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I've this circuit, I haven't connected the transistor yet, I want to check the proper function of the circuit before so I don't burn nothing.

enter image description here

What is the correct way to check the output signal, pins 5 and 7, I don't have a differential oscilloscope just a regular one and I'm not sure if is correct to connect it to the pins 5 and seven directly. What I've tried is to put a led with a series resistance of 820 ohm between at the output, pins 5 and 7, but the led doesn't turn on, although the pin seven has a signal because the led turn on if I connect it between 7 and ground.

How can I check, the if the circuit is generating the correct signal before I connect the transistor?

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  • \$\begingroup\$ What made you think that Pin 5 is an output? As per the datasheet (pg4) it is the "High Side Floating Supply Return". \$\endgroup\$ – brhans Jan 11 '16 at 21:15
  • \$\begingroup\$ The gate of the transistor is connected at pin 7 and the source at the pin 5, to make the transistor work must be a potential difference between them. \$\endgroup\$ – Luis Ramon Ramirez Rodriguez Jan 11 '16 at 21:26
  • \$\begingroup\$ How will the source or emitter of the power transistor be held down while C2 gets charged? In order to test the circuit you need to charge C2, you can use a resistor to 0V, then apply logic high to control input and observe that HO goes up. \$\endgroup\$ – user1582568 Jan 11 '16 at 21:48
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The IR2110 uses bootstrapping to create the supply for the high-side MOSFET so the high side driver won't work without both the low side MOSFET and a supply voltage in the 12V range (up to the hundreds of volts it is rated for).

You should be able to see the low side driver work without the MOSFETs, but it doesn't look like you have anything connected there.

When you have the two MOSFETs you have to drive the bridge low periodically to refresh the charge on C2 ('high' duty cycle cannot be 100%) otherwise there is no way to drive the gate of the high-side MOSFET above the positive rail.

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  • \$\begingroup\$ I'm just interested in the high side driver, I will use the driver in an asynchronous buck converter so I don't need to use the low side on the converter. Do I need to generate the low side signal and connect both mosfet in order to use the high side transistor? \$\endgroup\$ – Luis Ramon Ramirez Rodriguez Jan 11 '16 at 21:31
  • \$\begingroup\$ It would probably be best to use two MOSFETs that way you could get higher efficiency than with a diode, but you could also use a suitable DC-DC converter and use that rather than the bootstrap (don't use D1 but supply the high side circuit directly). \$\endgroup\$ – Spehro Pefhany Jan 11 '16 at 21:39
  • \$\begingroup\$ This if for a project at the university, I prefer to keep it simple at first and use it with the diode, also I don't have access to a better IC (complicated scenario in venezuela). \$\endgroup\$ – Luis Ramon Ramirez Rodriguez Jan 11 '16 at 21:44
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    \$\begingroup\$ If you use a diode at the bottom, this will work as the flyback from the inductor will charge C2, but you need to make sure C2 charges at the start. What is the output voltage of the converter? \$\endgroup\$ – user1582568 Jan 11 '16 at 21:54
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    \$\begingroup\$ @LuisRamonRamirezRodriguez Just a suggestion- you could post another question showing the full schematic of your proposed converter and asking about the questions raised here. This has deviated quite far from testing the chip to another design issue. \$\endgroup\$ – Spehro Pefhany Jan 11 '16 at 22:11

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