0
\$\begingroup\$

I decided to build My own PIC16F84A programmer,The Problem with 16F84A is the Voltage in MCLR pin (+13V). How we get this with a (-10v;+10v) Serial port. Yeah, There are alot of methods, but I choose the simple one. :=D

So I searched the net & I found this Simple one http://www.eeweb.com/blog/extreme_circuits/simple-universal-pic-programmer. Thankx Luke Weston for sharing.

Question 1 : I built the Circuit, but can't retrive the Vpp on the emitter of the BC337? is there a mistake ?

Question 2 : So i made the This circuit below to contorl the Vpp (+13v) with serial port signal (is two stage inverter), Is it correct ?

schematic

simulate this circuit – Schematic created using CircuitLab

If Signal from PC = -10v ===> Vpp = 0v

If Signal from PC = +10v ===> Vpp = +13v

first Modifications After Some Discussions:

schematic

simulate this circuit

\$\endgroup\$

1 Answer 1

Reset to default
1
\$\begingroup\$

This is not a good circuit for a variety of reasons. There are a lot of crappy circuits on the internet. In fact, the fraction of crappy circuits is higher on the internet than in real life. Those that know what they are doing and design a circuit to a particular requirement don't consider it a big enough deal to publish and often can't due to confidentiality issues anyway. It's the wannabes that can barely spell NPN that spend a week getting their LED to blink that then proudly want to show off their herculian accomplishement to the world.

Just a few problem off the to of my head:

  1. Is Q2 rated for 10 V reverse bias accross its B-E junction? That is certainly possible, but many small transistors aren't rated for that.

  2. Your passive pullup that drives Vpp will drop some voltage. It seems neither you nor whoever designed this circuit even looked at the programming spec. See the spec for Ihh in Table 5-1 AC/DC CHARACTERISTICS, section 5.0 PROGRAM/VERIFY MODE ELECTRICAL CHARACTERISTICS on page 13. Multiply that current times R4, then subtract that voltage from your 13 V supply. Your Vpp voltage is now out of spec.

  3. On the same table, notice the maximum Vpp rise time. Do you really think a 10 kΩ passive pullup can meet that? Maybe it can, but maybe not. Again, I don't like the passive pullup to drive Vpp high.

\$\endgroup\$
8
  • \$\begingroup\$ Hi Olin, I was thinking that i'm a newbie, but After Your answer I think i'm far to be a newbie .... lol. i'm going to read the programming spec, and you are right, there's no such free informations. With All my respects to you. \$\endgroup\$ Commented Mar 23, 2014 at 13:59
  • \$\begingroup\$ For The Q2 Emitter-Base Breakdown voltage is 5V. So, I planned to use a zener diode (5.1v) on parallal with the junction? \$\endgroup\$ Commented Mar 23, 2014 at 14:13
  • \$\begingroup\$ @user: A ordinary signal diode in reverse accross B-E of Q2 would do fine. 1N4148 is a common type that would work well. However, that doesn't address the Vpp passive pullup issues. \$\endgroup\$ Commented Mar 23, 2014 at 15:11
  • \$\begingroup\$ You are right, still thinking to solve Vpp problem(s) (Ihh = 200uA; Vpp Time rising = 8 us). \$\endgroup\$ Commented Mar 23, 2014 at 15:39
  • \$\begingroup\$ @user: I'd look into using a PNP to drive Vpp high from its collector. However, I wouldn't be trying to do this directly from serial port signals in the first place. \$\endgroup\$ Commented Mar 23, 2014 at 16:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.