IC Voltage regulator problem?

Anyone built the power supply from "electricity & electronics" by randy slone?

I just completed construction of this power supply and made the modifications on page 333 using the IC voltage regulators (nte956 and nte957). My input voltage is about 36v however the output voltage from the IC is about 28v.

The previous circuit that did not use the IC's gave a variable output of about +- 3.8 to 36 volts DC. Now I'm getting about +- 1.2 to 28. I'm wondering that if incorporating the IC's to improve the voltage regulation under load is the cause of this? Is this the expected outcome?

Here is a diagram of the positive side of the regulated power supply.

36V raw +dc__________________NTE956__________________________________Regulated V Pos. | | | | | | | | + |_____220 ohm 2W 10uF 35VDC 4700uF 50W | | - 5Kohm 2W potentiometer 10uF 35VDC | | | | | | Circuit Common Circuit Common Circuit Common
Reply to
Loading thread data ...

Your circuit can't possibly produce a +36 volt output due to a couple of reasons. (1) The NTE956, which is the equivalent to the standard LM317 voltage regulator, needs a 3 volt minimum differential between input and output voltages. Given your +36V input, the best that this regulator can produce is +33 volts. (2) Use the equation Vout = 1.25 (1 + (R2/R1)) + R2(Iq) to calculate the resistor values. As you can see by the equation, your resistance values will produce Vout of 29 volts... close to what you are getting. (R1 = 220 ohms and R2 is the 5K pot in your diagram).

In order to increase the Vout of your circuit, you'll have to either increase the total resistance of the pot or decrease the resistance of R1. Use the formula to calculate values to suit. Hint... it's a lot easier to calculate for a smaller R1, because of the limited selection of values for common pots.

Dave M
MasonDG44 at comcast dot net  (Just subsitute the appropriate characters in 
 Click to see the full signature
Reply to

You can figure it out yourself. The voltage regulator's mission in life is to make sure there is about 1.25 V across the 220 ohm resistor. In order for that to be the case (assuming there is negligible current from the IC common terminal), then you have a voltage divider formed by the 220 ohm resistor and 5K pot. Comes out to a bit over 28 V.

--- sam | Sci.Electronics.Repair FAQ Mirror:

formatting link
Repair | Main Table of Contents:
formatting link

+Lasers | Sam's Laser FAQ:
formatting link
| Mirror Sites:
formatting link

Note: These links are hopefully temporary until we can sort out the excessive traffic on Repairfaq.org.

Important: Anything sent to the email address in the message header above is ignored unless my full name is included in the subject line. Or, you can contact me via the Feedback Form in the FAQs.

Reply to
Sam Goldwasser

Thank you both very much for your help! I am very new to this and was just going back and reviewing the voltage gain equations for a non inverting voltage amplifier and transistor amplifier circuits that still have my head spinning. Thank you again as this certainly has helped.

Tai West

Sparta wrote:



Reply to

ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.