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Measuring voltage drop across a diode in a switching circuit

FreshmanNewbie

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I have the below circuit where the input voltage is 8V and a PWM pulse is given to the base of the Q101 transistor.

Circuit.PNG

Circuit Conditions:

-The 8V is the voltage source. SBATT is connected to a load which draws a minimum current of 80mA and a maximum of 100mA.

-The temperature of the circuit , which I have measured is not 150degC. The temperature is around 85degC

-The diode is not a schottky diode.

From what I have measured,

The voltage at the anode of the diodes D102 and D103 is constant like around 8V. But when I measure the voltage at the cathode of the D103 and D103, I get a PWM waveform whose peak value is 7.7V.

Voltage measurements are done using oscilloscope.

My questions :

Why is there a 0.3V voltage drop across the diode? Shouldn't it be typ. 0.7V?
 

betwixt

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Why is there a 0.3V voltage drop across the diode? Shouldn't it be typ. 0.7V?
Only when the diode is conducting and passing enough current. If the voltage can't drop below 7.7V after the diodes it probably means there is already that much from storage in the capacitors or leakage from 'SBATT'.

Consider what would happen if you removed everything after the diodes, would you still expect to see 0.7V across them?

Why are you using two 1A diodes in parallel when the load is only 0.1A ?

Brian.
 

FreshmanNewbie

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The peak value is 7.7V max

- - - Updated - - -

Thank you for the answer. Could you please explain in simple terms when you said "If the voltage can't drop below 7.7V after the diodes it probably means there is already that much from storage in the capacitors or leakage from 'SBATT'." - What is the storage that you are mentioning here?

"Consider what would happen if you removed everything after the diodes, would you still expect to see 0.7V across them?" - If I removed everything after the diodes, I would get a voltage of 8V-0.7V (typical) = 7.3V. Am I right?


And Actually, the 8V source, comes for two different voltage sources which are supposed to be identical in value. So, when one source is OFF, the other 8V source can power the circuit. That's why.
 

FvM

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The peak value is 7.7V max
That's voltage with zero output current - irrelevant.

Instead of assuming an idealized current independent 0.7V voltage drop, you want to look for the actual diode I/V characteristic.
 

KlausST

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Hi,

, I would get a voltage of 8V-0.7V (typical) = 7.3V. Am I right?
0.7V is by far not a fixed value.
It depends on a lot of parameters
* device type
* internal (and external) series resistors
* current
* temperature
....

So you say you have a scope.. thdn ypu can see all this ... and can see where the "0.3V" are coming from.
Did you get this value from the scope? How exactly did you get this value (timing, maeasurement circuit, test conditions)?

Klaus
 

betwixt

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"Consider what would happen if you removed everything after the diodes, would you still expect to see 0.7V across them?" - If I removed everything after the diodes, I would get a voltage of 8V-0.7V (typical) = 7.3V. Am I right?
So if you did this: (forgive the crude graphics)
+8V ------ Diode ------- (no connection)
0V ----------------------------------------
And you place a meter across the diode you would expect to see 0.7V? Try it.

Brian.
 

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