Voltage drop for Linear IC 78XX series - Page 1
Forum section: electronics / projects

author: jloewenheim
author_rank: Newbie
timestamp: February 27, 2022, 06:14:48 am
content: A very novice question, but here goes: I'm looking to replace an inline resistor/Zener regulated dual power rail (for powering OpAmps) with a 78/7915 regulator circuit. My issue is that the input power source is 48VDC, which is too high for those ICs. Is it better to use an inline dropping resistor, and 2 regulators per rail in series (say a 24v then the 15v), or just the 15v? Also, how do I calculate the appropriate resistor value?

author: daqq
author_rank: Super Contributor
timestamp: February 27, 2022, 06:35:30 am
content: An series resistor would not work - the regulator would still see the full voltage at its input before powering what its supposed to power. Unless you are suggesting a voltage divider or a resistor+zener diode preregulator.There are voltage regulators that accept a large voltage at its input.

author: bdunham7
author_rank: Super Contributor
timestamp: February 27, 2022, 07:02:18 am
content: I depends a lot on how much current output you need. The LM78/79xx regulators can't take 48VDC, but the LM317HV- and LM337HV- can do so directly and you just need to add your own divider resistors. Another way I've seen it done is to use a zener diode as a voltage dropper before the regulator--you'd need a 15V or so zener for this and it would only work for low-current applications because of power dissipation.

author: srb1954
author_rank: Super Contributor
timestamp: February 27, 2022, 07:30:34 am
content: Quote from: jloewenheim on February 27, 2022, 06:14:48 amA series resistor generally won't work very well if you have a changing load on the output of your regulators. You have to size your resistor small enough to not drop too much voltage when there is maximum current demand on the regulators and large enough to drop sufficient voltage to keep below the maximum specified input voltage for the regulators when there is minimum current draw. These two conditions might be mutually incompatible if you have a wide variation in current drain.If you have a relatively small regulator current drain with little variation you can use a series dropping resistor and shunt Zener regulator on the regulator input. In this case the series resistor has to be low enough to ensure that the voltage drop across the resistor isn't too low when the regulator is experiencing maximum load while the Zener diode has to dissipate maximum power when the load regulator load is at its minimum. This configuration can be very inefficient if the regulator load current is larger or subject to wide variations but does have the advantage that it provides an extra level of protection for your regulators against high voltage transients on your incoming supply.For higher current drains or a wide variation in the current drain a more efficient arrangement is to have a series Zener diode to drop the incoming supply to acceptable levels. The cost for a suitable Zener can be higher as it to be sized to have sufficient power rating to safely carry the maximum current that can occur when the regulator O/P is shorted. Although the series Zener arrangement is more power efficient it doesn't provide any protection against transients on your incoming supply.A third option is to not use the standard 7815/7915 regulators but use a regulator rated for at least the full 48V input voltage. These devices tend to be more expensive or are limited to relatively low output currents.To come up with a concrete answer to your problem you first need to specify out the minimum and maximum load currents for your circuit and the minimum and maximum range of input voltage from your supply.

author: timenutgoblin
author_rank: Regular Contributor
timestamp: February 27, 2022, 12:46:21 pm
content: Quote from: jloewenheim on February 27, 2022, 06:14:48 amIf you're planning to power your OpAmp with +/- 15VDC using 78/7915 voltage regulators then you need to provide at least +/- 18VDC to power the regulators. If your power source is 48VDC then you need to split the voltage into two halves to provide +/- 24VDC to power the regulators.You could use a pair of electrolytic capacitors, connected in series, and then connect them to the 48VDC power source to provide +/- 24VDC. It is important that the capacitors are matched in capacitance and voltage rating otherwise this will cause a charge/voltage imbalance. The junction point of the two electrolytic capacitors can function as a virtual ground for your 78/7915 voltage regulators.Depending on the current flowing through the regulators, you may need a heatsink.

author: oz2cpu
author_rank: Frequent Contributor
timestamp: February 27, 2022, 08:38:13 pm
content: make a pre-regulator of resistor, zener, emitter follower transistor,and then a 78xxnow the input voltage can be made within correct range,never go all the way to the edge of maxyou can see in the datasheet what that is, for the types you use

author: jloewenheim
author_rank: Newbie
timestamp: February 28, 2022, 04:34:17 am
content: Thanks to all of you for the input, I will update when i finally dig in and get working on this. Im thinking I will likely go the Resistor/Zener route pre-regulator or the higher voltage 317/337.

author: macboy
author_rank: Super Contributor
timestamp: February 28, 2022, 03:25:45 pm
content: Quote from: timenutgoblin on February 27, 2022, 12:46:21 pmA capacitive divider can't work with DC, only with AC.For a pre-regulator,. I'd use a simple emitter-follower regulator built from a resistor, a zener and a transistor or MOSFET. Specific components and values depend on load current requirements which haven't been given by the OP. Those same components could be made into a tracking pre-regulator.

author: TimFox
author_rank: Super Contributor
timestamp: February 28, 2022, 03:51:43 pm
content: Another possibility to reduce the voltage before a 78xx regulator is to use a Zener diode in series with the 78xx input, with a power rating sufficient for the power dissipated by the Zener voltage and maximum load current. To ensure that the voltage at the 78xx is not excessive at power-on, a reasonable resistor to ensure current through the Zener (resistor from 78xx input to 78xx common), such as 10 k, is a good idea. The voltage after the Zener may not be constant, and the Zener will not improve the power supply ripple, but that is what the 78xx is for. Thetotalpower dissipated in the circuit will be the same as for a series resistor circuit, or for a linear regulator itself that can tolerate the high input voltage.

author: srb1954
author_rank: Super Contributor
timestamp: February 28, 2022, 09:22:01 pm
content: Quote from: TimFox on February 28, 2022, 03:51:43 pmThe required Zener power rating has to be based on themaximumcurrent drawn when the regulator O/P isshortedso that the Zener doesn't exceed its maximum junction temperature under fault conditions.Failure to adequately size the Zener could result in a cascade of failures. The overheated Zener would most likely short out applying the full 48V input voltage to the regulator, which could, in turn, fail with a input to output short thus applying the full 48V to all downstream circuitry possibly destroying everything.

author: TimFox
author_rank: Super Contributor
timestamp: February 28, 2022, 09:27:28 pm
content: Yes, I agree. I should have written "worst-case load current on the Zener".

author: bdunham7
author_rank: Super Contributor
timestamp: February 28, 2022, 10:05:14 pm
content: Quote from: srb1954 on February 28, 2022, 09:22:01 pmA couple of tips from someone that has recently overhauled something with this exact design:1) The LM78xx regulators effectively have foldback current limiting, so the shorted output is not the worst case for the zener.2) A low-value resistor--one that doesn't matter too much at normal current--can be helpful since it will take an increasing share of the burden as current goes up.3) If you want the device to last, you need to look at a lot more than just the power rating of the zener. Even at a fraction of the maximum power, you are not safe long term unless you carefully consider the thermal effects, and DO-41 or similar packages are hard to heat sink. Almost all of the heat will be transmitted out through the leads (actually one of them, I can't remember which) and will burn up boards and even the zener itself unless you have at least a decent copper pad on the circuit board to take some heat. The best method I've seen is to use those old-fashioned solder turrents to stand off the zener and provide a nice chunk of copper. On the device I recently repaired, which didn't have room for solder turrets, I replaced 1W zeners with 5W and extended the legs so that the zener was up off of the board.

author: timenutgoblin
author_rank: Regular Contributor
timestamp: March 01, 2022, 04:57:26 am
content: My original suggestion was similar to the circuit below except using a 15V zener instead of a 5V zener. Also the resistor value would need to be revised. I think the advantage of this circuit is that it increases the maximum input voltage rating of the regulator. You could then connect another 7815 regulator at the output to get +15VDC.

author: bdunham7
author_rank: Super Contributor
timestamp: March 01, 2022, 06:07:38 am
content: Quote from: timenutgoblin on March 01, 2022, 04:57:26 amIt increases the output voltage, but does not increase the maximum safe input voltage. It may work some of the time with >40V input, but under certain conditions, such as overload or thermal shutdown, it will blow up.

author: not1xor1
author_rank: Frequent Contributor
timestamp: March 01, 2022, 09:19:13 am
content: these are from an ST microelectronics datasheetthis works like a power zener in series with the input of the regulator chip:this is a preregulator and would also reduce the output ripple (just put a capacitor in parallel with the zener):