I am new to this and have searched Hi and Low on how to install a diode. What I want to do is quite simple.

I have a Circuit that is 5 vdc 200mA. I have two sources of 5+ vdc that I want to isolate.

I need to put in two Diodes.

What type and how do I install them. I looked at a lot of you tube videos but I did not find the very basic installation.

 

Welcome to AAC!

The polarity is correct. 1N4001 would work.

What is your intent? How closely matched are the output voltages of the supplies?

You're going to have a voltage drop across the diodes.

 

The diodes are shown connected correctly.
For lowest voltage drop, you want to use Schottky diodes rated for at least 1 amp, such as a 1N5817.
Their voltage drop should about 0.3V @ 200mA.

For a lower voltage drop than that you would need to go to an ideal diode circuit, such as the LTC4411.

 

The two diodes will work assuming you can connect the returns of the two power supplies and the circuit returns all together.

I have no idea what isolated means in this case.

 

As shown, the two diodes will isolate the two 5 V sources from each other. When the switch is closed, both sources might supply current to the load, depending on the output voltage of each supply and the forward voltage drops of the two diodes. If you want one supply always to dominate when both are connected, adjust its output up to 5.1 V.

As noted, the voltage drop across standard rectifiers is over 10% of a 5 V source. Schottkys are better, and they are available as a single package dual-diode. This might be more convenient.

ak

 

Welcome to AAC!
View attachment 123648
The polarity is correct. 1N4001 would work.

What is your intent? How closely matched are the output voltages of the supplies?

You're going to have a voltage drop across the diodes.

On paper; the volt drop for silicon diodes is 0.7V, but it creeps nearer to 1V if you get close to its maximum current rating.

Shottky barrier diodes have lower drop at around 0.2V, Reverse voltage rating starts at 20V and gets expensive past 60V, They can be damaged by static charges.

1N5817 would probably be OK, but the diode that gets switched out of circuit could do with a high resistance bleed in parallel to dissipate any static charge that might accumulate. SB diodes are naturally a little bit leaky - so about 47k probably wouldn't make any noticeable difference.

 

N5817 would probably be OK, but the diode that gets switched out of circuit could do with a high resistance bleed in parallel to dissipate any static charge that might accumulate.

I don't understand that.
The input is connected to a voltage source and the output goes to the load.
How does the diode get "switched out" and accumulate a static charge?

 

I don't understand that.
The input is connected to a voltage source and the output goes to the load.
How does the diode get "switched out" and accumulate a static charge?

I guess you must've missed the switch in the TS's diagram.

SB diodes are susceptible to static - a small risk, but its there.

 

I guess you must've missed the switch in the TS's diagram.

Somehow I did.
I had the circuit in my mind without the switch.

 

My Intent is to use a PLC that has a pull down resistor internally. I suspect that there is a single power for the 5 vdc.(both the Out and the 5 vdc on the PLC but I do not know) The 5 vdc on the PLC is just a positive output and is used with a switch to go to the DIN. This is the pull down resistor. I do not know what the resistor "is" as it is inside sealed box. I have been told that I could get the 5+ vdc from an Output to pulldown the resistor but I should put a diode on the 5 vdc of the PLC. I was thinking that I should "protect" the Out on the PLC from reverse voltage as well (that was my isolate).
There is 5+ vdc on the hot side of the toggle 100% of the time. The 5+ vdc from the Out is normally momentary but could be left "on" by error. One would not normally use the toggle and the Out at the same time, but Murphy is alive and well. I think the voltage drop would not be an issue as I think it could be from 3 to 5 vdc for the resistor.

 

Are you sure, you don;t mean a pull-up resistor is in the PLC? You can usually convert a TTL output to an open collector output with a diode.

If you knew the zero output and the 1 output, you;d know something more than you know now.
Supppose the 1 ouput is 5 and the 0 output is 0.6 V, then you would not have a pull down resistor.

I think you can use the potentometer trick to find that resistor value.