Help modifying a circuit?

[Highpower]

Hello group.

I have a problem here I'm hoping someone can help me out with. Electronics isn't one of my strong points, and I'm trying to put together a home version of an item that was discontinued by the manufacturer.

I found this schematic on the net and put one together as drawn. It will output 300mV as it is supposed to, but I'm getting a current of 48mA. I need to maintain the voltage at 300mV, but I would like to increase the max current to 200mA.

I have tried replacing R6 and R3 with 6.8 Ohm resistors, and now get 100mV across R5 with the output leads shorted. Is there anything (simple) I can do to boost the current to 200mA?

Thank you for looking.

 

[(*steve*)]

By my calculations, you don't have R6 in your circuit.

If that is right, reducing R3 to 10 ohms should do the trick.

Beware of Q3 getting warmer than previously (I have not checked its characteristics). It should be in good thermal contact with Q2 and Q3 for the best temperature stability. (All connected to the same heatsink would be good)

You may have to reduce the value of R1 and R2 if the circuit cannot maintain 300mV under load.

What's this for anyway?

Now I've also checked the characteristics. Seems OK... but those reduced resistor values for R1 and R2 look like a good idea.

 

[thebluepuppy]

just lower the resistance. thats what i would do.i would test the math first dont want to get those capacitors to juicy they might pop

 

[Highpower]

Thank you for the reply Steve. I do have R6 in the circuit. As I stated in my original post, I have replaced both R6 and R3 with 6.8 Ohm resistors. That raised my voltage across R5 from 48mV to 100mV.

I don't have the means to make a PCB, so this is just a simple perfboard construction. No heat sinks. I have attached another view of the circuit.

RE: The value of R1 and R2. Do I need to keep the value of these 2 resistors equal? I'm afraid I don't know how to figure the values to limit the output of the circuit.
I am looking for 300mV and 200mA (max) at the output leads.

This is basically a low voltage power supply for an electrolytic cell used to remove copper and lead deposits from inside gun barrels. The Outers company discontinued making their "Foul-Out" system, so I have to try to make my own version.

One other note: I did add an extra 100 Ohm resistor and LED between the +3V and -3V input after the switch, to show when it was switched on - if that makes any difference.

 

[(*steve*)]

OK, you need to make R6 and R3 together add up to about 10 ohms. To detect > 200mA via the LED, R6 should be about 3 ohms (3.3 may be the nearest value.) that means R3 should be about 6.8 ohms.

The next problem is that R5 will drop 200mV at 200mA, and that's way too much considering the output voltage is supposed to be 0.3 of a volt. It needs to be reduced to 0.1 ohms. This will mean 1mV across it corresponds to 10mA (so it will read 20mV at 200mA).

R1 and R2 don't need to be the same resistance. The fact that 6.8 ohm resistors for R3 and R6 only gave you 100mA output leads me to believe that you need more base current, and that means a lower R1 and R2. perhaps you could reduce them by a factor of 4 or so. A higher current through R1 and R2 will help the circuit to achieve a more stable output voltage.

The LED and resistor you have added won't make a great deal of difference.

 

[Highpower]

The next problem is that R5 will drop 200mV at 200mA, and that's way too much considering the output voltage is supposed to be 0.3 of a volt. It needs to be reduced to 0.1 ohms. This will mean 1mV across it corresponds to 10mA (so it will read 20mV at 200mA).

Thanks again. I will change R6 to 3.3 Ohms.

If I change R5 to 0.1 Ohms, how many watts will it need to handle in order to protect the circuit in the event the output leads get shorted?
This does happen occasionally, and therefore the need for the short indicator.

Once I know I have the value of R5 correct, then I will work on reducing the resistance on R1 and R2.

 

[(*steve*)]

If I change R5 to 0.1 Ohms, how many watts will it need to handle in order to protect the circuit in the event the output leads get shorted?

Well, firstly, it doesn't protect the circuit at all. It just provides a means of measuring the current.

R3 (and R6) limit the current by dropping almost all of the battery voltage at the required current.

The power is determined by either I^2R or V^2/R or VI. Whilst they all give *exactly* the same answers, you can use these to determine the power dissipated in different cases...

Assuming that the circuit can maintain 0.3V across R4, the max power that the circuit could ever provide through R5 is 0.3^2/0.1 = 0.9W, but practically, we know that the maximum current could possibly be about 300mA (if R3 + R6 add up to around 10 ohms) so the power is I^2R, or 0.3^2*0.1, or less than a hundredth of a watt.

R2 and R6, under short circuit conditions could dissipate (together) up to 0.9W, so making each of them a 1W resistor would be wise.

It's a little trickier to determine the maximum dissipation of the transistor, but it is likely to be around 0.25W

This does happen occasionally, and therefore the need for the short indicator.

The circuit is likely to deliver more than 200mA when shorted (that is, if you want to retain something like 0.3V under loads approaching it)

There are other ways to build a current limit, one of which requires little more than another transistor and a resistor of about 2.7 ohms. However it may require changes to your circuit that are marginally more difficult to implement than changes in component values.

Once I know I have the value of R5 correct, then I will work on reducing the resistance on R1 and R2.

Before you do this, measure the voltage across R4 and the current through R5 for various loads (ranging from zero to a short circuit. What would be nice would be 0mA, 10mA, 20 ma, 40ma, 80mA, and whatever you get at short circuit.

Then you have something to compare the results to after you change the circuit and test it at (say) 0, 10, 20, 40, 80, 160mA, and short circuit.

 

[Highpower]

Outstanding! I'm starting to see light at the end of the tunnel now.

There are other ways to build a current limit, one of which requires little more than another transistor and a resistor of about 2.7 ohms. However it may require changes to your circuit that are marginally more difficult to implement than changes in component values.

What type of transistor would that be, and where would it go in the circuit? Since I have to order more new resistors anyway, I might as well pick up the transistor as well to save on shipping costs.

 

[(*steve*)]

You could do something like this:

Rs and the additional transistor provide a current limit.

Rs is chosen so that 0.6/Rs = i.

If you want the maximum current to be 200mA, choose a value of Rs that is 0.6/0.2 ohms (3 ohms). 2.7 would be the closes value you can easily get.

R6 has to be a higher value than Rs, otherwise the "short circuit" LED will never light up (note that I have simplified the circuit diagram to exclude that part of the circuit, but that's just because I'm lazy)..

 

[Highpower]

Rs is chosen so that 0.6/Rs = i.

If you want the maximum current to be 200mA, choose a value of Rs that is 0.6/0.2 ohms (3 ohms). 2.7 would be the closes value you can easily get.

Just to make sure I am understanding this correctly, if Rs = 2.4 ohms - then the maximum current would then be 250mA? Is that right?

On a side note - I was thinking about purchasing a resistor substitute box for future testing purposes. Are the inexpensive slide switch boxes worth having over time for a hobbyist? I'm sure a thumb wheel decade box would be the way to go if I was going to use it often. But on the other hand, I don't like having economy grade tools fail on me if it sits too long without being used. Your thoughts?

I really wish I would have bought a solder-less bread board to mock this up on before hand. I made the mistake of believing the circuit I found would duplicate the output of the factory unit, but it doesn't even come close on the current as drawn. Lesson learned. I think I'll add that bread board to my order now for future use.

And Steve,
I would hardly call answering all my newbie questions being "lazy". Far from it.
Thank you again.

 

[(*steve*)]

Just to make sure I am understanding this correctly, if Rs = 2.4 ohms - then the maximum current would then be 250mA? Is that right?

Yep. It's not exact, but it would be close. It does depend on the transistor characteristics, but close enough is probably good enough for this circuit

On a side note - I was thinking about purchasing a resistor substitute box for future testing purposes. Are the inexpensive slide switch boxes worth having over time for a hobbyist? I'm sure a thumb wheel decade box would be the way to go if I was going to use it often. But on the other hand, I don't like having economy grade tools fail on me if it sits too long without being used. Your thoughts?

I've never had one. I just have a box of resistors and try them out

The problem wit these substitution boxes is their potentially high inductance due to long leads, and low power ratings for the low resistance resistors. If you get one, make sure you can replace resistors if you damage them, and preferably are able to replace low value resistors with higher power versions.

I really wish I would have bought a solder-less bread board to mock this up on before hand. I made the mistake of believing the circuit I found would duplicate the output of the factory unit, but it doesn't even come close on the current as drawn. Lesson learned. I think I'll add that bread board to my order now for future use.

DEFINITELY get yourself a solderless breadboard. It's an invaluable tool for trying stuff out. It can also make it relatively easy to move the circuit over to veroboard or matrix board.

And Steve,
I would hardly call answering all my newbie questions being "lazy". Far from it.
Thank you again.

I just hope you can read my scribbled drawings. (Other people here do far better than I in that regard)

 

[Highpower]

Well it looks like I'm getting nowhere fast.

I've got the breadboard set up with the following changes:
R1 -- 750 ohm, 1.0 watt
R2 -- 500 ohm. 0.75 watt Trimpot
R6 -- 3.0 ohm, 1.0 watt
R3 -- 6.8 ohm, 0.5 watt
R5 -- 0.1 ohm, 1.0 watt

I've adjusted R2 to give 0.3 V at the output leads.
I've got 300 mV across R4 with no load.

(Current through R5) (Voltage across R4)
20.45 mA 238.9 mV
29.30 mA 226.1 mV
44.30 mA 210.1 mV
68.40 mA 187.3 mV
74.00 mA 181.6 mV

I'm getting a maximum of 170 mA measured between the output leads (shorted?) without R5 in the circuit.

The other thing is, the short LED stopped working. The max voltage I am getting out of the collector on Q4 is 1.39 V.
Edit to add: Just discovered that R6 has to be a minimum of 5 ohms to get the short LED to work. Going backwards now...

I have not added the current regulator (Q5) to the board.