Pre regulator voltage dropping

Started by Unlikekurt, January 06, 2021, 07:51:39 PM

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Unlikekurt

We have all faced it before; you're going to regulate but the B+ is a wee bit higher than you'd want and you'd like to spare your active regulator the dissipation burden and avoid dealing with too much thermal calculation and solder pours etc etc.
This scenario presented itself to me yet again today and I figured I'd ask what are people's preferred approach(es)
- Get a transformer with a secondary closer to what you'd want (this one is out)
- Add another regulator to bring it closer (also out for this one)
- Series zener to dump a handful or two handfuls of volts
- An appropriately sized resistor to burn away the unneeded power

Any takers?

ElectricDruid

What are the actual numbers? What have you got and what do you need?

I don't think there's a one-size-fits-all solution that would make sense every time. Probably each of those approaches makes sense in some situation or other.

Rob Strand

QuoteAn appropriately sized resistor to burn away the unneeded power
If there's no dropout risks and you can put an upper limit on the load current I'd go for the resistor.
The main idea is to shift power dissipation so no point shifting it to an active device with poorer
reliability.    If the amount of power dissipation was so high the resistor needed a heatsink and
you already had a heatsink for the main transistor then maybe consider putting in a second
active device mounted on the same sink.

If there is a risk of drop outs or or high variation in the load current then pre-regulator is better.

Remember a resistor will drop the input voltage and the line regulation will cause a voltage
shift.   The pre-regulator doesn't have that issue it makes things better for the second regulator.

For small amounts of "help" you can use a diode.   The voltage drop is relatively independent of
current so you don't get the voltage variations.

The decision all hinges on fine points.
Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

Unlikekurt

It is more of a curiosity.  I want to drop about 11 volts at .33w.   I'll likely move down to 10v and .25w and use a half watt resistor.  But I did start considering a 10V / 1w zener in series for space saving purposes.

merlinb

Quote from: Unlikekurt on January 06, 2021, 09:03:09 PM
I want to drop about 11 volts at .33w. 
Sounds like it's not worth it. Why not raise your regulator output voltage by 11 volts instead?

antonis

I'd use 2 X 680R (0.6W or 0.4W for space saving) metal film resistors in parallel..
"I'm getting older while being taught all the time" Solon the Athenian..
"I don't mind  being taught all the time but I do mind a lot getting old" Antonis the Thessalonian..

Rob Strand

I don't have a problem with the zener.   The amount of power your regulator is using seems quite small.   You might need to add a cap at the input of the regulator for stability.

In general the zener's not going offer much more than the resistor.   I might help regulation (if that's important) if the load current varies widely.

I'd say the series resistor method is more common, checkout this SWR amplifier for example,
https://elektrotanya.com/PREVIEWS/63463243/23432455/swr-sound/swr_550x_750x_amp_sch.pdf_2.png

There is actually a completely different approach to reduce power dissipation.  It works best when the load is constant.   In this case you need to consider the *minimum* load current.     What you do is place a resistor between the input and output.   The resistor supplies some of the load current and that reduces the load on the regulator and the dissipation.    It doesn't work with light loads because there is nothing to stop the resistor pulling the regulator output upto the input rail voltage.   This trick was *very* common on CRT monitors before say about 1985, back when the monitors had real transformers and linear regulators and it allowed them to use different regulator topologies.
Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

Unlikekurt

Merlin - Can't.  The active components downstream won't stand for it.
Rob/Antonis - I'm more than likely going to go the series resistor route.

However, this morning I had an idea.  Why not try to reduce it on the AC side of the rectifier?  This would eliminate the need for the additional capacitor at the regulator input.  However, it isn't working out quite as I'd expected.

Here is an example:
It is a bridge rectified bipolar supply.
let's call it 20-0-20
It produces around +/-26.4VDC, which seems to follow the math (*1.414 - (1.4))
I did some math to figure out where I'd want to be and inserted 150 Ohm resistors between each secondary leg and the rectifier.
The AC dropped to 13.7-0-13.7 which is just about what I'd anticipated
However, the rectified DC isn't making sense to me.  I end up with -14.7/+15.7
I'd been expecting closer to +/-18V (13.7*1.414 (-1.4)

Anything silly I'm overlooking.  And yes, there is a slightly larger current draw on the negative supply than on the positive.  But still, where are my extra couple of volts disappearing to?






Rob Strand

#8
QuoteThe AC dropped to 13.7-0-13.7 which is just about what I'd anticipated
However, the rectified DC isn't making sense to me.  I end up with -14.7/+15.7
I'd been expecting closer to +/-18V (13.7*1.414 (-1.4)

Anything silly I'm overlooking.  And yes, there is a slightly larger current draw on the negative supply than on the positive.  But still, where are my extra couple of volts disappearing to?


Small transformers have poor regulation and the DC regulation is worse than the AC regulation.   Measuring the AC voltage with a DC rectifier load can complicate things as well.

The regulation of small transformers, with an AC resistive load at full current, is normally in the 10% to 25% zone.   The DC regulation might end up at 15% to 38%.     IIRC, that's assuming DC load is operating the transformer at full power.

If the filter caps are small you might be seeing drops on the DC voltages due to ripple.

Another factor is if the transformer is rated at 1A *AC* the maximum DC load is only about 0.5A to 0.6A.  So if you are operating your DC at the 1A, the same as the AC rating, the transformer is actually operating outside of it's rating and the observed regulation is worse than expected.   

There's lot's of possible causes and you need to work through each one to find the problem(s).
Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

PRR

> rectified DC isn't making sense to me.  I end up with -14.7/+15.7
I'd been expecting closer to +/-18V  (13.7*1.414 (-1.4)


All that Rob said and: The 1.414 is for SINE into a peak catcher. When you stick a big resistor before a peak catcher the sines go flat.

And it may not save a heap of heat.

In general: work to control how much power comes in, or allow ample room for that power to get out. This may conflict with "I have to use this on-hand transformer!!", true. But I've generally been unhappy after jamming 5 pounds of stuff in a 5 pound bag.
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Unlikekurt

It definitely is ineffective, that's for sure!  I've never liked carrying around a full duffel bag either.  Thanks for everyone's feedback.
There isn't really a resolution to this thread as it was more of a speculative thing.

amptramp

Voltage dropping with a resistor has some advantages.  When the current is low, the voltage drop is low and the regulator operates from a higher voltage.  When the current is high, the voltage to the regulator is low and this tends to keep the regulator at a constant power dissipation.  Resistors can handle power application into the capacitor at the input to the regulator whereas diodes may not be able to handle the surge current.

If you have a switching regulator, it is common to put an L-C filter at the input but beware: a switching regulator is a negative resistance device.
If you increase the input voltage, the input current is reduced to maintain the same power to the load.  If you reduce the voltage, the current increases, which is why most switching regulator IC's have a low-voltage shutdown to protect the series pass elements.  Connect this negative resistance to an L-C filter on the input and you have an oscillator or at least bad transient response.  To a first approximation, the impedance of the filter is SQRT(L/C).  If the positive impedance of this filter is higher than the magnitude of the negative impedance of the regulator (which is output voltage over output current), it will oscillate.  If you have nearly this condition, you will not oscillate but you will have ringing in the transient response.