how does the ESD protection diodes work

[arr_baobao]

Hi guys,

just saw this ESD protection diode circuit, can anyone explain how does the 2 series ESD protection diodes work to protect the circuit from transient events thanks

 

[jiripolivka]

Hi guys,

just saw this ESD protection diode circuit, can anyone explain how does the 2 series ESD protection diodes work to protect the circuit from transient events thanks

View attachment 129863

The two-diode clamping circuit will not allow voltages exceeding the positive or negative lead to get to protected device input.
What is missing is the signal-source resistance which will define a maximum input current for a voltage exceeding the above indicated limits. For such current the diodes must be specified.

 

[Audioguru]

When the input voltage rises to 0.7V higher than Vcc then the top diode becomes forward biased and conducts. Or if the input goes more than -0.7V below (-) or Gnd then the bottom diode becomes forward-biased and conducts. Then the transient is clamped from rising higher if the power supply has a low internal resistance.

 

[arr_baobao]

When the input voltage rises to 0.7V higher than Vcc then the top diode becomes forward biased and conducts. Or if the input goes more than -0.7V below (-) or Gnd then the bottom diode becomes forward-biased and conducts. Then the transient is clamped from rising higher if the power supply has a low internal resistance.

I understand that when there is a spike or rise in data line higher than Vcc, the top diode conducts and volt flow into the path to protect the downstream IC/circuitry. But how does the voltage flow into another Vcc, which is a probably a power supply. Just dont understand the theory behind this.

thanks.

 

[jiripolivka]

I understand that when there is a spike or rise in data line higher than Vcc, the top diode conducts and volt flow into the path to protect the downstream IC/circuitry. But how does the voltage flow into another Vcc, which is a probably a power supply. Just dont understand the theory behind this.

thanks.

What theory? The input voltage is defined by the output voltage from a signal source having a source impedance say R. Then when the signal voltage exceeds Vcc (plus junction voltage, typically 0.7 V), the R with the diode forward resistance forms a voltage divider. Diode forward resistance is much smaller than R (say 10 Ohms against 100 Ohms), so the input voltage to the device is reduced 9 times when the diode opens.
This is why I wrote first that the signal source impedance R is missing from the circuit.

 

[Audioguru]

I understand your question. A power supply is usually with a series pass transistor that does not clamp when the top diode feeds it a positive transient. But if the power supply is a zener diode then it would clamp the transient.
ICs usually have a supply bypass capacitor so that they do not oscillate or ring. That capacitor would take time to charge and limit a short duration transient.