LM2903-Q1: LM2903-Q1

guozhu Lyu

Part Number: LM2903-Q1
Other Parts Discussed in Thread: LM2903B, LM339, TLV1811

Why the red circled part are 10M and the other resistors are 10k, if we change the 10M to 10k, what will happen and why?

over 1 year ago

0 guozhu Lyu over 1 year ago

Prodigy 100 points

And another question is, when we use LM2903B to simulate as an OVP circuit, if R8 is 10k, an unexpected pulse occured at MOS_G, but if R8 is 46.4k, no such pulse occured. Why?

0 Clemens Ladisch over 1 year ago

Guru 270600 points

These circuits use different resistor values to force different voltages at the two inputs. If you use identical resistors, the two inputs will have the same voltage, and the comparator output might oscillate.

0 Clemens Ladisch over 1 year ago in reply to guozhu Lyu

Guru 270600 points

As far as I can see, N+ and N− have the same voltage at that time, so the output is random. Consider adding hysteresis.

I'd guess that the 46.6 kΩ resistor does not allow enough base current for Q1.

0 Paul Grohe over 1 year ago

TI__Guru 52726 points

Hello Guozhu,

The idea is to place a small differential voltage across the inputs to keep the output in a known state.

There is about 25nA of bias current flowing out of the input and towards ground on the LM339 type bipolar device. See section 2 of the appnote.

The 10MEG resistor is used to generate a voltage drop from that bias current. 10Meg * 25nA = 250mV across the resistor. That places the 250mV on the negative input, which keeps the output low.

If you want the output high, swap the resistors.

0 guozhu Lyu over 1 year ago in reply to Clemens Ladisch

Prodigy 100 points

hello，Clemens Ladisch，Thanks for your reply. I increased the R8 from 10k to 100k, and found when R8 is above 20k, Ib no pulse current to cause Ic pulse current, and that pulse current will charge the M1 gs. It seems LM2903B output has a leakage curent during the startup if R8 is too low? While the N+ always higher than N- at that time.