# Law revision: I = (V/R)*~3.4 ..sometimes

Discussion in 'The Projects Forum' started by masked, Jul 31, 2010.

Jul 1, 2010
48
1
Evening all,

I'm playing with the circuit below where I want to limit the current across the "Out" chamber leads to 2.63mA.

Fortunately, I have three 9v batteries connected to Q7$_e$, so my voltage there is 27 minus .7 across Q7, for a total of 26.3V.
Since I want the Q7$_b$ current to be 26.3uA (a million times smaller), I use Ohm's Law to determine that I need a 1M resistor at R7.
This works on paper, and in 2 simulators.

However, when I build the actual device, I am getting about 9mA across the chamber leads. Since my voltmeter only reads amps to 3 decimal places, I can't examine the low values at Q7$_e_b$.

SInce the value is 3.4 times higher, it's easy to replace the 1M resistors with 3.3M resisitors. When I do that, I get the correct current in both polarities.. but that also destroys my voltage. So I'm trying to understand what could be going on in order to fix it properly.

Any ideas?

2. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,991
1,115

You're lucky that you put such a "big" resistors values.
And of course this circuit isn't property design.
The main current limiter is R9 and current gain of a Q3, Q5, Q7.
You could even remove R7, R8.
So you need redesign your circuit

And mayby you can tell what you wont to achieve, then we can help you.

Jul 1, 2010
48
1
Yeah, the design probably needs work; It's my first project, and that's why I'm on the forums And, you're right, I should list the desired results in each new thread.
The output I want is 27v limited to 2.63mA across the two leads to my chamber with polarity reversing every 60 seconds.

I'm confused about your statements though, because that's not how this circuit has been operating in the real world or in the simulators.
*If* I remove R7, R8 as you suggest, *then* R9 would be the main limiter (along with R5,R6) and Q3,Q4,Q5,Q6 all become forward-active rather than saturated.
But, as is, the circuit works when R9 is completely removed with no change in output.
I just recently added R9 because a tiny bit of voltage was leaking across the LED by Q1. Q1 was in forward output mode because the +9v supply was a tiny bit higher than the 8.3v output from my 555. R9 fixed the "leak" by lowering Q1$\small{_e}$ so that Q1 remains off when the 555 output is high.

From my design, and tested results, changing R7 and R8 are the only factors limiting current.
Regarding Q3,Q5,Q7, here's my plan, can you tell me if it's correct:
I can limit the amps across Q7$\small{_{ec}}$ to 2.63mA by limiting Q7$\small{_b}$ to 2.63uA. (because of the Q7 beta of 100.)
The same beta is used for Q5 and Q3, so that Q5 needs an available base current of 2.63uA to switch the 2.63mA, and Q3 needs 2.63nA to switch 2.63uA.
So as long as R5 is supplying (2.36mA + 2.63nA) then there will be enough current to drive Q3 and Q5.

Again, this works experimentally, on paper, and in simulators. What's confusing me is why the real output is ~9mA with the circuit as drawn, and ~2.36mA when I replace R7 and R8 with 3.3M.

What should I be testing next to figure it out?

Thanks,

Last edited: Aug 1, 2010
4. ### Markd77 Senior Member

Sep 7, 2009
2,803
594
You can't really trust the HFE (beta) of a transistor to be 100. It varies with current and possibly temperature. There are also large differences between samples of the same type of transistor.

Jul 1, 2010
48
1
THANK YOU!

That may be exactly what I was looking for. I was at a loss for what to do, but now I can just take the transistors out and test them individually with currents high enough for my meter to register.

Although, this brings me back to an earlier post's (unanswered) question:
"Is regulating current via the transistor base the right thing to do?"

What's the right way to limit my current? Just buy a CLD with the proper rating?