It seems like a really cheap and dirty way to have a half-assed regulator Well I know it won't keep a steady voltage like a regulator, but it could keep it from going over or under a cetain range, which would seem more favorable than having the rails be completely unregulated. Contrary to that logic, I've seen plenty of unregulated supplies... Is there something terrible about clamping?
Why haven't I ever seen a Diode Clamp in older unreg. PS?
- Thread starter Ethan
- Start date
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For instance, it might make sense to use in a dc heater supply, to limit the voltage from going over say around 7V:
IN----R---*---OUT
|
|
V
-
|
|
6.3V
(The V and the - are supposed to be the anode and cathode of the diode respectively...)
Since it just seems too cheap and easy not to do as a simple safegaurd, I was wondering if there is anything particularly bad about doing so.
IN----R---*---OUT
|
|
V
-
|
|
6.3V
(The V and the - are supposed to be the anode and cathode of the diode respectively...)
Since it just seems too cheap and easy not to do as a simple safegaurd, I was wondering if there is anything particularly bad about doing so.
Family Hoof
Well-known member
Are you talking about using a reverse biased zener diode, and having the zener voltage be your clamp point? (i.e. 12V zener diode = 12V output. input V must remain greater) Quick and dirty regulation for sure. There's a name for this type of regulator which I forget. While I was taught that it had been used decades ago, I've never seen an existing example. When I used this method in a 10V supply once I was only able to get about 4 or 5% load regulation.
[EDIT] Nevermind, I looked at the diagram again. You mean just plain clamping, right? I've never seen this commerically implemented either, although my textbook spent an awful lot of pages on it.
[EDIT] Nevermind, I looked at the diagram again. You mean just plain clamping, right? I've never seen this commerically implemented either, although my textbook spent an awful lot of pages on it.
BYacey
Well-known member
The problem with this kind of regulation is it uses quite a bit of power (heat) in order to maintain some form of regulation. In a heater circuit that is capable of supplying a few amps of current it would have to dissapate quite a bit of power in the event of 2 or 3 volts over zener point. This takes a pretty hefty zener diode.
It's a regular diode, not a zener. I was skimming through AOE, and found a cool example, figure 1.90 "Diode Limiter." It's two diodes conversly parallel, hanging off the rail and the other end to ground. It states that this circuit will limit the range of voltage swing to the forward drop of one diode (about 0.6V?).
But BYacey has a point, it could only be used where the upper range of the swing would be easily dissapated by the diode(s). So, if one were to use this circuit, I guess it would be best to use in low voltage low current applications? I suppose you COULD just spend a few bucks and get a regulator :roll: , but I just thought it was kinda cool.
But BYacey has a point, it could only be used where the upper range of the swing would be easily dissapated by the diode(s). So, if one were to use this circuit, I guess it would be best to use in low voltage low current applications? I suppose you COULD just spend a few bucks and get a regulator :roll: , but I just thought it was kinda cool.
NewYorkDave
Well-known member
Ethan,
I think you're looking at a signal application and trying to apply it to a power supply application. Reverse-biased diodes are used for signal clamping all the time. On opamp input stages, you'll often see reverse-biased diodes going from the input line to the + and - power supply rails. This ensures that the input signal never swings greater than the rail voltage + 0.6V. It helps to protect the input device from damage.
You also see this on power amp outputs, to shunt "flyback" from reactive loads to the supply rails, to avoid breakdown of the output transistors.
There are some instances where you see regular silicon diodes used reverse-biased in a power supply. The most common example is across a voltage regulator IC. This will shunt any overvoltages on the supply line of 0.6V or more greater than the unregulated input voltage across the IC rather than through the IC, avoiding output stage breakdown. Also, a second reverse-biased diode is sometimes seen between the regulated supply and ground for reverse-polarity protection.
In a power supply, a zener diode, gas arrestor tube or whatever used as a brute force clamp to short out overvoltages is often called a "crowbar." Since a crowbar's whole function is to short the supply if it rises above a certain voltage, it's strictly a safety device and should not be used for "regulation" in the usual sense. And it should be preceded by a fuse or fusible link in series with the supply line, since the whole idea of shorting out the supply is to interrupt it by causing a fuse to blow! (If the overvoltage condition is more than momentary, that is).
Zeners are also used for regulation. This type of regulation is known as "shunt regulation" and while it's not as good as an active regulator circuit using a series pass element, it's "good enuff" in some cases, especially in low power circuits that draw a constant current and don't need very tight regulation. When used as a regulator, as opposed to a crowbar, a zener needs a series resistor to form an "inverted-L" attenuator, with the resistor acting as the upper leg of the attenuator and the zener as the bottom leg. The output voltage is taken from the junction of the two.
Zeners need a certain amount of current (a few mA, I don't recall the exact "typical" figure) passing through them at all times to maintain their zener voltage. The series resistor is calaculated thus:
R = (Vin - desired Vout) / (Iload + Izener)
OK, I just got up off my ass and checked one of my books. It gives 5mA as a "typical" value of zener current. So let's say we want to take 15VDC unregulated and spit out 12VDC regulated. Our circuit draws 20mA at 12VDC.
R = (15 - 12) / (.02 + .005) = 3 / .025 = 120 ohms.
Under normal operation, the resistor will dissipate (15 - 12)sq / 120 = .075W. If the supply were shorted, the resistor would have to dissipate 15sq/120=1.8watts. So we'd probably use a 5W resistor if we DIDN'T want it to blow. If we DID want it to blow, we'd use, say, a 1W resistor (flameproof, of course!).
But wait, what happens if no load is connected? The zener will draw whatever current is necessary to maintain about 12V across its terminals, in other words, the whole 25mA will pass through the zener. Therefore, the zener should have a rating of
P = .025 * 12 = 0.3W. So we'd use a 500mW zener.
Zeners are noisy. The zener in our hypothetical power supply should be bypassed by a filter cap, and ideally kept away from the signal circuits.
This has been a fun mental exercise. But in the real world, I'd probably reach for an LM317 :green:
I think you're looking at a signal application and trying to apply it to a power supply application. Reverse-biased diodes are used for signal clamping all the time. On opamp input stages, you'll often see reverse-biased diodes going from the input line to the + and - power supply rails. This ensures that the input signal never swings greater than the rail voltage + 0.6V. It helps to protect the input device from damage.
You also see this on power amp outputs, to shunt "flyback" from reactive loads to the supply rails, to avoid breakdown of the output transistors.
There are some instances where you see regular silicon diodes used reverse-biased in a power supply. The most common example is across a voltage regulator IC. This will shunt any overvoltages on the supply line of 0.6V or more greater than the unregulated input voltage across the IC rather than through the IC, avoiding output stage breakdown. Also, a second reverse-biased diode is sometimes seen between the regulated supply and ground for reverse-polarity protection.
In a power supply, a zener diode, gas arrestor tube or whatever used as a brute force clamp to short out overvoltages is often called a "crowbar." Since a crowbar's whole function is to short the supply if it rises above a certain voltage, it's strictly a safety device and should not be used for "regulation" in the usual sense. And it should be preceded by a fuse or fusible link in series with the supply line, since the whole idea of shorting out the supply is to interrupt it by causing a fuse to blow! (If the overvoltage condition is more than momentary, that is).
Zeners are also used for regulation. This type of regulation is known as "shunt regulation" and while it's not as good as an active regulator circuit using a series pass element, it's "good enuff" in some cases, especially in low power circuits that draw a constant current and don't need very tight regulation. When used as a regulator, as opposed to a crowbar, a zener needs a series resistor to form an "inverted-L" attenuator, with the resistor acting as the upper leg of the attenuator and the zener as the bottom leg. The output voltage is taken from the junction of the two.
Zeners need a certain amount of current (a few mA, I don't recall the exact "typical" figure) passing through them at all times to maintain their zener voltage. The series resistor is calaculated thus:
R = (Vin - desired Vout) / (Iload + Izener)
OK, I just got up off my ass and checked one of my books. It gives 5mA as a "typical" value of zener current. So let's say we want to take 15VDC unregulated and spit out 12VDC regulated. Our circuit draws 20mA at 12VDC.
R = (15 - 12) / (.02 + .005) = 3 / .025 = 120 ohms.
Under normal operation, the resistor will dissipate (15 - 12)sq / 120 = .075W. If the supply were shorted, the resistor would have to dissipate 15sq/120=1.8watts. So we'd probably use a 5W resistor if we DIDN'T want it to blow. If we DID want it to blow, we'd use, say, a 1W resistor (flameproof, of course!).
But wait, what happens if no load is connected? The zener will draw whatever current is necessary to maintain about 12V across its terminals, in other words, the whole 25mA will pass through the zener. Therefore, the zener should have a rating of
P = .025 * 12 = 0.3W. So we'd use a 500mW zener.
Zeners are noisy. The zener in our hypothetical power supply should be bypassed by a filter cap, and ideally kept away from the signal circuits.
This has been a fun mental exercise. But in the real world, I'd probably reach for an LM317 :green:
Thanks Dave for that detailed explanation... After reading a bit more on it, I sat back and realized...When would I ever use this when I can shove an LM317 in there and have a lot more versatility!
Thanks! Learn summin' new every day.
Thanks! Learn summin' new every day.
NewYorkDave
Well-known member
Actually, it's good that you're considering unconventional ways of doing things instead of just reaching for the easy ready-made solution that everybody else uses. Even if you end up reaching the conclusion that the "standard" solution really is the better way, you've achieved a level of understanding that you wouldn't have had if you'd just taken the cookbook approach like most DIYers do.
This is why AoE has its "bad circuit" examples. It's just as important to be able to explain why a circuit won't work well, as it is to understand a circuit that does.
This is why AoE has its "bad circuit" examples. It's just as important to be able to explain why a circuit won't work well, as it is to understand a circuit that does.
