Will a Zener Diode work for me here? 4

[railking]

I need to insert a component or a VERY simple circuit intoa project that will limit DC voltage to 18 volts (I could use 16 or 17 volts too, but nothing more than 18 volts).

I don't want a voltage regulator because the voltage wil be running small motors and must vary from 0V to 18V, but if it goes over 18V it will damage some delicate circuitry I am using.

I am designing the circuit to handle up to 10 amps under load.

I was thinking of using a zener diode with a 17V or 18V rating but I am unsure if this will work.

Will an 18V Zener allow 0-18V to pass and then clamp the voltage at 18V?

If this is true, then anything over 18V would be disipated as heat, right?

If the likely maximum voltage that could ever ba applied to this cicuit is 20.5 volts from batteries, then it would be dissipating 2.5 volts at 10 amps (25 watts) right?

If this is all feaible, then I have more questions about what end of the zer gets connected to the battery and what end goes to my protected circuit device.

 

[logbook]

Using a distributor to find parts is a good way to go. I tried RS components

http://rs

http://www.com/

searching on "regulator" and going to the table for adjustable, linear, positive regulators, the maximum current rating available is 7.5A. You must understand that these are high power levels for small-signal electronics guys. Probably nobody would use a linear regulator above these levels because the power loss would be so excessive.

There are two questions: one, how can you solve the problem, and two how would I solve the problem? Now me, being a "clever clogs" might use a power MOSFET in series with the power rail. When the input voltage was less than say 17V the power MOSFET would be on continuously. As the voltage increased the power MOSFET would be switched, reducing the overall output voltage. This is a pulse width modulated scheme, a buck regulator. There would need to be a 15A inductor (choke) in series with the power MOSFET and a 15A diode to provide the flyback for the inductor. There would be a switched mode controller chip to drive the power MOSFET. You will be able to find this sort of circuit in National Semiconductor application notes for their "simple switchers".

http://www.national.com/appinfo/power/

These 10A components , the MOSFET, choke and diode are all chunky components and the power MOSFET would fail within a few seconds if you failed to get enough heatsinking on it. Once the junction reaches 200degC the MOSFET has had it (fried; dead). This scheme will get the most out of the battery because you will only lose say 300mV across the MOSFET/choke when it is not switching. However, the two diodes in series approach, although it loses 1.6V continuously, is very easy for a non-specialist. The LT1083 data sheet (although it can only handle 7.5A) operates with 1V headroom, so with under-voltage conditions it will run with a little less than 1V drop, the amount being a bit uncertain because the manufacturer doesn’t characterise the regulator for this use. Maybe it is only half a volt drop; you would have to try one.

However, having talked around the problem, I have now come up with an even better solution for both of us. Why not use the double diode as the over-voltage protection. When it is not needed switch it out of circuit by short-circuiting across it. This shorting link can be done with a relay contact or with a P channel power MOSFET. All you now need to do is to switch the relay or MOSFET when the voltage is low. This requires a reference voltage, a resistive divider, and a comparator. That is not much circuitry and the efficiency will be excellent.

Suppose we pick a 5V reference. Two resistors form a divider chain from the input. At 17V input the output of the divider chain is 5V. A comparator detects the input is slightly too high and switches the relay off or the power MOSFET off.. Job done. This should be enough to get you started.

Beware of one tricky point. When you switch in the extra diodes, the voltage will drop by say 1.6V. If the voltage was just too high it would then become 1.6v too low. The comparator would therefore keep switching and would be unable to "decide" if it should be on or off. The answer is to include a lot of hysteresis in the switching point. This consists of a resistor giving positive feedback to the comparator (making a schmitt trigger).

I have deliberately used some technical terms which you can look up to find the detailed information you will need.

 

[MacGyverS2000]

Quite a few of the voltage regulator datasheets show basic circuits which use the regulator as a true regulator, and not the current carrying device. This generally involves a handful of external components and one or more BJT/FET packages. I would suggest this as a possibility... several FETs in parallel will be more than enough to handle the 10A necessary, with little voltage drop or wasted power.

 

[logbook]

MacGyver2000,
the ones I have seen use an NPN pass transistor for a positive regulator. We need to use a PNP pass transistor (or P channel MOSFET) in order to get a low drop. Can you give us a specific example of such a regulator?

Using an emitter follower in the regulator is easy. Using a common emitter is much more difficult because the loop gain is poorly defined and getting stability in the loop is harder.