# Strange Diode bridge rectifier identification and function

#### grim22x7

Joined Apr 1, 2016
5
Hello,

I have a strange circuit that I am trying to trouble shoot. It's a pretty old design, although the parts themselves are relatively new. We're a contract manufacturer so we built their design. The customer returned this part to us because it wouldn't work. This section of the circuit seems to be functioning differently than what we've seen on other boards. We don't currently have another board to compare against. I don't want to start replacing parts until I can hopefully get a good idea of it's function. It is 6 diodes, 4 arranged in the standard bridge rectifier configuration with the top connected to a voltage and the bottom connected to ground. The last two diodes are connected across the middle where you might normally see a load. You can see it in left side of the image attached (hopefully, sorry it's fuzzy, it looked fine when I snapped it)

If anyone knows the name of this type of circuit so I can look it up or can explain how this circuit works and why it might be used, that would be greatly appreciated. Frankly this board was probably originally designed while I was still in elementary school. I'm lost, to me all paths point to ground. And yes the voltage on top of R7A is 240VDC.

#### #12

Joined Nov 30, 2010
18,216
Positive and negative voltages go through 4 diodes to ground.
At about 1 ma, that is a voltage regulator or a peak clipper with a forward voltage drop of 4 x (0.715 volts).
There might be something to do with speed, but no frequency is stated.
The op-amp U5B has a design limit around 5KHz.

Edited to fix my misteak.

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#### RichardO

Joined May 4, 2013
2,271
I have not done an extensive analysis but I think it is a voltage limiter/clipper circuit. It think the positive and negative voltages will clip at different values.

I am sure someone will do a simulation to prove me wrong here.

#### wayneh

Joined Sep 9, 2010
16,500
Positive voltages go through 4 diodes to ground and negative goes nowhere.
The voltage on C13 is limited to no more negative than 4 diode drops below ground.

#### #12

Joined Nov 30, 2010
18,216
The voltage on C13 is limited to no more negative than 4 diode drops below ground.
You caught me. It's bipolar. Both positive and negative voltages are clamped at 4 diode drops.

#### wayneh

Joined Sep 9, 2010
16,500
I figured the odds of there being 4 useless diodes was pretty low. Mistakes happen, but that would be pretty bad.

#### RichardO

Joined May 4, 2013
2,271
Both positive and negative voltages are clamped at 4 diode drops.
Yeah. That does look right to me now. At first glance it did not.

#### #12

Joined Nov 30, 2010
18,216
Yeah. That does look right to me now. At first glance it did not.
This is a rare circuit, and the way it's drawn makes your eyes loop.

Joined Nov 5, 2010
215
That's cute. Normally a bi-directional 4-diode-drop clipper would use 8 diodes, but this configuration uses only 6.

#12

#### wayneh

Joined Sep 9, 2010
16,500
That's cute. Normally a bi-directional 4-diode-drop clipper would use 8 diodes, but this configuration uses only 6.
So I guess the "normal" one is wrong? (By wrong, I mean inefficient use of components.)

Joined Nov 5, 2010
215
So I guess the "normal" one is wrong? (By wrong, I mean inefficient use of components.)
Yes, it's less efficient. I know two diodes are cheap, hardly noticed in a big BOM. But if it's in a consumer product that gets produced by the millions, every little cost savings counts. I'm always on the lookout for ways to trim unnecessary components and costs from circuits.

#### grim22x7

Joined Apr 1, 2016
5
I sort of expected that it was something like this but I was having some trouble wrapping my head around the way it functioned.

Digikey says that the bav99 diode has a max 1.25V forward voltage drop. So I should expect a voltage of 4x1.25v or 5V across C13. I'm definitely not seeing that. So I'll probably have to take some of those off and see if they are functional out of the system. Thankfully, I'm not working on Saturday so I'll have to take another look at it Monday. I appreciate your help and hope to be able to volunteer something in the future. I'll also keep you updated as I continue to work on this board.

#### wayneh

Joined Sep 9, 2010
16,500
So I should expect a voltage of 4x1.25v or 5V across C13. I'm definitely not seeing that.
No, no, no. You would expect zero. It is clipped to NO MORE THAN ±4 diode drops. Where it lies in between those extremes depends on the signal. A sine wave would center at zero.

#12

#### #12

Joined Nov 30, 2010
18,216
Digikey says that the bav99 diode has a max 1.25V forward voltage drop.
No, no, no. I already did that for you.
At about 1 ma, that is a voltage regulator or a peak clipper with a forward voltage drop of 4 x (0.715 volts).
This circuit does NOT guarantee that the resulting voltage will be both positive and negative 2.86 volts, one at a time, or both at once. It guarantees that the peak voltage remaining will NOT be more than 2.86 volts in either polarity, entirely depending on what the input signal is doing.

You gave us nothing about what the input signal is, so we can not guess how often or how fast. We can only say how much.

#### ian field

Joined Oct 27, 2012
6,539
I have not done an extensive analysis but I think it is a voltage limiter/clipper circuit. It think the positive and negative voltages will clip at different values.

I am sure someone will do a simulation to prove me wrong here.
Its an AC clipper at 4x Vf - either applied polarity goes through 2 of the bridge diodes, as well as the 2 in series within the bridge.

#### grim22x7

Joined Apr 1, 2016
5
My apologies. I misunderstood, when you said clipper I was thinking more along the lines of a shunt regulator. This clipper as I understand will only "shave off" any voltage higher than +/- the 4xVf drop is that correct?

The input is a half rectified sine wave from a transformer which is filtered to yield a measured 240VDC with about 1.4V ripple. This part of the circuit seems to be working correctly. This high voltage is used to trigger an igniter of some kind. (we only have the board and a simple test fixture and not the full assembly) The output of the op-amp is fed back into a controller which is supposed to light some LEDs when prompted but wont. We've tried replacing the controller (socketed IC easy out and easy in) which led us to the input circuit being at fault. Power to the op-amp is fine +/-12V

I'll have to come back to this later when I can get the board and docs in front of me. As always I appreciate the help.

#### #12

Joined Nov 30, 2010
18,216
when you said clipper I was thinking more along the lines of a shunt regulator.
It depends on how you think. To me, they are both the same. Either a clipper or a shunt regulator will dump any voltage in excess of its rating but it's usually called a clipper in AC circuits and a shunt regulator in DC circuits. Samey, samey. Apply AC to a pair of shunt regulators, one attached backwards to the other, and you will get the same results as this bi-polar clipper. It's just trying to keep the 240 volts from blowing the brains out of the amplifier chip.

#### ian field

Joined Oct 27, 2012
6,539
My apologies. I misunderstood, when you said clipper I was thinking more along the lines of a shunt regulator. This clipper as I understand will only "shave off" any voltage higher than +/- the 4xVf drop is that correct?

.
According to Fairchild; the BAV99 can handle 200mA.
The popular TL431 which has among others; the name "shunt regulator" maxes out at 100mA.

Its sort of inbetween clipper and shunt regulator, but the context here suggests clipper to me.

#### grim22x7

Joined Apr 1, 2016
5
I guess the difference is the expected input of each. In a clipper you want the input to stay below the voltage clip where as with the shunt regulator the input is usually higher.

I connected up the O-scope this morning and started poking around. My initial thought that this circuit was nonfunctional seems to have been more an issue with using the right tool for the job. A DMM wasn't showing me everything. And my not have been sensitive enough.

At C13 I have about 60mVrms, it's a relatively clean sine wave,negligible offset. C14 similar but 37Vrms. the output of the first opamp is 70mVrms. Which is about as expected. That output passes through a offset circuit which adds about 1.6Vdc to the signal. And that is conveyed to the controller chip exactly.

When I push the button to trigger the LEDs (that button changes the input) which pushes the input to the circuit (at C13) to 1Vrms and that yields an output at the controller of a 1.7Vrms full sine wave riding on 1.6Vdc. That all seems fine. Which is unfortunate since there is very little else to verify. I'm thinking that there might be an issue with the controller.