# How is the switch current related to output current in a DC-DC converter?

I have designed and implemented a DC-DC converter with the following specs:

• Input Voltage: 5V
• Input Current (max): 3A
• Output Voltage: 12V
• Output Current (max): 1A
• DC-DC Controller: MC34063
• Inductor: 15uH/5A max
• Schottkey diode, 550mV Vf 1A max
• 50KHz switching frequency

I have tied the Drive and Switch collector to form a darlington pair. The device outputs 12V, but the controller gets very hot (way beyond spec of 70C) even at 0.5A at the output.

The device says it has a maximum switch current of 1.5A, and I've incorrectly thought this is the maximum output current rating.

How do I figure out what the maximum output current will be, if I'm specified the maximum switching current?

• What caveman said | Iin x Vin x efficiency = Iout x Vout ALWAYS. | For boost as shown Vout "stands on Vin" and gives you Vin "free lunch". You need another (12-5) + Vdiuode + losses. Switch will be on about 60%-70% of time - say 60%. Vswitch avg on = I_convert/60% so 34063 falls far short of task. | You can drive an external bipolar or mosfet with it and get essentially unlimited power as far as IC ratings are concerned. Bipolar - can use IC to drive. FET - need simple cheap gate driver - ask if interested. Jan 7, 2015 at 7:53

The MC34063 datasheet actually tells you on page 11: $$I_{pk}(switch) = 2I_{out(max)}(\frac{t_{on}}{t_{off}}+1)$$