# Bridge rectifier not giving an expected output.

Discussion in 'General Electronics Discussion' started by BBROYGBVGW, Mar 8, 2013.

1. ### BBROYGBVGW

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Mar 8, 2013
I am building a project where I need to put, into a data logger, a voltage from a, variac controlled, mains connected transformer’s, secondary winding tapping, which will be proportional to the secondary output. It has to be reasonable accurate but the logger will only input with d.c current or voltage, however there are facilities in the software to apply formulae to the input, i.e. divide by the square root of 2 for example.
My idea is to use a bridge rectifier, which should result in a d.c. voltage to the value of the peak to peak value of the a.c. voltage. Then apply the above equation to the input to achieve the RMS equivalent.
The a.c voltage is low from 0.5V to 8V. I’ve built a small bridge rectifier and, as there will be no load, I’ve connected a 10k resistor in parallel with a 0.1 µF capacitor, across the d.c. output.
The d.c voltages I am reading from this setup are not what I would have expected. There is no d.c output below 0.6 a.c. input but this must be due to the threshold switching of the diodes (any suggestions for a type with a lower threshold would be welcome). The higher voltages are linear but reduced by a factor of between 2 and 1.5, i.e. 2.5V a.c. gives an output of 0.8V d.c. Can anyone suggest why this isn’t 3.5 d.c as I would have expected? I don’t have a scope unfortunately, as this would probably identify what is going on.

2. ### GonzoEngineer

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Dec 2, 2011
Your diodes have a voltage drop, usually around .7 volts.

I would use an "RMS to DC Converter."

Google AD536. It is a simple IC to implemet what you want.

3. ### BBROYGBVGW

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Mar 8, 2013
Ok thanks,
So the diode voltage drop would have a significant impact at my low voltages but I have wound the voltage above the range I want to use and the reduction factor is only reduced to the order of 1.3; for example with an a.c. input of 13V the output is giving 10.29V d.c. It dosen't seem to make sence.

I see that for my application it would be more suitable to use the RMS to d.c converter and I’ve had a look at the data sheet for this device. I suppose I can use the standard recommended connection arrangement but I don’t seem to find a recommended voltage ramge for the power supply. I have a 24V d.c. readily available would that be sufficient? (Edited input: I've just seen that the input voltage range IS given at 5 to 36V so my 24V will be fine.)
Also, how do I calculate the value of the averaging capacitor, for the application I want to use it for?

Last edited: Mar 8, 2013
4. ### GonzoEngineer

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Dec 2, 2011
If you have a diode bridge, you are going to have voltage drop across 2 diodes, hence the
1.3V and not just 7V. (The .7V is not an exact number, and depends of the specifiec diode you are using.)

As far as the supply voltage for the IC, I would recommend 12V. The current draw is not much, so you could use a standard 7812 Voltage regulator in a TO-220 package and not need a heat sink.

As far as the component values, I cheated and used a digital oscilloscope to obtain the true RMS value, versus what I was putting in. Now, my application uses varing frequencies and duty cycles, so it was critical. But your situation, I assume, will be a fixed 50 or 60 Hz, so you should be able to calculate the proper number you should be getting.

I hope this answes your questions. Feel free to ask more, and I will do a little more research this weekend when I am not working.

Cheers!

5. ### GonzoEngineer

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Dec 2, 2011
Also, I forgot you to mention that if you are measuring a bipolar signal, you will want to have both + and - 12 volt supplies.
You can purchase cheap DC to DC converters to get these voltages with a 24VDC input.

I forgot about this because I only use it to measure unipolar signals (20-60kHz.)

If you live in the US, I have bothe the AD536 and a suitable DC to DC converter you can have if you just reimburse me for shipping. (I am always out of beer!)

Wait a minute, I just read your post where you said you put 13V in......what is the range of the input voltage?????

Last edited: Mar 8, 2013
6. ### woodchips

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Feb 8, 2013
An RMS converter seems something of an overkill.

Are you sure you are using a 0.1uF smoothing cap? With the 10k load resistor this gives a time constant of 1ms, not really much use with a 50Hz waveform. Use something like 4.7uF giving 50ms, a compromise between voltage droop and response time.

Use Schottky diodes to reduce the forward voltage drop to 0.3V each, but with the low forward current should be a bit less.

Don't forget that you will get a nominal 1.414 x AC transformer voltage measured by a voltmeter actually as DC across the cap, will need scaling to arrive at RMS.

7. ### BBROYGBVGW

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Mar 8, 2013
Do you think that the diodes volt drop can account for the difference I’m finding, at the high end? The highest I’ve tested and measured is 24V a.c. which gives a 17.82V d.c. output. That’s 16V less than the 34V I thought I should expect, i.e. 24 x root2.
As for the value of the averaging capacitor, you’re right the frequency will remain at 50Hz so I can predict the expected output. Are you saying I should experiment with different value capacitance until I get the correct output? Any idea what sort of ballpark value should I be starting with?

8. ### BBROYGBVGW

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Mar 8, 2013
The signal will be a regular Sinewave, only likely to vary with amplitude. So I think my 24V supply should be ok shouldn't it. Is there an imperative reason to bring thisdown to 12V when the device has a supply voltage range up to 32V?

Thanks for the offer of sending, it's very kind of you but I'm in the UK; sorry to hear about your beer situation
The range of voltage will be 0 to ~ 12 Volts a.c but that is only because that is the input range of the data logger. The range can be determined by wherever the tapping is taken from on the tf/mr secondary windings.
I just quoted that to demonstrate that the volt drop across the diodes is not the cause of the odd result.

9. ### BBROYGBVGW

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Mar 8, 2013
Ok thanks for the info on the type of diodes; I'll give then a try.

Does that mean that, with a time constant of 1ms, the capacitor will only fill in the trough for 1ms, when the trough, from peak to peak, is 10ms (if my calculation is correct)?
Ok I’ll try a larger capacitor, as you suggest, and let you know the results.

10. ### BBROYGBVGW

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Mar 8, 2013
Ok then, I’ve used some different diodes that I had lying about, that have a forward volt drop of 0.6V; not quite the 0.3V suggested but in the right order of things, for the time being. I’ve also replaced the smoothing capacitor for a 4.7uF. The 10k resistor for a load is the same.
If I plot a curve of the a.c. input, against the d.c. output voltages I get a straight line but at a slightly different angle to an imaginary line of like for like voltages, i.e. an ideal curve of the same value of a.c voltage in as the d.c. voltage out; these lines intersect at 3V. The actual line was tested up to 50V a.c. input and it continues in a straight line relationship up to this value.
I can obtain the formula for this curve and apply it in the data logger to scale the readings to measure the correct value. However I really would like to know what is going on before I can be confident to rely on this. What happened to the 1.414 x a,c voltage, d.c. voltage that I would expect to be getting?

I’ve had another suggestion to use an op amp in an integrating circuit to covert the sinusoidal input to a d.c. output with a smoothing capacitor to take out the cosine ripple.
Anyone got any ideas for the type of op amp to use to try this and is it likely to be successful?

11. ### Harald KappModeratorModerator

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Nov 17, 2011
1.41*Vfms is the peak voltage of the DC signal. If you don't add sufficient smoothing capacitors, as woodchips has suggeested, your meter will not see the peak value bt only the average value of the dc. Depending on how exactly the meter measures DC, your meter will display something else than 1.41*Vrms.

To eliminate the voltage drop across the diodes, I suggest a different approach to the task:
1) take the AC output of the transformer (signal to be measured) and bring it into a mangeabnle small signal range (e.g. +-1V) using a resistive divider.
2) Use an OpAmp based precision recitifier to get a DC signal.
3) measure the DC signal.

12. ### BBROYGBVGW

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Mar 8, 2013
I've attached a copy of the curves plotted as referred to in my last posting, just to help explain what I’m saying.
My apologies for it not being very neat but this was originally only for my own information.

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13. ### BBROYGBVGW

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Mar 8, 2013
Sorry I didn’t see your post until when I uploaded my last one.

Thanks for the advice; this has been suggested to me by someone else too but I wasn’t sure how to put it together.

Which method out of those in your link of Precision Rectifier Circuits do you recommend? The Precision Bridge Rectifier for Instrument Applications or the Full Wave Rectifier with transfer characteristics, seems the most suitable.
I have a duel Op Amp, which is a LM 107. It is an audio Amp and as the frequency used will be 50Hz, it seemed a suitable choice.
Advice would be welcome.
Thanks

14. ### Harald KappModeratorModerator

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Nov 17, 2011
I think the full wave rectifier is appropriate.

15. ### BBROYGBVGW

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Mar 8, 2013
Ok thanks; would that be with the LM107?

What sort of values for R1, R2, and RL? Do they determine the gain factor?

16. ### Harald KappModeratorModerator

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Nov 17, 2011
You can use any opamp. The values of the resistors define the gain. Look up the respective equations in an opamp tutorial.

17. ### BBROYGBVGW

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Mar 8, 2013
Ok, I know I must be missing something very fundamental here but I have a dual LM107. I’m assuming this means there are two op-amps built into one package or chip. The data sheet shows one op-amp using pins 4 (input -), 5 (input+) and 10 output. 11 and 6 are the power supply pins; where do I connect for the pins of the second op-amp? Is it 2, 3 & 12? Why are these not shown as the second connections?

18. ### Harald KappModeratorModerator

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Nov 17, 2011
If it were a single opamp, you should connect the inputs of the unused opamp to GND or VCC using a pull-down or pull-up resistor (e.g. 10kOhm). Leave the unused output open.

As it is, the LM107 is a single opamp. The unused pins are, well, unused (N = Not Connected). Leave them unconnected.

19. ### BBROYGBVGW

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Mar 8, 2013
Thanks Harald.
So I need another one for the design we are looking at.
It has 14 pins and I was told, when I bought it, that it was a quad op-amps; but I suspected that was incorrect. If it means anything, it has the following, writen on it:: -
LF34 ?N
64208020
64 CHN

I might as well get a dual, as you said any op-amp will do.
Anything cheap and cheerful that comes to mind?

20. ### Harald KappModeratorModerator

11,513
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Nov 17, 2011
This could be an LF347. That part is a quad opamp, although not the most modern part (tha datasheet dates from 1994). Which doesn't mean it is a bad part. It may be well suited to your problem.
Connect the unused inputs as has been said, leave unused inputs open.

Next time you need an opamp (or any other component) I suggest you first evaluate what component you need, then go and buy it. Fitting an existing component (be it freshly over the counter or from your junk bin) may require tweaking the circuit to fit the component. Imho the other way round makes much more sense.

What does your current circuit schematic look like? Care to share?