Power Test Board04 Jul 2016
The power circuit for the GotT was designed to be a close to 'bulletproof' as was reasonably possible while still being fairly efficient in terms of power loss and heat.
It is designed to output a constant +3.3V DC at up to about 500mA and includes protection against reverse voltage, overcurrent, overvoltage, and over-temperature.
The key protection components of the power supply are designed to be relatively easy to replace should they fail as they should ultimately sacrifice themselves to protect the rest of the circuit and whatever it happens to be powering.
Let's have a look at the schematic for the power test board:
A few of the labels and test points have been stripped from the above schematic to make understanding it a bit more clear.
We will follow the flow of conventional current in this circuit from left to right.
The first component we encounter is a regular old DC barrel power jack.
This is the same type of jack found on many household electronics that use an AC adapter, often called a power brick, wall wart, or plug pack depending upon where in the world you might happen to be.
From the schematic symbol we can infer that
PIN 1 is the center pin of the jack, which in this case happens to be positive. While this is the most common configuration we could have just as easily wired it up the other way around.
What this means is that we will need to use an AC adapter that is also center positive.
Most AC adapters (decent ones in any case) will be labeled with a small diagram similar to one of the below symbols:
Note that, sadly, there is no standard mandating that adapters have any particular polarity so take care in ensuring you find a suitable adapter.
If you are uncertain you can always use a multimeter to check.
There is also no standard for the size of the center pin *sigh* but about 2 mm seems to be fairly common; your mileage may vary depending upon your location.
Ideally, we'll want to use an AC adapter with output in the 9V-12V DC range that can supply at least 500mA of current.
The power supply should have a label that looks something like this:
It is also important to check for something indicating that the output is DC.
In the example above, the output is labeled as 'DC12V'.
This symbol is also often used to indicate DC output:
In practice, these ratings should be treated only as approximations and should not be used to power a circuit directly.
AC adapters like this are known as unregulated because if the input voltage or load changes, the output will change accordingly.
There is nothing to regulate or hold the voltage at any particular level.
While you might feel that the AC power coming out of your wall is stable, you would generally be incorrect.
Fortunately, we happen to be building a regulated power supply so an unregulated input (within reason) should be just fine.
You may also have noticed what looks to be an extra terminal on the DC barrel jack connected to Ground.
Some DC jacks include a switch that either opens or closes when something is plugged into them thus allowing for this action to be detected by some additional circuitry if desired.
In our case we aren't particularly concerned with this so we can just connect it to Ground.
The next component in series is a switch.
Nothing fancy this time around, just a simple SPST, ON/OFF, slide switch.
The only real requirement here is that it beefy enough to handle whatever we might throw at it.
The slide switch used on the power test board here is rated for 125V DC (which is way overkill but should provide great isolation when OFF) and 500mA which should be just about right.
Worst case scenario is that we destroy our switch, which is relatively easy to replace with another one due to its through-hole design.
It could also simply be shorted if we feel like connecting and disconnecting the DC barrel jack repeatedly which is only a minor inconvenience.
Connected to our switch we find a P-Channel MOSFET.
This serves as our reverse voltage protection.
Sure, we could use a big-old diode here but the voltage drop and thus heat loss across it could be quite significant and its reverse voltage rating would need to be fairly high.
We could use a Schottky diode for a lower forward voltage but many Schottkys tend to have a high reverse leakage current and the power loss is still fairly significant. The MOSFET, with its low RDS(on) is a more elegant, albeit slightly more expensive, solution.
As discussed in the Lights Test Board post, it was stated that, generally speaking, in a P-Channel MOSFET conventional current flows from
This MOSFET looks to be hooked up backward though with the positive voltage on the
DRAIN. What's up with that?
- The truth is that while 'generally speaking' most MOSFETs will be hooked up this way (so it is still a useful rule to remember) most MOSFETs will actually conduct equally well in either direction, as long as the gate to channel potential (VGS) is enough to keep the device ON.
There is another reason for hooking the MOSFET up backward though.
If you recall that every MOSFET has an inherent body diode then the reason for this configuration will become apparent.
In a P-Channel MOSFET the body diode's forward direction is from
This means that a voltage at the drain will undergo a voltage drop across this diode in the forward direction (typically about -1V or so).
Note that when the MOSFET is ON this voltage drop actually gets reduced to RDS(on) as the MOSFET channel conducts in parallel to the body diode but it is important to consider when transitioning the MOSFET from OFF to ON.
We also learned that an enhancement mode P-Channel MOSFET can be turned ON by pulling its
GATE voltage lower than its
SOURCE voltage meaning that it's VGS(th) is negative. This is to say that a VGS less than VGS(th) will turn the MOSFET ON.
The pull-down resistor on the
GATE keeps the
GATE from floating by pulling it to 0V.
So if we were to apply +12V at the
DRAIN we should see about +11V at the
Since VGS = VG - VS:
VGS = 0V - 11V = -11V
Assuming that this is less than the VGS(th) (and still within the maximum allowable negative voltage range of the MOSFET) the MOSFET will turn ON.
If we were to hook our power supply up backward, say by using a supply that is center pin negative by accident, what would happen?
Well, nothing! The body diode will prevent current from flowing backward through the circuit.
But for the sake of argument, let us assume the MOSFET was actually, somehow, already ON with the voltage reversed.
GATE would now be near +12V and the
DRAIN would be at 0V.
Sine the MOSFET is in on this scenario, the voltage drop across the body diode is negligible, so the voltage at the
SOURCE would also be near 0V.
VGS = +12V - 0V = +12V
Since +12V is well above VGS(th) (which, again, must be negative for a P-Channel MOSFET to turn ON), the MOSFET will be OFF.
In fact, it is virtually impossible for the MOSFET to turn on if the voltage source is connected backward.
In the case of p–n junctions, such as those in MOSFETs, this state of little or no current flow is known as reverse bias.
The Zener diode between the
GATE and the
SOURCE helps with overvoltage protection. When the
SOURCE voltage is less than the reverse breakdown voltage of the Zener diode, it does absolutely nothing. Because we cannot say for certain that the input voltage will be exactly +12V (pesky unregulated AC adapters) we need a way to ensure that we remain within the MOSFETS maximum VGS rating.
By choosing a Zener diode with a breakdown voltage that is slightly less than the maximum VGS of our MOSFET we can ensure that any higher voltages are effectively clamped to the Zener breakdown voltage.
This is, in essence, a quick and dirty voltage regulator.
Faulty Flow (ALT: Drop It Like It's Hot)
Next is a special type of fuse known as a PTC resettable fuse, sometimes referred to by the trade-names PolysWitch (TE) or PolyFuse (Littlefuse).
Like a fuse, a PTC fuse protects against overcurrent conditions with the additional caveat that it automatically resets itself after the condition has passed.
When such an overcurrent condition occurs the PTC fuse heats up and its resistance goes from often less than one ohm to several hundreds or thousands of ohms.
When this occurs, the fuse is said to be in a tripped state. While they might seem quite fancy, PTC devices were first conceived of and patented by Bell Labs in 1939.
There are a few key parameters that need to be considered when choosing a resettable fuse:
- Operating voltage: The maximum voltage the fuse can withstand while tripped without damaging it.
- Hold Current: The maximum amount of current the fuse can tolerate without tripping.
- Trip Current: The minimum current at which the fuse will always trip.
Because the Zener diode next to the P-Channel MOSFET offers some overvoltage protection, we'll need to make sure that our fuse's operating voltage is at least as large as the Zener voltage of the diode.
This will help ensure that the fuse will never see a voltage higher than it is rated for.
Hold current and trip current can be a little confusing so let's look at an example.
Let us assume that the maximum current we want our power supply to draw is around 490mA.
We might choose a PTC fuse with a hold current of 500mA.
If our fuse has a trip current of 1A we can say with certainty that our fuse will trip when the current draw of our power supply is somewhere between 500mA and 1A.
The exact point at which the fuse will actually trip is dependent upon a number of factors including the ambient temperature, magnitude of the current draw, and the particular fuse in use.
It is also important to note that PTC fuses do not trip all at once like their melting counterparts.
Since they are resistive devices, PTC fuses can take some time to reach their maximum resistance. This can be anywhere from 1/10th of a second to a couple of seconds.
While certainly not ideal, the goal here is simply to stop the flow of current before any expensive components after the fuse heat up enough to fail.
At last we arrive at the interesting piece of the puzzle, the regulator.
In a standard Linear Regulator the resistance of the regulator varies in accordance with the load to maintain a constant output voltage.
Depending upon the input voltage, this can also result in a significant amount of power loss in the form of heat.
By contrast, a Switching Regulator like the one in use for this circuit operate by rapidly switching ON and OFF.
The Duty Cycle controls how much charge is transfered to the load in a very efficient manner that results in very low power loss to heat.
The particular regulator used in this power supply has a fixed +3.3V DC output (other fixed and adjustable versions are available) and can source up to 500mA of current.
Moreover, the regulator used is known as a Step-Down or Buck regulator for its ability to efficiently reduce higher input voltages to its output voltage.
Since our input voltage will be consistently higher than +3.3V this is what we require.
By contrast, a Step-Up or Boost regulator is capable of producing a higher output voltage than its input voltage, often at some loss to the amount of current it can supply compared to its input.
Buck-Boost Regulators are devices which combine these two types of regulators and are capable of maintaining a constant output voltage with inputs both higher and lower than the output.
These are typically more expensive and more complex devices.
The capacitor between the input of the switching regulator and Ground is a Bypass or Filter capacitor.
Its purpose is to allow any AC component of the supply voltage to pass to Ground.
It also acts as a little reserve of current to smooth out any dips in the supply voltage by releasing its charge should the voltage sag for some reason such as a momentary change in load resistance, which often occurs when turning devices on, or a small dip in the supply voltage due to draw from other devices connected to it.
GND pin is pulled to Ground as is the Enable (
EN) or ON/OFF pin.
EN pin can be used to enable or disable the regulator using logic level signals and is ON when below a voltage specified in the datasheet, otherwise it is OFF.
While the datasheet may specify that this pin can be left floating to enable the regulator it is always good practice to force a known state, in this case by connecting it to Ground if not controlled by another external device.
The Schottky diode, inductor, and capacitor on the output of the switching regulator are the most critical components to its operation.
Together they allow for the appropriate feedback to keep the regulator stable and provide clean, continuous output.
The best resource for determining the appropriate values of these components is virtually always the regulator's datasheet.
If the datasheet doesn't provide much or any information regarding proper selection of these components your best bet is to find a different part.
An output capacitor is required to filter the output and provide regulator loop stability.
Together with the inductor, the output capacitor forms a Low-pass Filter.
For the output capacitors on switching power supplies, the ESR value is the most important parameter.
Because capacitors (and inductors) are not ideal components (which is to say that they do not behave perfectly and exhibit characteristics other than capacitance or inductance) they are often treated as though they are ideal components in series with a resistor.
This practical resistance is ESR.
ESR is measured as an AC resistance at a given frequency, often 100 kHz.
Generally, low ESR electrolytic or solid tantalum capacitors are a good bet.
The ESR rating of capacitors often gets smaller as the voltage rating goes up so it's often necessary to find a capacitor rated for 50V or more to obtain a desirably low ESR rating.
However, if the ESR of the output capacitor is too low, an unstable feedback loop can result in an oscillation of the output so be sure to check the datasheet for acceptable ranges.
Buck regulators, like the one we're using here, require a diode to provide a return path for the inductor current when the switch turns OFF.
Typically, this diode must be 'fast' and should be located close to the regulator using short leads and traces.
Due to their very fast switching speeds and low forward voltage drop, Schottky diodes are often used for this application, especially when the output voltage is relativity low (around 5V or less).
- Typical diodes have a forward voltage drop of ~.7V compared to .15V-.48V for Schottkys
- 'Fast' refers to the reverse recovery time of the diode when it switches from the conducting to the non-conducting or blocking state (for instance when the voltage across it changes sign).
Regular diodes often have reverse recovery times of 100 to several hundred nanoseconds (1/10,000,000 of a second) which may seem very fast.
Because of their unique construction Schottky diodes often have a switching time of around 100 picoseconds or 1/10,000,000,000 of a second!
Switching regulators have two basic modes of operation: continuous and discontinuous.
The difference between the two types relates to the inductor current, whether it is flowing continuously, or if it drops to zero for a period of time in the normal switching cycle.
In many cases the preferred mode of operation is the continuous mode.
This mode offers greater output power, lower peak switch, inductor, and diode currents, and can have lower output ripple voltage.
However, the continuous mode requires larger inductor values to keep the inductor current flowing continuously, especially at low output load currents and high input voltages.
It's nice to know when something is on.
If our power supply were connected to something with nice blinky LEDs this would be easy to see but in the case of a standalone test board might not always be the case.
The solution is to simply add an indicator LED to the output of our power supply to tell us that it's working.
Blue is a pretty color and makes our electronics look expensive in an arbitrary, conceited sort of way.
Blue LEDs tend to have a significantly higher forward voltage than their red or green counterparts; usually in the 2.85V range.
To calculate the value of the current limiting resistor we can use the formula:
R = (VS - VLED)/ILED
- R is the resistance of the current limiting resistor in Ohms (Ω): The term we are attempting to solve for
- VS is the source voltage in volts (V): +3.3
- VLED is the voltage drop across the LED or its forward voltage in volts (V): +2.85
- ILED is the current through the LED, measured in Amperes (A): .005 (5mA) Taken from the typical forward current (IF) in the datasheet
Inserting the above terms into the equation yields:
R = (3.3 - 2.85)/(.005) = 90Ω
While we could probably locate a 90Ω resistor it might be a bit difficult.
Back in 1952 the IEC decided to standardize resistor values.
While you can certainly find virtually any specific value if you look hard enough it is much easier to use one of these ubiquitous and cheap common values.
The standard practice for current limiting resistors is simply to round up to the next highest commonly available value.
These values are chosen such that they are approximately equally spaced on a logarithmic scale which is just complicated enough that it's difficult to do in your head.
With enough practice, recognizing common values gets to be second nature.
In our case, a 100Ω resistor is a good choice and should provide adequate protection to our LED even if the rated resistance is off by 10%.
It is handy to be able to test our power supply without actually connecting it to anything that might be expensive.
A Dummy Load is essentially a big, fat resistor that simulates whatever might be connected to the output of our power supply without having to worry too much about what happens to it.
We know from Ohm's Law that:
V = I*R or I = V/R or R = V/I
If we want to put a 485mA load on out power supply (a bit less than its 500mA rating) we can calculate the resistance required thusly:
R = 3.3/.458 ≈ 6.8Ω
A resistor with a rating of ±5% would give us a resistor anywhere from 6.46Ω to 7.14Ω
Plugging these values back into the equation I = V/R yields:
I = 3.3/6.46 ≈ .511A or 511mA (which is just above what our supply is rated for but not unreasonably so.)
I = 3.3/ 7.14 ≈ .462A or 462mA
So it will probably be a good idea to ensure that our resistance is a bit higher than 6.46Ω by measuring it with a multimeter prior to use as a dummy load.
We can also calculate the power that the resistor will need to dissipate by using the formula:
P = V*I = 3.3 * .485 ≈ 1.6W
So we will want to ensure that our resistor is rated to handle at least 1.6 watts.
A large, wirewound resistor is perfect for this sort of application.
All of the design files and BOM for this board can be found on the GotT-Translation-Circuit GitHub page and PCBs may be ordered here: