# Why would this (simple) over voltage protection circuit raise the output voltage?

as a beginner I have recreated this circuit: I found it in this guide, and I was hoping it would work as intended but instead it gave me some weird results. I connected power to the circuit and probed the input voltage (2.5V), and then subsequently I probed for the output voltage. It gave back readings of many different voltages, but all higher than the input (3.6V-4.0V, rarely but sometimes higher). I have no idea why this would happen (no inductors or anything like that).

If you look in the guide, it lists the specific components; I had to subsitute a few:

1. I used the S8550 instead of the FMMT718
2. I used a 2.7V zener diode similar to this, instead of the 1N4740A

I think perhaps I should have chosen a transistor that had a lower saturation voltage (like the guide says), but besides that I don't see these components doing any harm.

So if anyone can help me troubleshoot this it would be much appreciated.

P.S.

Bonus Question:

I understand the principle behind the circuit and nearly everything seemed to make sense to me. The only thing I didn't get is why R4 is there. Is it even needed? I think maybe the author of the guide accidentally kept it in from the first circuit he drew, or maybe it has a purpose and function I cannot understand. Thanks all in advanced!

• Also, I am unsure if this circuit would be okay to use for practical purpose? I hope to use it to charge my ultra capacitor. It seems almost too simple, and perhaps it is more for teaching the basic concept rather than for actual use? Or maybe that's just the beauty of it.. but in any case, if someone can give a overall recommend / thumbs up to the circuit it would be much appreciated. I'd rather use something else than blow my capacitor on first charge (I'm glad I checked the output voltage before using it haha) – Bureto Apr 13 at 5:06
• Have you thought about V*I power and temperature rise? – Tony Stewart Sunnyskyguy EE75 Apr 13 at 5:13
• Why don't you simulate the circuit, it will be easy to see what's wrong? – anrieff Apr 13 at 5:26
• Btw, I have a 2kF 2.7V supercap which I charge with an ordinary 3VDC supply + series schottky diode, it's simple and it works. – anrieff Apr 13 at 5:32
• @Bureto For most of the charge cycle, you probably should want a constant current control circuit that operates well at an output voltage near ground (which the capacitor will be near, at first.) The voltage will rise linearly. Then, near the end of the charging cycle you want to switch over to voltage control to avoid over-charging. So I don't really like the circuit you've picked for this. There's no current limit other than the base current and the PNP beta. – jonk Apr 13 at 5:55

R4 is needed to form the reference voltage with D2 by providing the bias current. The intersection of the V_in, R4 with the I-V curve of the Zener diode determines the bias point of that zener. If R4 was not present, D2 is only connected to the base of the BJT through R5, which would require negative base current.

Regarding your observation of a higher voltage at the output than at the input. From a theoretical standpoint, it should not be higher. Troubleshooting is all about breaking the problem down into smaller blocks and checking your measurements against your expectations.

Here's what I would do to troubleshoot the system.

1. Disconnect everything from the input, check V_in, does it match your power supply setting?

2. Double check the orientation, connections, and component values of all components just as a sanity check.

3. Measure the net between R4 and D2 while disconnected from the R5, Q2, Q1. What this is doing is checking where the reference level is at. This should be near 2.5 V assuming your input is 2.5.

4. Measure the base of Q1 and R6. This should be at 0 V as Q2 is off and R6 is pulling to ground.

If any of these voltages are off, you can look deeper into that area and try to determine what is going wrong.

Regarding charging supercapacitors. You can charge supercapacitors in many ways, constant current, constant power, etc. The design of the charging circuit should protect against overcurrent and reverse voltage.

A simple Shottkey diode and current limiting resistor is will work. This isn't the fastest way, but it gets the job done. Reference [1] shows an example of this circuit in Figure 2a. Size the current limiting resistor such that V_in / R is less than the maximum current rating of your supercapacitor.

If you want the overvoltage protection, you can put the Shottkey diode in front of your overvoltage circuit and the resistor on the output. The Shottkey will protect the overvoltage circuit as well. Before you do this, solve whatever problems you're having with the overvoltage circuit first. Don't add more complexity to troubleshoot.

• Thanks Mike. Yeah, I guess I should have said I also had a resistor at the output to limit current. From what I read that reduces max efficiency to 50% but that's no problem for me haha. I'll make sure to add a Shottkey diode, I forgot about reverse voltage. Anyways, I guess I better go troubleshoot the system! Thanks. – Bureto Apr 13 at 15:29
• Originally, I doubted this answer regarding the need for R4. At first glance it looks like zero current through Q2's base until the input voltage is high enough to overcome VBE+Vz. I simulated it. We have to remember that the zener is not a perfect device. It begins to conduct current long before it's published Zener voltage. The specified zener voltage is for a particular current. If R4 is omitted, Q2 begins to turn on at around 2V_in as the Zener starts conducting. This produces a soft knee in the circuit's output voltage starting at a voltage much less than the desired value. – Randy Nuss Apr 13 at 15:34

Although you've picked an answer, your problem does not seem to make sense.

1) You claim to have an input of 2.5 volts.

2) You claim to have an output of 3.4 to 4 volts.

The only thing I can think is that your circuit is oscillating at high frequency, and your meter is not handllng it well.