Using the high-low side driver IR2110 - explanation and plenty of example circuits

Using the high-low side driver IR2110 - explanation and plenty of example circuits

  January 20, 2013    Tahmid

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Update: Please refer to this updated post for bootstrap cap sizing: https://tahmidmc.blogspot.com/2023/12/sizing-ir2110-high-side-bootstrap.html

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In many situations, we need to use MOSFETs configured as high-side switches. Many a times we need to use MOSFETs configured as high-side and low-side switches. Such as in bridge circuits. In half-bridge circuits, we have 1 high-side MOSFET and 1 low-side MOSFET. In full-bridge circuits we have 2 high-side MOSFETs and 2 low-side MOSFETs. In such situations, there is a need to use high-side drive circuitry alongside low-side drive circuitry. The most common way of driving MOSFETs in such cases is to use high-low side MOSFET drivers. Undoubtedly, the most popular such driver chip is the IR2110. And in this article/tutorial, I will talk about the IR2110.

You can download the IR2110 datasheet from the IR website. Here's the download link:

www.irf.com/product-info/datasheets/data/ir2110.pdf

First let’s take a look at the block diagram and the pin assignments and pin definitions (also called lead assignments and lead definitions):

Fig. 1 - IR2110 block diagram (click on image to enlarge)

 Fig. 2 - IR2110 Pin/Lead Assignments (click on image to enlarge)

Fig. 3 - IR2110 Pin/Lead Definitions (click on image to enlarge)

Notice that the IR2110 comes in two packages – 14 pin through-hole PDIP package and the 16-pin surface mount SOIC package.

Now let's talk about the different pins.

VCC is the low-side supply and should be between 10V and 20V. VDD is the logic supply to the IR2110. It can be between +3V to +20V (with reference to VSS). The actual voltage you choose to use depends on the voltage level of your input signals. Here’s the chart:

Fig. 4 - IR2110 Logic "1" Input Threshold vs VDD (click on image to enlarge)

It is common practice to use VDD = +5V. When VDD = +5V, the logic 1 input threshold is slightly higher than 3V. Thus when VDD = +5V, the IR2110 can be used to drive loads when input “1” is higher than 3 point something volts. This means that it can be used for almost all circuits, since most circuits tend to have around 5V outputs. When you’re using microcontrollers the output voltage will be higher than 4V (when the microcontroller has VDD = +5V, which is quite common). When you’re using SG3525 or TL494 or other PWM controller, you are probably going to have them powered off greater than 10V, meaning the outputs will be higher than 8V when high. So, the IR2110 can be easily used.

You may lower the VDD down to about 4V if you’re using a microcontroller or any chip that gives output of 3.3V (eg dsPIC33). While designing circuits with the IR2110, I had noticed that sometimes the circuit didn’t work properly when IR2110 VDD was selected as less than +4V. So, I do not recommend using VDD less than +4V.

In most of my circuits, I do not have signal levels which have voltages less than 4V as high and so I use VDD = +5V.

If for some reason, you have signals levels with logic “1” having lower than 3V, you will need a level converter / translator that will boost the voltage to acceptable limits. In such situations, I recommend boosting up to 4V or 5V and using IR2110 VDD = +5V.

Now let’s talk about VSS and COM. VSS is the logic supply ground. COM is “low side return” – basically, low side drive ground connection. It seems that they are independent and you might think you could perhaps isolate the drive outputs and drive signals. However, you’d be wrong. While they are not internally connected, IR2110 is a non-isolated driver, meaning that VSS and COM should both be connected to ground.

HIN and LIN are the logic inputs. A high signal to HIN means that you want to drive the high-side MOSFET, meaning a high output is provided on HO. A low signal to HIN means that you want to turn off the high-side MOSFET, meaning a low output is provided on HO. The output to HO – high or low – is not with respect to ground, but with respect to VS. We will soon see how a bootstrap circuitry (diode + capacitor) – utilizing VCC, VB and VS – is used to provide the floating supply to drive the MOSFET. VS is the high side floating supply return. When high, the level on HO is equal to the level on VB, with respect to VS. When low, the level on HO is equal to VS, with respect to VS, effectively zero.

A high signal to LIN means that you want to drive the low-side MOSFET, meaning a high output is provided on LO. A low signal to LIN means that you want to turn off the low-side MOSFET, meaning a low output is provided on LO. The output on LO is with respect to ground. When high, the level on LO is equal to the level of VCC, with respect to VSS, effectively ground. When low, the level on LO is equal to the level on VSS, with respect to VSS, effectively zero.

SD is used as shutdown control. When this pin is low, IR2110 is enabled – shutdown function is disabled. When this pin is high, the outputs are turned off, disabling the IR2110 drive.

Now let’s take a look at the common IR2110 configuration for driving MOSFETs in both high and low side configurations – a half bridge stage.

 Fig. 5 - Basic IR2110 circuit for driving half-bridge (click on image to enlarge)

D1, C1 and C2 along with the IR2110 form the bootstrap circuitry. When LIN = 1 and Q2 is on, C1 and C2 get charged to the level on VB, which is one diode drop below +VCC. When LIN = 0 and HIN = 1, this charge on the C1 and C2 is used to add the extra voltage – VB in this case – above the source level of Q1 to drive the Q1 in high-side configuration. A large enough capacitance must be chosen for C1 so that it can supply the charge required to keep Q1 on for all the time. C1 must also not be too large that charging is too slow and the voltage level does not rise sufficiently to keep the MOSFET on. The higher the on time, the higher the required capacitance. Thus, the lower the frequency, the higher the required capacitance for C1. The higher the duty cycle, the higher the required capacitance for C1. Yes, there are formulae available for calculating the capacitance. However, there are many parameters involved, some of which we may not know – for example, the capacitor leakage current. So, I just estimate the required capacitance. For low frequencies such as 50Hz, I use between 47µF and 68µF capacitance. For high frequencies like 30kHz to 50kHz, I use between 4.7µF and 22µF. Since we’re using an electrolytic capacitor, a ceramic capacitor should be used in parallel with this capacitor. The ceramic capacitor is not required if the bootstrap capacitor is tantalum. Update: Please refer to this updated post for bootstrap cap sizing: https://tahmidmc.blogspot.com/2023/12/sizing-ir2110-high-side-bootstrap.html

D2 and D3 discharge the gate capacitances of the MOSFET quickly, bypassing the gate resistors, reducing the turn off time. R1 and R2 are the gate current-limiting resistors.

+MOSV can be up to a maximum of 500V.

+VCC should be from a clean supply. You should use filter capacitors and decoupling capacitors from +VCC to ground for filtering.

Now let’s look at a few example application circuits of the IR2110.

 Fig. 6 - IR2110 circuit for high-voltage half-bridge drive (click on image to enlarge)

 Fig. 7 - IR2110 circuit for high-voltage full-bridge drive with independent switch control (click on image to enlarge)

In Fig. 7 we see the IR2110 being used to drive a full bridge. The functionality is simple and you should understand it by now. A common thing that is often done is that, HIN1 is tied/shorted to LIN2 and HIN2 is tied/shorted to LIN1, enabling the control of all 4 MOSFETs from 2 signal inputs, instead of 4 as shown below in Fig. 8.

 Fig. 8 - IR2110 circuit for high-voltage full-bridge drive with tied switch control - control with 2 input signals (click on image to enlarge)

 Fig. 9 - Using the IR2110 as a single high-voltage high-side driver (click on image to enlarge)

In Fig. 9 we see the IR2110 being used as a single high-side driver. The circuit is simple enough and follows the same functionality described above. One thing to remember is that, since there is no low-side switch, there must a load connected from OUT to ground. Otherwise the bootstrap capacitors can not charge.

 Fig. 10 - Using the IR2110 as a single low-side driver (click on image to enlarge)

 Fig. 11 - Using the IR2110 as a dual low-side driver (click on image to enlarge)

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If you've had failures with IR2110 and had driver after driver, MOSFET after MOSFET get damaged, burn and fail, I'm pretty sure that it's due to you not using gate-to-source resistors, assuming of course that you designed the IR2110 driver stage properly. NEVER OMIT THE GATE-TO-SOURCE RESISTORS. If you're curious, you can read about my experience with them here (I have also explained the reason that the resistors prevent damage):

http://tahmidmc.blogspot.com/2012/10/magic-of-knowledge.html

For further reading, you should go through this:

http://www.irf.com/technical-info/appnotes/an-978.pdf

I have seen in many forums that people struggle with designing circuits with IR2110. I too had a lot of difficulty before I could confidently and consistently build successful driver circuits with IR2110. I have tried to explain the application and use of IR2110 thoroughly through explanation and plenty of examples and hope that it helps you in your endeavors with IR2110.

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Ferrite Transformer Turns Calculation for High-Frequency/SMPS Inverter

Ferrite Transformer Turns Calculation for High-Frequency/SMPS Inverter

  December 23, 2012    Tahmid

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On different forums, I often find people asking for help in calculating the required turns for a ferrite transformer they are going to use in high-frequency/SMPS inverters. In a high-frequency/SMPS inverter, the ferrite transformer is used in the step-up/boost stage where the low voltage DC from the battery is stepped up to high voltage DC. In this situation, there are really only two choices when selecting topology – push-pull and full-bridge. For transformer design, the difference between a push-pull and a full-bridge transformer for same voltage and power will be that the push-pull transformer will require a center tap, meaning it will require twice the number of primary turns as the full-bridge transformer.

Calculation of required turns is actually quite simple and I’ll explain this here.

For explanation, I’ll use an example and go through the calculation process.

Let’s say the ferrite transformer will be used in a 250W inverter. The selected topology is push-pull. The power source is a 12V battery. Output voltage of the DC-DC converter stage will be 310V. Switching frequency is 50kHz. The selected core is ETD39. Remember that the output of the transformer will be high frequency AC (50kHz square wave in this case). When I refer to an output of high voltage DC (eg 310VDC mentioned above), this is the DC output obtained after rectification (using ultrafast recovery diodes configured as bridge rectifier) and filtration (using LC filter).

During operation, the battery voltage does not stay fixed at 12V. With high loads, the voltage will be less than 12V. With low loads and near-fully charged battery, the voltage may be higher than 13V. So, it must be kept in mind that the input voltage is not constant, but is variable. In inverters, the battery low-cut is usually set at 10.5V. So, we’ll take this as our lowest possible input voltage.

Vinmin = 10.5V

The formula for calculating the number of required primary turns is:

 

For our push-pull transformer, this will be one-half the required number of turns.

N_pri means number of primary turns; N_sec means number of secondary turns; N_aux means number of auxiliary turns and so on. But just N (with no subscript) refers to turns ratio.

For calculating the required number of primary turns using the formula, the parameters or variables that need to be considered are:

• Vin_(nom) – Nominal Input Voltage. We’ll take this as 12V. So, Vin_(nom) = 12.
• f – The operating switching frequency in Hertz. Since our switching frequency is 50kHz, f = 50000.
• B_max – Maximum flux density in Gauss. If you’re used to using Tesla or milliTesla (T or mT) for flux density, just remember that 1T = 10⁴ Gauss. B_max really depends on the design and the transformer cores being used. In my designs, I usually take B_max to be in the range 1300G to 2000G. This will be acceptable for most transformer cores. In this example, let’s start with 1500G. So B_max = 1500. Remember that too high a B_max will cause the transformer to saturate. Too low a B_max will be under utilizing the core.
• A_c – Effective Cross-Sectional Area in cm². You will get this information from the datasheets of the ferrite cores. A_c is also sometimes referred to as A_e. For ETD39, the effective cross-sectional area given in the datasheet/specification sheet (I’m referring to TDK E141. You can download it from here: www.tdk.co.jp/tefe02/e141.pdf  ), the effective cross-sectional area (in the specification sheet, it’s referred to as A_e but as I’ve said, it’s the same thing as A_c) is given as 125mm². That is equal to 1.25cm². So, A_c = 1.25 for ETD39.

 

So now, we’ve obtained the values of all required parameters for calculation Npri – the number of required primary turns.

Vin_(nom) = 12                                        f = 50000                              B_max = 1500                          A_c = 1.25

Plugging these values into the formula:

                        N_pri = 3.2

We won’t be using fractional windings, so we’ll round off N_pri to the nearest whole number, in this case, rounded down to 3 turns. Now, before we finalize this and select N_pri = 3, we better make sure that B_max is still within acceptable bounds. As we’ve decreased the number of turns from the calculated figure (down to 3.0 from 3.2), B_max will increase. We now need to figure out just how much B_max has increased and if that is still an acceptable value.

Vin_(nom) = 12                 f = 50000                     N_pri = 3                         A_c = 1.25

 

                                                                   B_max = 1600

The new value of B_max is well within acceptable bounds and so we can proceed with N_pri = 3.

So, we now know that for the primary, our transformer will require 3 turns + 3 turns.

In any design, if you need to adjust the values, you can easily do so. But always remember to check that B_max is acceptable.

• For example, if for construction difficulties, winding 3 turns + 3 turns becomes difficult, you may use 2 turns + 2 turns or 4 turns + 4 turns. Increasing number of turns won’t hurt – you’ll just be under utilizing the core. However, decreasing number of turns increases B_max, so just recheck to make sure B_max is okay. The range I’ve stated for B_max (1300G to 2000G) is just an estimate. It will work for most cores. However, with many cores, you can go higher to decrease the number of turns. Going lower will just be under utilizing the core, but may sometimes be required if number of turns is too low. 

 

• I’ve started off with a set B_max and gone on to calculate N_pri from there. You can also assign a value of N_pri and then check if B_max is okay. If not, you can then increase or decrease N_pri as required and then check if B_max is okay, and repeat this process until you get a satisfactory result. For example, you may have set Npri = 2 and calculated Bmax and decided that this was too high. So, you set Npri = 3 and calculated Bmax and decided it was okay. Or you may have started with Npri = 4 and calculated Bmax and decided that it was too low. So, you set Npri = 3 and calculated Bmax and decided it was okay.

Now it’s time to move on to the secondary. The output of our DC-DC converter is 310V. So, the transformer output must be 310V at all input voltages, from all the way up from 13.5V to all the way down to 10.5V. Naturally, feedback will be implemented to keep the output voltage fixed even with line and load variations – changes due to battery voltage change and also due to load change. So, some headroom must be left for feedback to work. So, we’ll design the transformer with secondary rated at 330V. Feedback will just adjust the voltage required by changing the duty cycle of the PWM control signals. Besides feedback, the headroom also compensates for some of the losses in the converter and thus compensates for the voltage drops at different stages – for example, in the MOSFETs, in the transformer itself, in the output rectifiers, output inductor, etc.

This means that the output must be capable of supplying 330V with input voltage equal to 10.5V and also input voltage equal to 13.5V. For the PWM controller, we’ll take maximum duty cycle to be 98%. The gap allows for dead-time.

At minimum input voltage (when Vin = Vinmin), duty cycle will be maximum. Thus duty cycle will be 98% when Vin = 10.5 = Vinmin. At maximum duty cycle = 98%, voltage to transformer = 0.98 * 10.5V = 10.29V.

So, voltage ratio (secondary : primary) = 330V : 10.29V = 32.1

Since voltage ratio (secondary : primary) = 32.1, turns ratio (secondary : primary) must also be 32.1 as turns ratio (secondary : primary) = voltage ratio (secondary : primary). Turns ratio is designated by N. So, in our case, N = 32.1 (I’ve taken N as the ratio secondary : primary).

N_pri = 3

N_sec = N * N_pri = 32.1 * 3 = 96.3

Round off to the nearest whole number. N_sec = 96.

Thus 96 turns are required for the secondary. With proper implementation of feedback, a constant 310VDC output will be obtained throughout the entire input voltage range of 10.5V to 13.5V.

Here, one thing to note is that even though I took 98% as the maximum duty cycle, maximum duty cycle in practice will be smaller since our transformer was calculated to provide 330V output. In the circuit, the output will be 310V, so the duty cycle will be even lower. However, the advantage here is that you can be certain that the output will not drop below 330V even with heavy loads since a large enough headroom is provided for feedback to kick in and maintain the output voltage even at high loads.

If any auxiliary windings are required, the required turns can be easily calculated. Let me show with an example. Let’s say we need an auxiliary winding to provide 19V. I know that the output 310V will be regulated, whatever the input voltage may be, within the range initially specified (Vinmin to Vinmax – 10.5V to 13.5V). So, the turns ratio for the auxiliary winding can be calculated with respect to the secondary winding. Let’s call this turns ratio (secondary : auxiliary) N_A.

N_A = N_sec / N_aux = V_sec / (V_aux + V_d). V_d is the output diode forward drop. Let’s assume that in our application, a schottky rectifier with a V_d = 0.5V is used.

So, N_A = 310V / 19.5V =15.9

N_sec / N_aux = N_A

N_aux = N_sec / N_A = 96 / 15.9 = 5.96

Let’s round off N_aux to 6 and see what the output voltage is.

V_sec / (V_aux + V_d) = N_A = N_sec / N_aux = 96 / 6 =16.0

(V_aux + V_d) = V_sec / N_A = 310V / 16.0 = 19.375V

V_aux = 19.375V – 0.5V = 18.875V (rounded off)

I would say that’s great for an auxiliary supply. If in your calculations you come to a voltage that is too far off the required target voltage and thus greater accuracy is required, take V_aux as something higher and use a voltage regulator.

For example, if in our previous example, instead of18.875V we had gotten 19.8V but needed more accuracy, we could've used 24V or thereabouts and used a voltage regulator to give 19V output.

So, there we have it. Our transformer has 3 turns + 3 turns for primary, 96 turns for secondary and 6 turns for auxiliary.

Here’s our transformer:

Calculating required number of turns for a transformer is actually a simple task and I hope that I could help you understand how to do this. I hope this tutorial helps you in your ferrite transformer designs. Do let me know your comments and feedback.

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