# Current Limiter, can someone troubleshoot this circuit?

so I am trying to limit the current output with x2 3904 transistors and a sensing resistor. This is currently what I have, but am not sure why it doesn't work. That current supply is going to be a photo-resistor (or something along those lines). The line going out is hooked up to a constant 10V DC. So right now with this simulation I am varying the current from 0-10mA, however I want it to limit at 3mA, how would I achieve this? Is this circuit even wired correctly? Any help is greatly appreciated. I also don't exactly know how to choose the value for R5 (in this circuit).

PS. I don't know if it's needed, but the output voltage should be around 6V. And ouput is the lower of the 2 wires going to the right of the screen.

• Sorry, I'm just trying not to over complicate this. There is a 10V DC supply going in, that's all. And in the output there will be a load. I can change the resistor value accordingly and what not, I just want to get the concept down of this current limiter. What formulas does it use? Why and how does it work? Are there other methods? – Maty Apr 10 '15 at 22:51
• Output is supposed to let 0 through 3mA through, if the photo resistor lets more than 3mA through, I want to short it, but still get a constant 3mA output. – Maty Apr 10 '15 at 22:55

There is another way to implement a current sink when you have a regulated voltage available. It often works well enough. You can consider using it since you have 10V available. The base current will be IC/beta. Choose R2 and R3 so that the current through them is 10x the base current or more.

It is a good idea to keep Vemitter at around 0.5V or more. But the collector of Q1 can't be lower than Vemitter, so keep that in mind. I have used this for driving blue or white LED's from 3.3V or from a Li battery.

I think this is more-or-less what you have in mind.

If you change RL from 0 ohms to about 2.5K, the current will stay pretty constant at about 3mA.

The resistor value R1 sets the current- it's the Vbe of Q1 divided by the desired current. Vbe is typically around 0.6 to 0.7V, so R1 should be 200-233$\Omega$, so I used a standard E24 value (5% resistor).

You should be able to run the simulation and try different values for RL. simulate this circuit – Schematic created using CircuitLab

• I'm not sure how this circuit would work without a current source. – Maty Apr 11 '15 at 0:28
• The current source is RL. If you have a real current source (reasonable compliance voltage) and it exceeds about 3mA the collector voltage of Q2 will rise from less than 1V to limit the current. – Spehro Pefhany Apr 11 '15 at 0:32
• Oh ok, I am running the simulation but I am getting very unique graphs... I'm confused as to what R2 does. If the voltage before R1 gets greater than 0.7, then that will open up Q1 and let the excess current through, correct? But isn't the path of least resistance still RL? – Maty Apr 11 '15 at 0:45
• R2 is bias to turn Q2 on and allow the current to flow. Above 3mA, Q1 steals base current from Q2 and the current is limited. – Spehro Pefhany Apr 11 '15 at 0:47
• Ah, I see. So as long as RL is under 10k resistance (because of R2), this should work? So my output line would come out of RL? That resistor's current can never go beyond 3mA or else the current will increase R1 voltage to >0.7V (because Ic = Ie) and that will turn on Q1 and like you mentioned, Q1 will steal the current from Q2 base because it's grounded. Now how does this still keep current flowing through RL? Wouldn't Q2 be off? – Maty Apr 11 '15 at 0:55

The circuit you have drawn will output a constant ~600μA. It's sometimes referred to as a "$V_{BE}$ over $R$" current source, because Q3 and Q4 act in a feedback loop to maintain the voltage across R5 to be a $V_{BE}$, which of course appears across R5, and the resulting current flows through Q3. Q4's $V_{BE}$ is fairly invariant to the value of I1 (being only logarithmically dependent on it), so the output current is pretty constant.

You might instead consider a circuit like this: Below 3mA, too little voltage is across R1 for Q1 to turn on, so Q1 remains off and the I1 current is simply replicated by the Q2/Q3 current mirror. Above 3mA, however, Q1 turns on and shunts the extra current away from R1 and instead straight to ground.

For even better clamping action, you can gain up Q1's current, like so: It should be noted that the clamping threshold is strongly temperature-dependent (because $V_{BE}$ is strongly temperature-dependent). If you want accuracy, you should use an op-amp circuit and a dedicated voltage reference chip. Otherwise, evaluate the clamping threshold over the useful temperature range to ensure it's acceptable.

• Where is the output in this circuit? – Maty Apr 11 '15 at 0:30
• The current drawn into Q3's collector from the 10V supply. Same as your circuit. – Zulu Apr 11 '15 at 0:31
• I think I understand the role of R1 and Q1, like you mentioned there is too little voltage (>0.7V), for Q1 to turn on, so that's not doing anything. However, how is Q2 on? In Q3, there is current supplying the base and the Vbe is >0.7, so that's on, right? But what's making Q2 Vbe >0.7? And why do you have Q2/Q3 mirrored, what does that do? – Maty Apr 11 '15 at 0:50
• I1 supplies current to Q2, turning it on. Q2 is a "diode-connected" transistor, which just means it automatically balances its $V_{BE}$ voltage to whatever is necessary for it to conduct I1. When you take this voltage and apply it to the base of Q3, then Q3 also has exactly the right $V_{BE}$ voltage to conduct I1. That's how a current mirror works, and that's what a current mirror does: you feed a current into one side (the diode-connected transistor), and it replicates the same current on the other side. The resistors R3 and R4 improve its accuracy. – Zulu Apr 11 '15 at 3:16