Control on/off state of IC with arduino

ScottWang

Joined Aug 23, 2012
7,397
The DC to DC converter only two pins, if you treat it as a DC relay, then it can be using the n-ch or p-ch mosfet to drive it.

If you already bought it then you can try it in two ways with n-ch or p-ch mosfet.
 

Thread Starter

jpborunda

Joined Apr 9, 2014
55
Take an Ohmmeter. Measure between IN- and OUT-. If the resistance is ∞, then you could use a low-side switch...
Ok so I just did the following measurements:
Vin(-) and Vout(-) = 0Ω
Vin(+) and Vin(-) = 5kΩ.
Vout(+) and Vout (-)= 121MΩ
Vin(+) and Vout(+) = 0Ω

Can you explain what does it mean that there is no resistance ? I mean, why would an infinite resistance mean that I CAN make low-side connection?

From what you guys explained, it seems like it can only be connected as a high side switch.
 

MikeML

Joined Oct 2, 2009
5,444
Ok so I just did the following measurements:
Vin(-) and Vout(-) = 0Ω
Vin(+) and Vin(-) = 5kΩ.
Vout(+) and Vout (-)= 121MΩ
Vin(+) and Vout(+) = 0Ω

Can you explain what does it mean that there is no resistance ? I mean, why would an infinite resistance mean that I CAN make low-side connection?

From what you guys explained, it seems like it can only be connected as a high side switch.
The fact that there is a DEAD SHORT from Vin- to Vout- means that it is NOT a transformer-isolated switcher. This means that to switch it on/off with an external FET, the FET has to break the Vin+ line, not the Vin- line.

The best way to implement a high-side switch is to use a PFET, source to the Positive power supply, drain to the converter Vin+, and to turn on the PFET, the gate has to be driven low.

I think we have come full circle...
 

Thread Starter

jpborunda

Joined Apr 9, 2014
55
The fact that there is a DEAD SHORT from Vin- to Vout- means that it is NOT a transformer-isolated switcher. This means that to switch it on/off with an external FET, the FET has to break the Vin+ line, not the Vin- line.
Ok, I just wanted to know what it meant physically, so thanks for the explanation!

I've managed to obtain a PFET and tested the proposed circuit, it works as expected.

Thank's a lot for the help.
 
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