# Simple Shunt Voltage Limiter

Following up on my previous question, I tried this simple circuit to ensure that there is a minimum load on my power supply (to prevent the unloaded supply from rising above Vmax of a part). I actually built this and it works as I expect, but I wonder if there are any additions that should be made to improve its performance?

simulate this circuit – Schematic created using CircuitLab

My idea is, if the load (Rload) is disconnected or not drawing at least a minimum current, the supply rail will rise because of inadequate current through Rsource. When the voltage reaches 9.6 (Vzener + Vbe), Q1 will start to turn on, and will sink enough current (and turn on the "Overvoltage" LED) to keep the rail below about 9.6V.

If the load is drawing enough current, the zener/transistor circuit will be inactive.

Incidentally, I believe that just a zener alone would be sufficient, except that it would have to shunt all the excess current itself. By adding a higher-power transistor Q1, a lower-power zener can be used.

• To analyze this correctly you need to provide the V range of input and VA limit on source or ESR. Then the Vmax and I load min and Vmin at I load max. Can you define these? – Tony Stewart Sunnyskyguy EE75 Jan 1 '17 at 20:56
• I didn't want anybody to grind through the numbers for me; I was mostly hoping for "don't do that, it's unstable" or "this is OK, but you should have a pull-down resistor from base to ground." – Dave M. Jan 1 '17 at 21:00
• However, in my actual implementation, I have 24VAC (nominal), 60Hz going into a bridge rectifier & 22uF filter, and I'm feeding a load that usually draws between 0.5W and 2.0W, but can't survive more than 40VDC. (The load is actually a 33063-based buck converter to go from (no more than) 40V to 5V at 75-250mA. I'm assuming converter efficiency of ~80%. – Dave M. Jan 1 '17 at 21:06
• YOur design is sub-optimal for LED power, and Source loss. Better to use a heatink with a Darlington and a dummy R load with LEDs for the Zeners instead. due to current likely exceeding component rating in your design. Also note the centre tap ought to be connected as a half bridge. – Tony Stewart Sunnyskyguy EE75 Jan 2 '17 at 0:22

simulate this circuit – Schematic created using CircuitLab

A bridge is a voltage doubler unless you connect the centre tap also to 0V then it becomes a half bridge.

33R load is possible with 33R series then voltage is 50%. No load is always 50% more due to rms-pk (rt2=141%) xfmr loss 8~10% thus 141%+10% - diode drop~ 150% of RMS with no load.

Choose Vf of LED to tune active clamp.

• Wow, thanks Tony! I don't have access to the transformer center tap, but looking at it that way helps to understand how I got from "about 24V" to "holy shit, 80V?" I'll try to import some of your ideas, including learning more about Darlington devices. – Dave M. Jan 2 '17 at 2:41

As requested in your comment ...

Don't do that, unless your LED D2 can handle all the current that you need to sink to keep your voltage down, regardless of what current you throw at it.

If LED D2 is a standard 5mm LED, they are probably good to 20mA. Most small zeners would take that, and more, before letting out the magic smoke.

Perhaps put the LED in series with a lower voltage zener diode (LEDs have a pretty steep V/I function, often steeper than zeners!), and connect Q1 collector directly to rail. This will mean then that the LED has to only conduct 1/beta of the current being sunk. Bipolar transistors can be had that will sink amps and dissipate 10s of watts, with adequate heatsinking. To make the LED shine adequately, connect a resistor in parallel with C1, to sink a certain current at the 0.7v needed to turn on Q1, this will ensure the LED gets enough current to light when the circuit is limiting.

• Thanks! I had the LED in my test circuit to see whether it was shunting, but had thought about moving it elsewhere since my final implementation may need to shunt more current. Putting it in the zener path is a great idea, as is the pulldown to help it turn on. – Dave M. Jan 1 '17 at 21:35

In your circuit, the 2N3904 base-emitter junction is as risk. Its datasheet doesn't spec maximum allowed base current, but none of its graphs go above 20mA base current. With a 12v input, base current can rise very close to this limit.

I'd suggest using a beefier transistor - A 2N2222 or a 2N4401 might be a safer choice.
And if your 12V source could go a little higher at any point (possibly even a transient event) an even beefier transistor might be in order. The capacitor mitigates transients, but who knows how long a transient event lasts?

• I have tentatively selected an NXP BC56PAS (mouser.com/ds/2/302/BC54_55_56PAS_SER-840047.pdf) for that role. It allows Ib to 300mA, and Ic to 2A. However, even with the 3904, wouldn't the transistor's hFE protect it from excessive Ib? As Ib increases, Ic increases by a multiple. So most of the shunt current gets pulled that way, rather than through the zener and Ib. (Assuming I move the LED somewhere else so it doesn't get fried!) – Dave M. Jan 1 '17 at 22:33
• Yes, you're quite right that Ic should start flowing long before Ib reaches very high. But in testing similar current-limiting circuits like this, I've found many transistor failures - open base-emitter. You'll often see a few hundred ohms in series with the base in current limiters. Perhaps the base-emitter blows open before collector current turns on - I've never managed to capture the event to see how it happens. – glen_geek Jan 1 '17 at 23:58
• Thanks, that's the kind of info I was hoping to get...I don't have the experience to know about common failure modes, and I'm too old to wait around making mistakes to get the experience! – Dave M. Jan 2 '17 at 2:45