collect! 19 methods and techniques for converting 5V to 3.3V level

Tip 1, Use LDO regulator to supply power from 5V power supply to 3.3V system

A standard three-terminal linear regulator typically has a dropout voltage of 2.0-3.0V. They cannot be used to reliably convert 5V to 3.3V. Low Dropout (LDO) regulators with a dropout voltage of a few hundred millivolts are ideal for this application. Figure 1-1 is a block diagram of a basic LDO system with the corresponding currents labeled. As can be seen from the figure, the LDO consists of four main parts:

1. Turn on the transistor

2. Band gap reference source

3. Operational amplifier

4. Feedback resistor divider

When choosing an LDO, it is important to know how to differentiate the various LDOs. The device's quiescent current, package size and type are important device parameters. Depending on the specific application to determine various parameters, the optimal design will be obtained.

The quiescent current IQ of the LDO is the ground current IGND of the device when the device is operating with no load. IGND is the current used by the LDO for regulation. When IOUT>>IQ, the efficiency of the LDO can be approximated by dividing the output voltage by the input voltage. However, at light loads, IQ must be factored into the efficiency calculation. LDOs with lower IQ have higher light load efficiency. Increased light load efficiency has a negative impact on LDO performance. LDOs with higher quiescent current respond faster to sudden line and load changes.

Tip 2, Low Cost Power Supply System Using Zener Diodes

A low-cost regulator scheme using Zener diodes is detailed here.

A simple low-cost 3.3V regulator can be made with Zener diodes and resistors, as shown in Figure 2-1. In many applications, this circuit can be a cost-effective replacement for an LDO regulator. However, this regulator is more load sensitive than an LDO regulator. Also, it is less energy efficient because R1 and D1 always have power dissipation. R1 limits the current into D1 and the PICmicro? MCU to keep VDD within the allowable range. Since the reverse voltage of the Zener diode changes as the current through it changes, the value of R1 needs to be carefully considered.

R1 is chosen so that at maximum load—typically when the PICmicro MCU is running and driving its output high—the voltage drop across R1 is low enough that the PICmicro MCU has enough voltage to maintain operation. Also, at minimum load—usually when the PICmicro MCU is in reset—VDD does not exceed the power rating of the Zener diode, nor does it exceed the maximum VDD of the PICmicro MCU.

Tip 3, Lower cost power supply system using 3 rectifier diodes

Figure 3-1 details a lower cost regulator solution using 3 rectifier diodes.

We can also put several regular switching diodes in series and use their forward voltage drop to reduce the voltage going into the PICmicro MCU. That's even less than the cost of a zener diode regulator. The current consumption of this design is usually lower than that of circuits using Zener diodes

The number of diodes required varies according to the forward voltage of the diodes chosen. The voltage drop across diodes D1-D3 is a function of the current through these diodes. R1 is connected to prevent the voltage on the PICmicro MCU VDD pin from exceeding the maximum VDD value of the PICmicro MCU during hours of minimal load—typically when the PICmicro MCU is in reset or sleep state. Depending on the other circuits connected to VDD, the value of R1 can be increased, or R1 may not be required at all. Diodes D1-D3 are chosen so that at maximum load—typically when the PICmicro MCU is running and driving its output high—the voltage drop across D1-D3 is low enough to meet the minimum VDD requirement of the PICmicro MCU.

Tip 4, Using a Switching Regulator to Power a 3.3V System from a 5V Supply

As shown in Figure 4-1, a buck switching regulator is an inductor-based converter used to step down an input voltage source to a lower-amplitude output voltage. Output regulation is achieved by controlling the conduction (ON) time of MOSFET Q1. Since the MOSFET is either in a low-impedance state or in a high-impedance state (ON and OFF, respectively), high input source voltages are efficiently converted to lower output voltages.

By balancing the voltage-time of the inductor while Q1 is in these two states, the relationship between the input and output voltages can be established.

For MOSFET Q1, we have:

When choosing the value of the inductor, it is a good initial choice to make the inductor's maximum peak-to-peak ripple current equal to ten percent of the maximum load current.

When selecting the output capacitor value, a good initial value is to make the LC filter characteristic impedance equal to the load resistance. In this way, if the load is suddenly removed during full-load operation, the voltage overshoot can be within an acceptable range.

When choosing diode D1, choose a component with a current rating high enough to withstand the inductor current during pulse cycle (IL) discharge.

digital connection

When connecting two devices with different operating voltages, it is necessary to know their respective output and input thresholds. Once the threshold is known, the method of attaching the device can be chosen based on the other requirements of the application. Table 4-1 is the output and input thresholds used in this document. When designing your connections, be sure to refer to the manufacturer's data sheet for actual threshold levels.

Tip 5, 3.3V → 5V direct connection

The easiest and most ideal way to connect a 3.3V output to a 5V input is a direct connection. Direct connection needs to meet the following 2 requirements:

? VOH of 3.3V output is greater than VIH of 5V input

? VOL of 3.3V output is less than VIL of 5V input

One example where this approach can be used is connecting a 3.3V LVCMOS output to a 5V TTL input. From the values given in Table 4-1 it is clear that the above requirements are met.

VOH (3.0V) of 3.3V LVCMOS is greater than VIH (2.0V) of 5V TTL and

VOL (0.5V) of 3.3V LVCMOS is less than VIL (0.8V) of 5V TTL.

If these two requirements are not met, additional circuitry is required to connect the two parts. See Tips 6, 7, 8, and 13 for possible solutions.

Tip 6, 3.3V→5V Using MOSFET Converter

If the VIH of a 5V input is higher than the VOH of a 3.3V CMOS device, additional circuitry is required to drive any such 5V input. Figure 6-1 shows a low-cost two-component solution.

When choosing the resistance value of R1, two parameters need to be considered, namely: the switching speed of the input and the current consumption on R1. When switching the input from 0 to 1, account for the input rise time due to the RC time constant formed by R1, the input capacitance of the 5V input, and any stray capacitance on the board. The input switching speed can be calculated by the following formula:

Since the input capacitive reactance and stray capacitance on the board are fixed, the only way to increase the input switching speed is to reduce the value of R1. Reducing the resistance of R1 to obtain a shorter switching time is at the expense of increasing the current consumption when the 5V input is low. Typically, switching to 0 is much faster than switching to 1 because the on-resistance of the N-channel MOSFET is much smaller than that of R1. Also, when selecting an N-channel FET, the VGS of the selected FET should be lower than the VOH of the 3.3V output.

Tip 7, 3.3V→5V using diode compensation

Table 7-1 lists the input voltage thresholds for 5V CMOS, and the output drive voltages for 3.3VLVTTL and LVCMOS.

From the above table it can be seen that both the high and low input voltage thresholds of the 5V CMOS input are about one volt higher than the thresholds of the 3.3V output. Therefore, even if the output from the 3.3V system could be compensated, there would be little or no room for noise or component tolerances. What is needed is a circuit that compensates the output and increases the difference between the high and low output voltages.

Once the output voltage specification has been determined, it has been assumed that the high output drives a load between the output and ground, and the low output drives a load between 3.3V and the output. If the load for the high voltage threshold is actually between the output and 3.3V, then the output voltage is actually much higher because the mechanism pulling the output high is the load resistor, not the output transistor.

If we design a diode compensation circuit (see Figure 7-1), the forward voltage of diode D1 (typical value 0.7V) will make the output low voltage rise, and get 1.1V to 1.2V low voltage at 5V CMOS input. It is safely below the low input voltage threshold of the 5V CMOS input. The output high voltage is determined by a pull-up resistor and diode D2 to the 3.3V supply. This makes the output high voltage approximately 0.7V above the 3.3V supply, or 4.0 to 4.1V, well above the 5V CMOS input threshold (3.5V).

Note: For the circuit to work properly, the pull-up resistor must be significantly smaller than the input resistance of the 5V CMOS input to avoid a drop in output voltage due to the resistor divider effect at the input. The pull-up resistor must also be large enough to ensure that the current loaded on the 3.3-V output is within device specifications.

Tip 8, 3.3V→5V using a voltage comparator

The basic working of a comparator is as follows:

? When the inverting (-) input voltage is greater than the non-inverting (+) input voltage, the comparator output switches to Vss.

? The output of the comparator is high when the voltage at the non-inverting (+) input is greater than the voltage at the inverting (-) input.

To maintain the polarity of the 3.3V output, the 3.3V output must be connected to the non-inverting input of the comparator. The inverting input of the comparator is connected to the reference voltage determined by R1 and R2, as shown in Figure 8-1.

Calculate R1 and R2

The ratio of R1 and R2 depends on the logic level of the input signal. For a 3.3V output, the inverting voltage should be placed at the midpoint voltage between VOL and VOH. For LVCMOS output, the midpoint voltage is:

If the logic level relationship of R1 and R2 is as follows,

If the value of R2 is 1K, then R1 is 1.8K.

A properly connected op amp can be used as a comparator to convert a 3.3V input signal to a 5V output signal. This takes advantage of the comparator's property that the comparator forces the output high (VDD) or low (Vss) depending on the magnitude of the voltage difference between the "inverting" input and the "non-inverting" input.

Note: For the op amp to operate properly from a 5V supply, the output must have rail-to-rail drive capability.

Tip 9, 5V→3.3V direct connection

Typical 5V output VOH is 4.7 volts, VOL is 0.4 volts; and typical 3.3V LVCMOS input VIH is 0.7 x VDD, VIL is 0.2 x VDD.

When the 5V output is driven low, there is no problem because the 0.4V output is less than the 0.8V input threshold. When the 5V output is high, the VOH of 4.7 volts is greater than the VIH of 2.1 volts, so, we can directly connect the two pins without conflict, provided that the 3.3V CMOS output is 5 volt tolerant.

If the 3.3V CMOS input cannot tolerate 5 volts, there will be a problem because the input's maximum voltage specification is exceeded. See Tips 10-13 for possible solutions.

Tip 10, 5V→3.3V use diode clamp

Many manufacturers use clamping diodes to protect the device's I/O pins from exceeding the maximum allowable voltage specification. Clamping diodes keep the voltage on the pin from more than one diode drop below VSS or more than one diode drop above VDD. To use clamping diodes to protect the input, the current through the clamping diodes is still of concern. The current through the clamping diode should always be small (on the order of microamps). If too much current flows through the clamping diodes, there is a risk of component latch-up. Since the source resistance of the 5V output is usually around 10Ω, a series resistor is still required to limit the current flowing through the clamp diode, as shown in Figure 10-1. A consequence of using a series resistor is that the input switching speed is slowed because of the RC time constant formed on the pin (CL).

If a clamping diode is not available, an external diode can be added to the current flow, as shown in Figure 10-2.

Tip 11, 5V→3.3V Active Clamp

One problem with using a diode clamp is that it will inject current into the 3.3V supply. In a design with a high current 5V output and a lightly loaded 3.3V rail, this current injection can push the 3.3V supply voltage beyond 3.3V. To avoid this problem, a triode can be used instead, which directs excess output drive current to ground instead of the 3.3V supply. The designed circuit is shown in Figure 11-1.

The base-emitter junction of Q1 performs the same function as the diode in the diode clamp circuit. The difference is that only a few percent of the emitter current flows out of the base into the 3.3V rail, and the vast majority of the current flows into the collector, from which it flows harmlessly to ground. The ratio of base current to collector current, determined by the current gain of the transistor, typically 10-400, depending on the transistor used.

Tip 12, 5V → 3.3V Resistive Divide

A simple resistor divider can be used to step down the output of the 5V device to a level suitable for the input of the 3.3V device. The equivalent circuit of this interface is shown in Figure 12-1.

usually, The source resistance RS is very small (less than 10Ω), if the selected R1 is much larger than RS, then the influence of RS on R1 can be ignored. At the receiving end, the load resistance RL is very large (greater than 500 kΩ), if the selected R2 is much smaller than RL, then the influence of RL on R2 can be ignored.

There is a tradeoff between power consumption and transient time. To minimize the power dissipation requirements of the interface current, the series resistors R1 and R2 should be as large as possible. However, the load capacitance (combined by the stray capacitance CS and the input capacitance CL of the 3.3-V device) may adversely affect the rise and fall times of the input signal. If R1 and R2 are too large, the rise and fall times may be unacceptably long.

If the effects of RS and RL are neglected, the equations to determine R1 and R2 are given by Equation 12-1 below.

Equation 12-2 shows the formula for determining rise and fall times. To facilitate circuit analysis, the Thevenin equivalent calculation is used to determine the applied voltage VA and the series resistance R. The Thevenin equivalent calculation is defined as the open circuit voltage divided by the short circuit current. With the constraints imposed by Equation 12-2, for the circuit shown in Figure 12-1, the Thevenin equivalent resistance R should be determined to be 0.66*R1 and the Thevenin equivalent voltage VA should be 0.66*VS.

For example, suppose the following conditions exist:

? Stray capacitance = 30 pF

? Load capacitance = 5 pF

? ≤ 1 μs maximum rise time from 0.3V to 3V

? Applied source voltage Vs = 5V

The calculation to determine the maximum resistance is shown in Equation 12-3.

Tip 13, 3.3V→5V Level Shifter

Although level translation can be done discretely, an integrated solution is generally preferred. Level shifters are available in a wide range of applications: unidirectional and bidirectional configurations, different voltage translations, and different speeds allow the user to choose the best solution.

Board-level communication between devices (for example, MCU to peripherals) is most common via SPI or I2C?. For SPI, a unidirectional level shifter is appropriate; for I2C, a bidirectional solution is required. Figure 13-1 below shows both solutions.

simulation

The final challenge of the 3.3V to 5V interface is how to convert the analog signal so that it crosses the power barrier. Low-level signals may not require external circuitry, but systems that pass signals between 3.3V and 5V will be affected by power supply variations. For example, in a 3.3V system, an ADC converts a 1V peak analog signal with higher resolution than an ADC in a 5V system, because more of the ADC's range is used for conversion in a 3.3V ADC. On the other hand, the relatively high signal amplitudes in a 3.3V system may conflict with the system's lower common-mode voltage limitations.

Therefore, to compensate for the above differences, some kind of interface circuitry may be required. This section discusses interface circuits to help alleviate the problem of signals transitioning between different power supplies.

Tip 14, 3.3V→5V Analog Gain Block

When connecting from a 3.3V supply to 5V, an analog voltage boost is required. The 33 kΩ and 17 kΩ resistors set the gain of the op amp so that full scale is used at both ends. The 11 kΩ resistor limits the current flowing back into the 3.3V circuit.

Tip 15, 3.3V→5V Analog Compensation Module

This module is used to compensate the analog voltage from 3.3V to 5V conversion. The following is to convert the analog voltage powered by 3.3V power supply to be powered by 5V power supply. The 147 kΩ, 30.1 kΩ resistors and +5V supply on the upper right are equivalent to a 0.85V voltage source with a 25 kΩ resistor in series. This equivalent 25 kΩ resistor, three 25 kΩ resistors, and the op amp form a difference amplifier with a gain of 1 V/V. The 0.85V equivalent voltage source will shift any signal present at the input upward by the same amount; a signal centered at 3.3V/2 = 1.65V will also be centered at 5.0V/2 = 2.50V. The upper left resistor limits the current from the 5V circuit.

Tip 16, 5V→3.3V Active Analog Attenuator

This trick uses an op amp to attenuate signal amplitude from a 5V to 3.3V system.

The easiest way to convert a 5V analog signal to a 3.3V analog signal is to use a resistor divider with an R1:R2 ratio of 1.7:3.3. However, there are some problems with this approach.

1) The attenuator may be connected to a capacitive load, forming an undesired low-pass filter.

2) An attenuator circuit may need to drive a low impedance load from a high impedance source.

In either case, an op amp is needed to buffer the signal.

The required op amp circuit is a unity-gain follower (see Figure 16-1).

The output voltage of the circuit is the same as the voltage applied to the input.

To convert the 5V signal to a lower 3V signal, we just add a resistive attenuator.

If the resistor divider precedes the unity gain follower, it will provide the lowest impedance for the 3.3V circuit. Also, the op amp can be powered from 3.3V, which will save some power. Power dissipation on the 5V side can be minimized if X is chosen to be very large.

If the attenuator is after the unity gain follower, then there is the highest impedance to the 5V source. The op amp must be powered from 5V, the impedance on the 3V side will depend on the value of R1||R2.

Tip 17, 5V→3.3V Analog Limiter

Attenuation can sometimes be used as gain when passing a 5V signal to a 3.3V system. If the desired signal is less than 5V, then feeding the signal directly into the 3.3V ADC will result in a larger conversion value. The danger arises when the signal approaches 5V. Therefore, there is a need for a method of controlling voltage overshoot while not affecting the voltage in the normal range. Three implementation methods will be discussed here.

1. Use diodes to clamp overvoltage to 3.3V supply system.

2. Using Zener diodes, clamp the voltage to any desired voltage limit.

3. Use an op amp with a diode for accurate clamping.

The easiest way to do overvoltage clamping is exactly the same as connecting a 5V digital signal to a 3.3V digital signal. Use resistors and diodes to allow excess current to flow into the 3.3V supply. Resistor values must be chosen to protect the diodes and 3.3V supply without negatively impacting analog performance. This type of clamping can cause the 3.3V supply voltage to rise if the impedance of the 3.3V supply is too low. Even though the 3.3V supply has a nice low impedance, this type of clamping will make the input The signal adds noise to the 3.3V supply.

In order to prevent the input signal from affecting the power supply, or to make the input more calm when dealing with large transient currents, a slight change to the above method is used instead of a Zener diode. Zener diodes are usually slower than the fast signal diodes used in the first circuit. However, Zener clamps are generally more robust and do not depend on the characteristics of the power supply when clamping. The size of the clamp depends on the current flowing through the diode. This is determined by the value of R1. R1 may also be unnecessary if the output impedance of the VIN source is large enough.

If more accurate overvoltage clamping independent of the power supply is required, precision diodes can be obtained using op amps. The circuit is shown in Figure 17-3. The op amp compensates for the forward voltage drop of the diode so that the voltage is clamped exactly at the supply voltage at the non-inverting input of the op amp. If the op amp is rail-to-rail, it can be powered by 3.3V.

Since the clamping is carried out through the op amp, it will not affect the power supply.

The op amp does not improve the impedance that occurs in low voltage circuits, the impedance is still R1 plus the source circuit impedance.

Tip 18, Driving Bipolar Transistors

When driving a bipolar transistor, the base "drive" current and the forward current gain (Β/hFE) will determine how much current the transistor will sink. If the transistor is driven by a microcontroller I/O port, use the port voltage and the upper port current limit (20 mA typical) to calculate the base drive current. If using 3.3V technology, a lower value base current limiting resistor should be used instead to ensure sufficient base drive current to saturate the transistor.

The value of RBASE depends on the microcontroller supply voltage. Equation 18-1 shows how RBASE is calculated.

If a bipolar transistor is used as a switch to turn on or off a load controlled by a microcontroller I/O port pin, a minimum hFE specification and margin should be used to ensure full saturation of the device.

3V technology example:

For both examples, it is good practice to increase the base current margin. Driving a base current of 1 mA to 2 mA ensures saturation at the expense of increased input power dissipation.

Tip 19, Driving N-Channel MOSFET Transistors

Care must be taken when selecting an external N-channel MOSFET for use with a 3.3V microcontroller. The MOSFET gate threshold voltage indicates the device's ability to fully saturate. For 3.3V applications, select a MOSFET with an on-resistance rating for a gate drive voltage of 3V or less. For example, a FET with a rated drain current of 250 μA for a 100 mA load with a 3.3V drive will not necessarily provide satisfactory results with 1V applied to the gate-source. When transitioning from 5V to 3V technology, the gate-source threshold and on-resistance characterization parameters should be carefully checked, as shown in Figure 19-1. Slightly reducing the gate drive voltage can significantly reduce leakage current.

For MOSFETs, low-threshold devices are common, with drain-to-source voltage ratings below 30V. MOSFETs with a drain-source voltage rating greater than 30V typically have a higher threshold voltage (VT). As shown in Table 19-1, the threshold voltage of this 30V N-channel MOSFET switch is 0.6V. With 2.8V applied to the gate, this MOSFET has a nominal resistance of 35mΩ, making it ideal for 3.3V applications.

For the specifications in the IRF7201 data sheet, the gate threshold voltage minimum is specified as 1.0V. This does not mean that the device can be used to switch current at 1.0V gate-source voltage, because for VGS(th) lower than 4.5V, no specification is stated. The IRF7201 is not recommended for 3.3V driven applications requiring low switch resistance, but it can be used for 5V driven applications.