Circuit for converter AC 50Hz to 60Hz

Perhaps: (1) the motor can't handle a square waveform, or (2) the driving circuit can't handle the lagging phase presented by the motor. What frequency are you trying to drive the motor at? What kind of motor are you using? A motor designed for 60 Hz isn't likely to present adequate Xr to limit the current through it when driven at a significantly lower frequency.

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Dave M
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Never take a laxative and a sleeping pill at the same time!!

1) Motor can be handled the square ware, because I have tried a weaker performance motor in 120V design. And I use 120VAC to convert from 50Hz to 60Hz, the motor is working well. However, once I use 220V input and a 220V motor, the problem happened.

2) It is a AC motor called shaded-pole motor, the motor's speed is limited by input frequency. So, if I use 50Hz, the speed is 50 X 60 =

3000RPM. Now, I use 60Hz, the speed should be 60 x 60 = 3600RPM.

What's the +V supply to your FET drivers? I think you're not pulling up the high-side drivers hard enough, they're in their analog region, and they overheat.

Good luck! Rich

Welcome to the wonderful world of power electronics.

Try putting a small light bulb in series with HV while you are debugging.

Why R23? Are you trying to measure the motor current with it?

Keep in mind that the boostrap supply for the high side drivers does not work down to DC, although the UV detect should prevent any damage.

Best regards, Spehro Pefhany

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Motors tend to have an inrush current that's a multiple of the normal operating current. Specifications of the motor should tell you. Things are made worse by driving the motor with a 60Hz block. Guess you will need to limit the current during inrush time. For experimenting I'd make sure the circuit works well using a resistive load that takes an (arbitrary) five times the normal operating current of the motor. Then you can use a resistive load in series with the motor that can be switched off when the motor spins up. This way you'll have a chance to find out more about the motor and your circuit without blowing too many FETs. You may also try to start with the normal operating frequency of the motor and increase it later. Eventually you may need either heavier FETs or some PWM control or both.

petrus bitbyter

My FET Driver L6387 is using a 12V regulated DC supplier. I do not understand what you mean on "pulling up the high-side drivers hard enough". Can you tell me more detail?

Thanks!

Do you mean no need to add the resistor R23 on monitor current is better? The HV is a 220VAC rectified into DC already!

I have used a very high current FET. Actually, the motor is only operated into 1 to 2 amp maximum. So, I believe that the FET is high current enough for affording the current by inductive load. THANKS~

petrus bitbyter =E5=AF=AB=E9=81=93=EF=BC=9A

pg

In article , wrote: [....]

An induction motor looks like a messy RLC circuit when stopped and an even messier one when it is running.

The impedance of the moter can be quite low at high frequencies. The sharp edges on the waveform you are applying to it will cause large currents to flow in the MOSFETs. You are going to need inductors in the motor wires to pass the EMI tests so why not add them now.

No matter what you do, applying harmonics to an induction motor will make it run hot. If the motor is over rated for the application, you may not be bothered by this, but I suggest you try using the high for 120 degrees, open for 60 degrees, low for 120 degrees, open for 60 degrees waveform instead.

Elsewhere it was suggested that you put a lightbulb in the HV wire. I'll second that and suggest you put 3 or 4 sockets in parallel. This way you can inch up on it by screwing in more 25V bulbs.

If you have any DC component in the motor current, due perhaps to switching inbalances, the motor will consume much more current than normal.

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kensmith@rahul.net   forging knowledge

Thanks~

However, I don't understand what you mean on

"but I suggest you try using the high for 120 degrees, open for 60 degrees, low for 120 degrees, open for 60 degrees waveform instead."

I am using hardware implement only, how to control the on / off time for the FET?

The idea is like this:

--------------------------------- Q1 Q2 ! ----- ----- ! 0 0 --!D Q!--------!D Q!---- ! 1 0 6*F ------!> ! 6*F---!> ! ! ! 0 1 ! ! ! /Q!------+-- 0 0 ! ! ! ! ! ! ! R ! ! ! ! ! ----- ----- ! ! ! ! ! +---------------------- ! ! ! ! ! ! ----------- +------!---- ! ! ----- ! ! !AND >-- Pull up ! --!D Q!----------------!---- --------!> ! ! ! ! /Q!--+ ------!---- ------ ! !AND >-- Pull down -------------!----

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kensmith@rahul.net   forging knowledge

Thanks~ But you are using D-latch or D-Flip Flog? What is Q1 and Q2 mean?

Thank you!!! May I ask that the first one D-flip flog should be in Reset or in "Not(Reset)"? As 74LS74 is in NOT(RESET) rather than Reset.

And what is the Q1 and Q2 mean? I am sorry that I don't understand the truth table it mentioned. Thanks~

We're still talking about that 120V motor, right? (when you post, DON'T CLICK THE "REPLY" LINK! Click the "Show options" link, then click THAT reply link - that way google will quote context for you, so people know what you're talkng about.)

But even with that, it's a little out of my area of alleged expertise, but if you've got a Vdd of 120 V, but only 12V to drive the gate with, it's not going to turn on very hard. But I probably misunderstand the circuit - are they bootstrapped?

Thanks, Rich

You are already using the CD4013 / MC14023 flip-flop aren't you. I drew the schematic in terms of CD4013s. They have an active high reset. If you use 74XX74s you need to change the circuit to use the other signal as the reset.

The Q1 and Q2 are the two flip-flips. I'm showing the intended counting sequence for the divide by 3 circuit.

BTW: Please include context in your posts. This time I could go back up the thread and find out that it was my post you responded to. In other cases, it may not be posible, or I may get lazy.

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kensmith@rahul.net   forging knowledge

Thank you for your information. I just over look the components.

Yes, I will use MC14013 that can be active high on reset pin.

However, I want to know that the output of the AND gate. Since it is for driving FET high side and low side. May I know that how is the output sequence of the AND gate? I just want to use all N-MOS for output as it is more easy to find.

And for the Q1 and Q2 sequence table, I am still don't know which Flip-flog you are going to describle. Is the "Q" output of the Top-right one and the bottom one flip flog?

(Sorry about no quoted the reference message before, I will keep it and quote the message)

Ken Smith =E5=AF=AB=E9=81=93=EF=BC=9A