# Power Supply Modding.

Hi!

I recently got access to an old linear 22 volt 4 amperes current limited power supply from an old telephone station. Due to it’s relative simplicity, I decided to reverse engineer it and try and make it a variable lab bench power supply. I attach the original schematic and the modification I tried to this post. However, I have two problems: first of all, for some reason I can not lower the voltage below 7,8 V. The second is that the voltage changes rapidly if I rotate the initial path of the pot (from 8V to 23V aprox.) and then spents about 3/4 of the entire turn from 23V to 30V. I could really use some help, since I have tried changing the feedback resistor and potentiometer value and I always get the 7,8 V threshold.

PS: the circuit with the 7805 is the old one, the one with the 317 the modified one.

Thanks!

You are describing the actions of using a logarithmic pot instead of a linear pot. You probably got the pot from something with an audio volume control.

Measure the resistance from the center pin to one of the end pins. You will see the log resistance change.

Get a linear resistance pot.

Paul

I though that at first, but I changed the pot to another one I was completely sure was lineal and still had the same issue. However, even assuming that's true it does not explain the 8V threshold? Or it might be also because of the pot?

If you look at how to set the output voltage on an LM317, you will see that there needs to be a voltage divider, with the Adjust pin tied to the divider point. So, for 23V out, there needs to be enough resistance between the Out pin and the Adjust pin to draw the Maximum minimum load current. That is calculated like this:

1.2V/5mA = 240Ω – 1.2V is the worst case Reference Voltage, and 5mA is the worst case Minimum Load Current.

And the lower, voltage setting resistor is calculated as:

(23V-1.25V)/5mA = 4.4k – 1.25V is the Nominal Reference Voltage

So, the voltage divider is composed of a 240Ω resistor in series with a 4k4 resistor. And the Adjust pin would connect where those two resistors meet.

Notice, in your diagram, there is nothing resembling that 4.4k resistor. I.e., there is no voltage divider. The Adjust pin goes right to ground. AND, that variable resistor leg needs to be where the 4k4 resistor would go.

To do this correctly, the upper resistor needs to be a fixed resistor that will be something like:

(1.2V - ~0.7V)/5mA = 100Ω

And the lower resistor(s) will be a total of 4k4 – with some sort of adjustable component.

Also, you realize, of course, this power supply can never be adjusted to less than around ~1.25V - ~0.7V, depending on what the actual Reference Voltage is, and on what the actual VBE voltage is [and this will vary with temperature – thus slightly varying the output voltage].

True! I changed the value of the fixed resistor with a 120 ohm one at the pot with a 4k7 one (since I did not have the exact components) and it worked! Thank you so much!

Atomillo: True! I changed the value of the fixed resistor with a 120 ohm one at the pot with a 4k7 one (since I did not have the exact components) and it worked! Thank you so much!

you might want to look here.

The old analog stuff is heavy and clumsy. Going digital and recycling old laptop power supplies gives you much better results.

With any linear regulator, beware of power dissipation. That's the voltage dropped across the regulator x the current through the regulator. At lower voltages (more voltage dropped across the regulator) your transistor will get hotter (at the same current).

I took a quick look at the datasheet and the 2N3055 is rated at 15A so that's good. But, I didn't make any power calculations, and it can be difficult to calculate the power capability because it depends on your heatsink and ambient temperature.

You may be OK. But if you start with 22V and regulate-down to around 5V, and draw 1A or more, it set's off alarm bells in my head.

I dont worry about this. The original PSU worked at 22V 4 amps from 36 volts unregulated. I dont plan to draw more than 1.3 A of current.

DVDdoug: I took a quick look at the datasheet and the 2N3055 is rated at 15A so that's good.

The max current isn't the only rating to worry about - secondary breakdown is perhaps more likely to happen in high power circuits, so I'd first check the secondary breakdown (aka SOAR) graph. Commercial bench power supplies switch transformer windings to help with efficiency and also to avoid secondary breakdown becoming an issue (by keeping the max Vce low).

There are much better transistors available today than the 2N3055 BTW...

Does anyone even read datasheets anymore?

MarkT: The max current isn't the only rating to worry about - secondary breakdown is perhaps more likely to happen in high power circuits, so I'd first check the secondary breakdown (aka SOAR) graph. Commercial bench power supplies switch transformer windings to help with efficiency and also to avoid secondary breakdown becoming an issue (by keeping the max Vce low).

There are much better transistors available today than the 2N3055 BTW...

The worst case Power Dissipation will happen at the lowest voltage which will be approximately 0.55V:

[36V-0.68Ω1.3A-0.55V]*1.3A = **44.9W*

Before, the 2N3055 had to dissipate:

[36V-0.68Ω4A-22V]*4 = **45.1W*

So, in the spirit of honoring the original request of the OP, I used a spreadsheet to calculate the Max Current for different Output Voltages, taking into effect the Current Control Resistor [0.68Ω]:

| Output Voltage | Max Current | | - | - | | 0.55 | 1.3 | | 1 | 1.3 | | 2 | 1.4 | | 3 | 1.4 | | 4 | 1.5 | | 5 | 1.5 | | 6 | 1.6 | | 7 | 1.6 | | 8 | 1.7 | | 9 | 1.7 | | 10 | 1.8 | | 11 | 1.9 | | 12 | 2.0 | | 13 | 2.1 | | 14 | 2.2 | | 15 | 2.3 | | 16 | 2.5 | | 17 | 2.6 | | 18 | 2.8 | | 19 | 3.0 | | 20 | 3.3 | | 21 | 3.6 | | 22 | 4.0 |

So, the OPs intention to only use this supply at currents up to 1.3A is Right On The Mark!

I used the following formula, and then solved for ``` I ``` using the Quadratic Equation:

``````(30-0.68I-V[sub]A[/sub])I = P  where P = 45.1W
``````
``````[b]I[/b] = [(30-V[sub]A[/sub]) - √ [(30-V[sub]A[/sub])[sup]2[/sup] - 4*0.68*45.1]] / [2*0.68]
``````

Atomillo:
I dont worry about this. The original PSU worked at 22V 4 amps from 36 volts unregulated. I dont plan to draw more than 1.3 A of current.

Do you still require an adjustable current limitation?
Then I wish you a ton of luck to get the “off-rail” current sensing working in a reasonable range of accuracy. The old lab power supplies transfos had a separae winding to power the electronics in such a way that the current shunt was inside the rails.

I have built such a power-supply 45 years ago. That was an endless fiddling!

I am glad to have digital technique available just for a few bucks today.

Here (or elsewhere) you get all you need incl. digital displays for just 20 bucks:
DSP3005