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Gazoo said:If you have your meter on the 10AMP scale reading, then it will show .25ma's. This is the same as 250ma's.
Gazoo said:Well I really don't know...you have an autoranging meter and it looks like you have it all set up like you did before. I don't know why it reads .25ma's. Sorry.
roSSco said:The diode and cap are in PARALELL to the LD. If you think those items have NO EFFECT on the current that gets to the LD, you are mistaken.
I'll ask again: Do you want an estimate of the current going to the LD, or do you want to know exactly how much current is going through the diode?
knimrod, while many of the things you say above are true, they do not apply to the situation at hand.
chimo said:If you are using the meter in current mode in series with your battery, do not use the mA mode, use the 20A mode (with the probe plugged into the proper location). The mA mode introduces a larger resistance in the circuit and will cause the driver to drop out more quickly.
Note: when you insert the DMM in the circuit, you are affecting operating parameters of the circuit. A better way to measure the current going to the LD is to measure the voltage drop across the resistors between Vout and Vadj. Since you know the value of the resistors, you can calculate the current using: I = V/R
Why do it this way? The DMM has a very high impedance when measuring voltage. (Translation: it draws essentially no current and therefore does not affect the operation of the circuit)
knimrod said:[quote author=chimo link=1186966870/240#248 date=1200262825]If you are using the meter in current mode in series with your battery, do not use the mA mode, use the 20A mode (with the probe plugged into the proper location). The mA mode introduces a larger resistance in the circuit and will cause the driver to drop out more quickly.
Note: when you insert the DMM in the circuit, you are affecting operating parameters of the circuit. A better way to measure the current going to the LD is to measure the voltage drop across the resistors between Vout and Vadj. Since you know the value of the resistors, you can calculate the current using: I = V/R
Why do it this way? The DMM has a very high impedance when measuring voltage. (Translation: it draws essentially no current and therefore does not affect the operation of the circuit)
chimo said:[quote author=knimrod link=1186966870/240#250 date=1200268609][quote author=chimo link=1186966870/240#248 date=1200262825]If you are using the meter in current mode in series with your battery, do not use the mA mode, use the 20A mode (with the probe plugged into the proper location). The mA mode introduces a larger resistance in the circuit and will cause the driver to drop out more quickly.
Note: when you insert the DMM in the circuit, you are affecting operating parameters of the circuit. A better way to measure the current going to the LD is to measure the voltage drop across the resistors between Vout and Vadj. Since you know the value of the resistors, you can calculate the current using: I = V/R
Why do it this way? The DMM has a very high impedance when measuring voltage. (Translation: it draws essentially no current and therefore does not affect the operation of the circuit)
knimrod said:[quote author=chimo link=1186966870/240#251 date=1200270055][quote author=knimrod link=1186966870/240#250 date=1200268609][quote author=chimo link=1186966870/240#248 date=1200262825]If you are using the meter in current mode in series with your battery, do not use the mA mode, use the 20A mode (with the probe plugged into the proper location). The mA mode introduces a larger resistance in the circuit and will cause the driver to drop out more quickly.
Note: when you insert the DMM in the circuit, you are affecting operating parameters of the circuit. A better way to measure the current going to the LD is to measure the voltage drop across the resistors between Vout and Vadj. Since you know the value of the resistors, you can calculate the current using: I = V/R
Why do it this way? The DMM has a very high impedance when measuring voltage. (Translation: it draws essentially no current and therefore does not affect the operation of the circuit)