Made a boost converter that takes 5Vin and converts it to 12V, max 1.7A using a MC34063ABD IC. I selected an inductor of 22uH and Schottky diode that is rated for 2A max. When I apply power the whole thing draws 1.2A current without any load attached to the output and things get crazy hot.
When I measure Voltage output is in the order of mV so I'm thinking a short somewhere.
If I reverse the orientation of the diode, all is fine an dandy and I get 12V on the output and negligible current draw from the input UNTIL I start applying a load like a 12V fan which is supposed to draw 0.3A max. The fan spins fine but the current drawn on the input reads close to 1.7A and of course things get crazy hot again.
According to the data sheet for the diode, the cathode is marked with a white stripe. In the first case I had the cathode connected to the first PIN of the IC and inductor. In the second case I have it connected to power and the anode goes to the first pin of the IC and inductor.
I should mention I do not have a heatsink on top of the IC and I also do not have a current sense resistor in series with the inductor and the positive lead of the bulk cap on the input so I'm thinking that's should be the latter should be the cause for the large current draw, but I do not get why the orientation of the diode causes the behavior described.
https://assets.nexperia.com/documents/data-sheet/PMEG2020AEA.pdf
From what you've described, something in your 34063 circuit is mis-wired!
If the voltage output is close to zero, something is shorted or you have the diode reversed. If this is the case you might have killed your 34063.
Possibilities include a wrong (too small inductor), capacitor on the output not able to handle 15V, mistake in wiring.
Check your circuit wiring again. The cathode (dark band) connects to the output of the boost circuit.
If you take a pic of your circuit and post it with your schematic, it would help anyone to help you troubleshoot.
5V/22μH is 0.2A/μs or 2A peak in 10μs, or 4A peak in 20μs - corresponding to 50kHz switching frequency. Are these numbers what you expect?
This is a dumb chip with no duty cycle control, it will turn on for up to (almost) full oscillator cycle at a time. If this causes switch transistor rating to be exceeded, bummer.
Regardless of the switching device or controller, boost circuit topology is pretty much the same. For example here's one I quickly knocked up, note the orientation of the diode.
Also, note that the internal switch of the MC34063 is limited to Ipeak 1.5A. To get your 1.7A output, you will need to drive an external MOSFET.
My schematic attached. Bulk caps are 100uF 16V each.
My math for max current of 1.7A comes from (IOut = (Vin x Duty Cycle) / L x switching frequency x efficiency)):
Vin = 5
Duty Cycle = 0,706 or 12V/(5V+12V)
L = 22 x 10**(-6)
switching frequency = 100000
Efficiency = 0,9
My math for Vout comes from 1.25 ( 1 + (866K/100K)) = 12.075
I also put a discharge resistor of 100K in the circuit to discharge the output cap faster.
switching frequency = 1000000
There's your problem! MC34063 cannot run at 1MHz!
switching frequency = 1000000
There's your problem! MC34063 cannot run at 1MHz!
My bad, sorry that was a typo, it should have said 100000. I have corrected that.
Jeesus, mate, flip the drawing for good sake, make input on the left and output on the right.
Try connecting two diodes in series, to increase reverse voltage margin, maybe your diode is experiencing breakdown. Other people may say what voltages are developed in boost converter.
switching frequency = 1000000
There's your problem! MC34063 cannot run at 1MHz!
My bad, sorry that was a typo, it should have said 100000. I have corrected that.
Even the datasheet says maximum frequency 42kHz for the onboard oscillator.
https://www.onsemi.com/pdf/datasheet/mc34063a-d.pdf
switching frequency = 1000000
There's your problem! MC34063 cannot run at 1MHz!
My bad, sorry that was a typo, it should have said 100000. I have corrected that.
Even the datasheet says maximum frequency 42kHz for the onboard oscillator.
(Attachment Link)
https://www.onsemi.com/pdf/datasheet/mc34063a-d.pdf
Thanks! That's the confusing part to me. The first thing the data sheet lists about frequency is 100KHz. How are the two related?
Now, If I plug 42KHz as the frequency value into the formula I get 4.2A !!!
Even the datasheet says maximum frequency 42kHz for the onboard oscillator.
Thanks! That's the confusing part to me. The first thing the data sheet lists about frequency is 100KHz. How are the two related?
Now, If I plug 42KHz as the frequency value into the formula I get 4.2A !!!
I've never tried it, but if you ignore the onboard oscillator and connect an external oscillator, maybe it will work at 100kHz? But if you're going to the trouble of an external oscillator, you should probably consider a newer more modern alternative part.
But like I said, you should be able to drive an external MOSFET switch to handle the extra current.
I'll have to change the values a bit to account for IC current limitation being 1.5A and also add a current sensing resistor to limit it.
If I'm not mistaken, a 0.133ohm resistor should limit it to 1.5A considering a voltage drop of 0.2V.
i.e.: Rsc = Vdrop/Imax = 0.2V/1.5A = 0.133 ohm
Then I think if I'd use a 22uH inductor and the switching frequency is 100kHz (still not clear if it's supposed to be 100kHz or 45kHz max), then that should give me up to 1.3A.
Do these add up?
A current of 1.3A should be ok for what I need (i.e.: driving a bi-polar stepper that draws 1.2A)
Please calculate Ipk(switch), using the formula in the table Figure 17 in the data sheet.
Ipk(switch) must be less than 1.5A
Please calculate Ipk(switch), using the formula in the table Figure 17 in the data sheet.
Ipk(switch) must be less than 1.5A
I realized that if I strictly go by the switching frequency shown on DigiKey (35kHz), I am using a too small inductor which gives me too much current. Per Iout max formula, I should be using a larger inductor like 100uH to get 1.12A.
The data sheet says the switching frequency is 33kHz typical.
https://www.digikey.com/en/products/detail/stmicroelectronics/MC34063ABD-TR/1038906Now about Ipk(switch), that determines the max switching current this IC can handle, correct?
I'd like to check if I understand the formulas:
2 x Iout [(ton - toff) +1)] where ton - toff = (Vout + Vf - Vin) / Vin - Vsat
Vf is the voltage drop across the diode (my diode has a drop of 0.52V)
Vout is 12V
Vin is 5V
But how do I determine Vsat for my circuit? Table 6 lists two saturations depending on used configuration and I'm not clear if I'm using darlington configuration or not because I do have pins 1 and 8 connected (not to each other), as well as a resistor from pin 8 to VCC.
switching frequency = 1000000
There's your problem! MC34063 cannot run at 1MHz!
My bad, sorry that was a typo, it should have said 100000. I have corrected that.
Even the datasheet says maximum frequency 42kHz for the onboard oscillator.
(Attachment Link)
https://www.onsemi.com/pdf/datasheet/mc34063a-d.pdf
The way I read this, is that with a
1nF capacitor on pin 3 and 0V on pin 5 and an ambient temperature of 25 degree Celsius the output frequency will typically be 33KHz but can be as low as 24KHz or as high as 42KHz. The desired frequency depends on Ct (connected between pin 3 and ground) and depends on the needed on time (4.0 x 10−5 ton). Page 11 of the given datasheet.
So the given on page 1 is that it can operate to 100KHz seems to be valid.
But how do I determine Vsat for my circuit? Table 6 lists two saturations depending on used configuration and I'm not clear if I'm using darlington configuration or not because I do have pins 1 and 8 connected (not to each other), as well as a resistor from pin 8 to VCC.
Since this device uses bipolar junction transistors instead of field effect transistors there is a voltage drop over the collector emitter junction. When fully saturated the drop over the pin 1 transistor is typical 0.45V. This is with a current of 1A through the junction.
With pin 1 and 8 connected the drop is typical 1V again with a current of 1A through the junctions.
Your setup only uses the pin 1 transistor so Vsat will be ~0.45V.
To get better performance you might want to look into newer devices that do use MOSFET's. Will give better efficiency and allow higher switching frequencies. Something like
this one for instance.
Hello,
If the requirements are 5v input (constant, no sag at any current) and the output 12v at 300ma, the design looks very marginal no matter what you do.
We have to remember that the peak current is related to the output power and input and output voltages. The ideal duty cycle is related to the input and output voltage, and the peak current is related to the output power and the duty cycle, along with the efficiency. The efficiency plays a role too as it may only be 85 percent or even less.
The transistor drop could be as high as 1v maybe even higher. The drop across Rsc could be roughly 0.4 volts. We end up down to maybe 3.5v effective input voltage if that. If the duty cycle was 0.58 in the ideal case, it will have to be even much less which drives the peak current up even higher. It seems like the limit of 1.5 amps just may not be enough.
We haven't even looked at the minimum duty cycle given the limitations of a boost circuit with resistive losses, which may be greater than that required to get the right output, if that was even possible.
The 22uH did not sound high enough for this device right off the bat. If anything, maybe 250uH or greater.
We should go over this design more carefully. With all these limitations I'd be surprised if this ever works with the given input/output requirements. I guess there is a chance that things are not as bad as they read from the data sheet, but one build may work while the next one fails.
A good, simple simulation may help decide also.
So ok, 22uH is definitely out. Probably 100uH min, but the startup surge is going to be too high and may cause a problem. 200uH would be much better. 20uf output cap, 1000uf low ESR input cap. This might just work.
A drawback with this chip is not only the use of internal bipolar transistors, but also no slow start that came in with more modern chips. No slow start means higher input startup surge current and higher output surge voltage at startup.
Hello,
If the requirements are 5v input (constant, no sag at any current) and the output 12v at 300ma, the design looks very marginal no matter what you do.
We have to remember that the peak current is related to the output power and input and output voltages. The ideal duty cycle is related to the input and output voltage, and the peak current is related to the output power and the duty cycle, along with the efficiency. The efficiency plays a role too as it may only be 85 percent or even less.
The transistor drop could be as high as 1v maybe even higher. The drop across Rsc could be roughly 0.4 volts. We end up down to maybe 3.5v effective input voltage if that. If the duty cycle was 0.58 in the ideal case, it will have to be even much less which drives the peak current up even higher. It seems like the limit of 1.5 amps just may not be enough.
We haven't even looked at the minimum duty cycle given the limitations of a boost circuit with resistive losses, which may be greater than that required to get the right output, if that was even possible.
The 22uH did not sound high enough for this device right off the bat. If anything, maybe 250uH or greater.
We should go over this design more carefully. With all these limitations I'd be surprised if this ever works with the given input/output requirements. I guess there is a chance that things are not as bad as they read from the data sheet, but one build may work while the next one fails.
A good, simple simulation may help decide also.
So ok, 22uH is definitely out. Probably 100uH min, but the startup surge is going to be too high and may cause a problem. 200uH would be much better. 20uf output cap, 1000uf low ESR input cap. This might just work.
A drawback with this chip is not only the use of internal bipolar transistors, but also no slow start that came in with more modern chips. No slow start means higher input startup surge current and higher output surge voltage at startup.
But with a 200uH inductor, I'll get only 560mA max current, right? (considering 35kHz frequency)
switching frequency = 1000000
There's your problem! MC34063 cannot run at 1MHz!
My bad, sorry that was a typo, it should have said 100000. I have corrected that.
Even the datasheet says maximum frequency 42kHz for the onboard oscillator.
(Attachment Link)
https://www.onsemi.com/pdf/datasheet/mc34063a-d.pdf
The way I read this, is that with a 1nF capacitor on pin 3 and 0V on pin 5 and an ambient temperature of 25 degree Celsius the output frequency will typically be 33KHz but can be as low as 24KHz or as high as 42KHz. The desired frequency depends on Ct (connected between pin 3 and ground) and depends on the needed on time (4.0 x 10−5 ton). Page 11 of the given datasheet.
So the given on page 1 is that it can operate to 100KHz seems to be valid.
How can I calculate that? Or can I just use a 1nF cap to get ~33kHz frequency?
I've updated the schematic, I think pin 7 and 8 were not connected properly in my initial version and I was also missing the current sense resistor on pin 6.
How can I calculate that? Or can I just use a 1nF cap to get ~33kHz frequency?
The frequency also depends on the feedback. The specification given are with the feedback voltage being 0, and is only useful as a test. To calculate the needed capacitor you have to follow the formulas given on page 11 of the onsemi datasheet.
First you choose the power supply characteristics and then you calculate the components. The frequency is one of the characteristics. If you want it to be 33KHz you choose that. The ratio between the voltages then leads to the on/off time and based on the frequency you can calculate the needed on time, which is the base for the capacitor you need to use.
Hello,
If the requirements are 5v input (constant, no sag at any current) and the output 12v at 300ma, the design looks very marginal no matter what you do.
We have to remember that the peak current is related to the output power and input and output voltages. The ideal duty cycle is related to the input and output voltage, and the peak current is related to the output power and the duty cycle, along with the efficiency. The efficiency plays a role too as it may only be 85 percent or even less.
The transistor drop could be as high as 1v maybe even higher. The drop across Rsc could be roughly 0.4 volts. We end up down to maybe 3.5v effective input voltage if that. If the duty cycle was 0.58 in the ideal case, it will have to be even much less which drives the peak current up even higher. It seems like the limit of 1.5 amps just may not be enough.
We haven't even looked at the minimum duty cycle given the limitations of a boost circuit with resistive losses, which may be greater than that required to get the right output, if that was even possible.
The 22uH did not sound high enough for this device right off the bat. If anything, maybe 250uH or greater.
We should go over this design more carefully. With all these limitations I'd be surprised if this ever works with the given input/output requirements. I guess there is a chance that things are not as bad as they read from the data sheet, but one build may work while the next one fails.
A good, simple simulation may help decide also.
So ok, 22uH is definitely out. Probably 100uH min, but the startup surge is going to be too high and may cause a problem. 200uH would be much better. 20uf output cap, 1000uf low ESR input cap. This might just work.
A drawback with this chip is not only the use of internal bipolar transistors, but also no slow start that came in with more modern chips. No slow start means higher input startup surge current and higher output surge voltage at startup.
But with a 200uH inductor, I'll get only 560mA max current, right? (considering 35kHz frequency)
Hi,
What do you mean a max of 560ma?
Did you change the output spec from 12v at 300ma, to 12v at something even greater than 500ma?
How can I calculate that? Or can I just use a 1nF cap to get ~33kHz frequency?
The frequency also depends on the feedback. The specification given are with the feedback voltage being 0, and is only useful as a test. To calculate the needed capacitor you have to follow the formulas given on page 11 of the onsemi datasheet.
First you choose the power supply characteristics and then you calculate the components. The frequency is one of the characteristics. If you want it to be 33KHz you choose that. The ratio between the voltages then leads to the on/off time and based on the frequency you can calculate the needed on time, which is the base for the capacitor you need to use.
Are we referring to the same data sheet? the one I have is from STM and there are no formulas on page 11:
https://www.st.com/content/ccc/resource/technical/document/datasheet/03/f9/c4/3d/7f/eb/4c/5e/CD00001232.pdf/files/CD00001232.pdf/jcr:content/translations/en.CD00001232.pdf