# Change current of bandgap

Here is my bandgap,and i want to change the current "I" ,according to the formula:

I=$\frac{nk\times lnK}{qR}$, n,k and q are constant

So i can only change $lnK$ and R to adjustment the current,the calculation of $ln$ is too complicated ,so i choose to change the value of R,after i change it,the voltage before R1 change,but the current does not ,what do i miss for adjusting the current? • Pro tip: instead of drawing 8 PNPs in parallel there's an option "mult" which you can set to 8 and then one symbol gives you 8 transistors in parallel. – Bimpelrekkie Nov 11 '17 at 12:13
• actually i know that,thx @Bimpelrekkie – Shine Sun Nov 11 '17 at 12:21
• The resistor value should influence the current as it changes the current where the current through Q1 and R2 are equal. I suspect something else is going on. If you annotate the DC operating points and node voltages maybe things will become more clear. – Bimpelrekkie Nov 11 '17 at 12:27
• @Bimpelrekkie like this? – Shine Sun Nov 11 '17 at 13:17
• Uh, I only see the operating point info of the resistors. No node voltages which are the first thing I'd want to see. – Bimpelrekkie Nov 11 '17 at 21:43

## 1 Answer

You have a feedback loop (the two branches n the left) and an output stage (the branch on the right). The formula you show is for the feedback loop. K controls the voltage drop across the resistor in the loop. The output current can be changed by the PMOS current mirror ratio between the output and the loop branches. The resistor in the output branch will control the output voltage. In order to get a temperature independent output voltage you need to bring the output resistor, output diode in harmony with the loop resistor and diodes.