Diode switch - Electronic Circuits Projects Diagrams Free

prprog · 8th August 2008, 07:37 PM

I started reading about diode switching capabilities. I don't think I undertand. Based on some diagrams and text is the following circuit correct? (I probably draw the diode upside down???) Is there some point (control voltage) at which the diode will conduct and there will be 5v at the red dot? (refer to the attach diagram)

Trying to learn the basics. Help will be appreciated.

Thanks a lot.

crutschow · 8th August 2008, 07:46 PM

The voltage in is going to ground so it won't work as shown.

If you remove the ground from the input then it depends upon the load at the red dot. If it's just a high impedance voltmeter and the diode is silicon, then the voltmeter will read +5V with an input of about +5.6V. A silicon diode has about a 0.6 to 0.7V drop in the forward direction for low currents.

prprog · 8th August 2008, 08:47 PM

Quote:

Originally Posted by crutschow

The voltage in is going to ground so it won't work as shown.

If you remove the ground from the input then it depends upon the load at the red dot. If it's just a high impedance voltmeter and the diode is silicon, then the voltmeter will read +5V with an input of about +5.6V. A silicon diode has about a 0.6 to 0.7V drop in the forward direction for low currents.

Let me see if I understood. If I put +5.6V at the control input them I will be able to measure +5V at the red dots. Correct? Is the circuit correct now?

At less than +5.6V there will be no +5V at the red dots?

Thanks,

Mikebits · 8th August 2008, 09:03 PM

Have a read here. Good explanation

http://www.allaboutcircuits.com/vol_3/chpt_3/1.html

AllVol · 9th August 2008, 03:12 AM

Voltage at the red dot with no load should measure the input voltage minus the drop across the diode, which is around .7 volts. Example: Vin - Vdrop = Vout or 5 - .7 = 4.3 no load.

reese · 12th August 2008, 06:31 PM

Because there is no load at the output of the diode, you should measure the input voltage at the output. what you have here is an open circuit with no return path. If you use a resistor at the output, then you would measure around .7V if it's a silicon diode.

audioguru · 12th August 2008, 07:47 PM

The datasheet for the 1N4148 silicon diode shows that its forward voltage is only about 0.2V at a current of 0.5uA which is a 10M meter input with 5V across it.
The forward voltage is 0.37V at 10uA, 0.6V at 100uA, 0.625V at 1mA and 0.73V at 10mA. at 100ma the forward voltage is a little over 0.9V and at 800mA it is almost 1.5V.
The forward voltage reduces a little when the temperature of the diode increases.

reese · 12th August 2008, 08:02 PM

Typically you would use a resistor that would not exceed the max rating for the part. The 1N4148 (1N914) is rated up to 200mA continuous current. A circuit designed to draw 3-5mA is a good target. This will give a Vf of up to 700mV across the diode.

Reese

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8th August 2008, 07:46 PM	(permalink)
crutschow	The voltage in is going to ground so it won't work as shown. If you remove the ground from the input then it depends upon the load at the red dot. If it's just a high impedance voltmeter and the diode is silicon, then the voltmeter will read +5V with an input of about +5.6V. A silicon diode has about a 0.6 to 0.7V drop in the forward direction for low currents. __________________ Carl

8th August 2008, 09:03 PM	(permalink)
Mikebits	Have a read here. Good explanation http://www.allaboutcircuits.com/vol_3/chpt_3/1.html __________________ Forget injuries, never forget kindnesses.

9th August 2008, 03:12 AM	(permalink)
AllVol	Voltage at the red dot with no load should measure the input voltage minus the drop across the diode, which is around .7 volts. Example: Vin - Vdrop = Vout or 5 - .7 = 4.3 no load. __________________ All those who believe in psycho kinesis, raise my hand.

12th August 2008, 06:31 PM	(permalink)
reese	diode output Because there is no load at the output of the diode, you should measure the input voltage at the output. what you have here is an open circuit with no return path. If you use a resistor at the output, then you would measure around .7V if it's a silicon diode.

12th August 2008, 07:47 PM	(permalink)
audioguru	The datasheet for the 1N4148 silicon diode shows that its forward voltage is only about 0.2V at a current of 0.5uA which is a 10M meter input with 5V across it. The forward voltage is 0.37V at 10uA, 0.6V at 100uA, 0.625V at 1mA and 0.73V at 10mA. at 100ma the forward voltage is a little over 0.9V and at 800mA it is almost 1.5V. The forward voltage reduces a little when the temperature of the diode increases. __________________ Uncle $crooge

12th August 2008, 08:02 PM	(permalink)
reese	Typically you would use a resistor that would not exceed the max rating for the part. The 1N4148 (1N914) is rated up to 200mA continuous current. A circuit designed to draw 3-5mA is a good target. This will give a Vf of up to 700mV across the diode. Reese