# Max Vout of current monitor?

Discussion in 'General Electronics Discussion' started by eem2am, Jun 16, 2013.

1. ### eem2am

422
0
Aug 3, 2009
Hello,
I am using ZXCT1009 to monitor current flow from a 3V battery.
What is the maximum voltage that Vout of zxct1009 can be?
(i.e. the voltage at the "iout" pin.)

I.e supposing current from the "iout" pin is 1mA...then how big can i make the output resistor?.....would 1K8 be ok? Whats the highest resistance it could be?

ZXCT1009 Datasheet
http://www.diodes.com/datasheets/ZXCT1009.pdf

2. ### CDRIVEHauling 10' pipe on a Trek Shift3

4,960
652
May 8, 2012
It's a current output, so output voltage isn't specified. Therefore Eout = (Iout * Rout). The value of Rout isn't specified either, as it will depend on the voltage you want to develop across the Rout. At least that's my take on the specs.

Chris

3. ### eem2am

422
0
Aug 3, 2009
The datasheet on oage 5 speaks of a dropout voltage "VDP", but doesnt state what it is......there must be some working "headroom" required by the part?

If the " iout" resistor was 2.5mA and the v+ was 2.5V then the current into it could never be 1mA since this gives no "Headroom"?
But the datasheet doesnt state what is the headroom voltage?

4. ### Harald KappModeratorModerator

11,513
2,651
Nov 17, 2011
The "headroom" is the min. Vcc=2.5V (page 2, table electrical characteristics). This is the minimum voltage that will drop from the Vin pin to the Iout pin. Therefore Vout=Iout*Rout <=(Vin-2.5V) needs to be fulfilled.
The application information on page 4 tells you how to calculate the required resistors.