# Boost converter using TL5001

I am having a circuit for a TL5001 boost converter that outputs approximately 400 Volts.

I am trying to modify the circuit as to output a voltage between 20V- 100V. The output current will be less than 1A.

I am attaching the circuit and hope that I can get your advice on how to do the modification.

I already have a couple bare PCB of the full circuit.

• 9 V input as in 9 V battery? – winny Dec 17 '17 at 9:10
• You are wanting to boost the input voltage by a factor of 10 (for 100V output.) You must therefore provide 10 times as much current on the low side as you need on the high side. That's 10A on the low side if you really need 1A on the high side. You won't be getting that out of a 9V battery. – JRE Dec 17 '17 at 9:51
• For the input power I can actually use an external power supply. – Ihab Riad Dec 17 '17 at 9:55
• Please correct “1A” typo error – Tony Stewart Sunnyskyguy EE75 Dec 17 '17 at 15:55

this chip aims for 1V on the FB pin and wants about 100K of resistance or lower.

Replaceing R4 with 91K and R5 with a 500K potentiomer would get the voltage divider range needed but the resistance would be too high.

so reduce those and R3 by a factor of 5:

replace R3 with 2.0M R4 with 18K and R5 with a 100K potentiometer.

• That sounds like it might get the voltage correct, but what about the current? What would have to be changed in order to get up to 1A from that circuit? – JRE Dec 17 '17 at 9:47
• Is it possible that I can use a 10k ohm potentiometer for R5 but increase the value of R4. I was having in mind of using a digital potentiometer that I can control using a microcontroller. – Ihab Riad Dec 17 '17 at 9:59
• I think the answer to this last question is no as I will actually lose the range of 20-100V. Could you guys give me the equation to calculate the output voltage. Is it ok to use 6 Mohm resistance with the RT pin. – Ihab Riad Dec 17 '17 at 10:06
• Thanks Jasen, Could I use smaller values for R4 and R3 as to use R5 = 10kohm. – Ihab Riad Dec 17 '17 at 11:56
• I Jasen, I tried to use 1.1M for R3 and R5 was 10Kohm, R4 I tried a number of different values but what I noticed was the output became very unstable. Is 1.1M very small. – Ihab Riad Dec 19 '17 at 19:08

For 9V in and 400Vout using R3=10M with divider of 20k pot +R4=22k to FB feedback.You need 400 to 100 gain change so pot changes to 4x(20+22) 168k. = 250k pot // with a shunt R = 500K

Then no change to R4 if you permit <<20V min. I would choose a 5 turn pot.