversely proportional to each individual resistance.

A diode is not like a resistor, the resistance varies with voltage. As the voltage rises, the current rises disproportionately.

 

[NickS]

a little late and unnecessary but here is the picture from post 1 (slightly modified).

[FONT="Courier New"]                     
                    0v
                     |
                   __|__
                    / \ diode
                   /___\
                     |
                     |
          input------o-----------------output
                     |
                   __|__
                    / \
                   /___\
                     |
                     |
                    0v
[/FONT]