LM317 adjust pin current + general questions

[bitshift]

Hey everyone,

I've been building an adjustable power supply with the LM317T. The datasheet http://www.st.com/web/en/resource/technical/document/datasheet/CD00000455.pdf states the adjust pin current is typically 50uA but maximally 100uA.

When I measure the current using my multimeter, I measure 4.17mA. This is a lot more than I was expecting.

I've attached a picture of my circuit.

If I understand multimeter burden voltage correctly, then this could not be the cause because the burden voltage could only lower the current and not increase it.

This leads me to believe my understanding of the LM317 is incorrect.

From what I understand, in the basic configuration, the output is 1.25V higher than the voltage of the adjust pin both referenced to ground.

R1 is placed from the output to the adjust pin in order to satisfy the minimum load requirement. This is because there is always a 1.25V differential across R1 giving you a reliable current. If you're driving the adjust pin with an op amp, you can take R1 out but then you need to guarantee the minimum load current in another way.

R2 is for.... well I don't actually know. I think this is where the problem lies. What is the purpose of R2?

Now back to the current measurement on the adjust pin. Since the LM317 is a linear regulator I am going to assume ohms law applies. As a result the current on the adjust pin should change depending on the voltage applied to it. This seems to compute since I'm applying 5V to the adjust pin with a resistance of 1.2K which gives me the measured 4.17mA. So why does the datasheet state 50uA - 100uA on the adjust pin?

[MLXXXp]

You're measuring the total adjust circuit current, including that flowing through R1. For the adjust pin current you need to measure between the adjust pin and the junction of the two resistors. Move the resistor leads from column 6 to column 5, then place your ammeter between column 6 and column 5.

Since there will be a constant 1.25V across R1, there will be a constant current through it. Almost no current flows out of the adjust pin (50uA - 100uA), so the current through R2 will equal the constant current through R1. Therefore, there will be a constant voltage across R2. The voltage across R2 plus 1.25V will be the output voltage.

[bitshift]

You're measuring the total adjust circuit current, including that flowing through R1. For the adjust pin current you need to measure between the adjust pin and the junction of the two resistors. Move the resistor leads from column 6 to column 5, then place your ammeter between column 6 and column 5.

Since there will be a constant 1.25V across R1, there will be a constant current through it. Almost no current flows out of the adjust pin (50uA - 100uA), so the current through R2 will equal the constant current through R1. Therefore, there will be a constant voltage across R2. The voltage across R2 plus 1.25V will be the output voltage.

Perfect explanation thanks MLXXXp!

When I measure the the current from the adjust pin to the junction of the two resistors I get a nice and typical 48uA